JEE Main + AdvancedClass XIOscillations & WavesMechanical waves

Waves

How a disturbance travels through a medium without carrying matter with it — transverse and longitudinal waves, the progressive wave, wave speed, superposition, standing waves, beats and the Doppler effect

🎯 Overview Chapter hero + roadmap

🔬 Interactive 3D · One view of the whole chapter: a harmonic disturbance travels along a stretched string as a transverse wave, $y(x,t)=A\sin(kx-\omega t)$. Each particle of the string merely oscillates up and down about its rest position while the crest — the pattern, the energy and the phase — moves steadily to the right at the wave speed $v=\omega/k=\sqrt{T/\mu}$. Drag the tension slider to watch the wave speed change, and follow one marked particle to see that it never travels with the wave.

A wave is a disturbance that travels through a medium, carrying energy and information from one point to another without any net transport of the medium itself. Drop a pebble into a still pond and ripples spread outward in widening circles, yet a leaf floating on the surface only bobs up and down as each ripple passes — it is not carried along to the shore. Speak, and your voice reaches a listener across the room, but the air itself does not stream from your mouth to their ear. This is the central and at first surprising fact of wave motion: what moves is the pattern, the phase and the energy, while the particles of the medium merely oscillate about their fixed equilibrium positions. Waves are one of the great unifying ideas of physics, describing sound, light, ripples on water, seismic tremors and the vibrations of musical instruments with a single mathematical language. 🔉⇢

The chapter begins by classifying mechanical waves — those that need a material medium to travel through — into two kinds according to the direction in which the particles of the medium move relative to the direction the wave travels. In a transverse wave the constituents of the medium oscillate perpendicular to the direction of wave propagation, as on a plucked string or a shaken rope, where crests and troughs move along the string while each element moves up and down. In a longitudinal wave the constituents oscillate back and forth along the very direction in which the wave travels, producing compressions and rarefactions; sound in air is the most familiar example. Transverse mechanical waves need a medium that can support a shearing stress and so can travel only through solids and on the surface of liquids, whereas longitudinal waves, requiring only that the medium can be compressed, travel through solids, liquids and gases alike. 🔉⇢

To describe a wave quantitatively we picture a sinusoidal, or harmonic, wave and write its displacement relation. If $y(x,t)$ is the displacement of the element of the medium at position $x$ at time $t$, a wave travelling in the positive $x$-direction is written $y(x,t)=A\sin(kx-\omega t+\phi)$. Here $A$ is the amplitude, the maximum displacement of a particle from equilibrium; $k=2\pi/\lambda$ is the angular wave number, set by the wavelength $\lambda$, the distance between two consecutive points in the same phase; $\omega=2\pi\nu=2\pi/T$ is the angular frequency, set by the period $T$ and frequency $\nu$; and $\phi$ is the initial phase. The quantity $(kx-\omega t+\phi)$ is the phase of the wave, and points of equal phase move together. A single snapshot in time shows the wave's shape in space, repeating every wavelength; watching a single particle in time shows simple harmonic motion, repeating every period. 🔉⇢

From the displacement relation follows the speed of a travelling wave. A point of constant phase satisfies $kx-\omega t=\text{constant}$, so differentiating gives the wave speed $v=\omega/k$, which combines with $\omega=2\pi\nu$ and $k=2\pi/\lambda$ to give the fundamental relation $v=\nu\lambda$: the wave speed equals the frequency times the wavelength. It is vital to distinguish this wave velocity, at which the pattern moves, from the particle velocity $\partial y/\partial t=-A\omega\cos(kx-\omega t)$, the speed at which an individual element of the medium oscillates. The two are quite different quantities: the wave may race along at hundreds of metres per second while each particle merely jiggles a fraction of a millimetre about its rest position. 🔉⇢

What actually fixes the wave speed is not the source but the medium. For a transverse wave on a stretched string the speed is $v=\sqrt{T/\mu}$, where $T$ is the tension and $\mu$ the linear mass density (mass per unit length): stretch the string tighter and the wave travels faster, load it with mass and it travels slower. The form of this result — a restoring-force factor over an inertia factor, under a square root — recurs for every mechanical wave. For a longitudinal wave the relevant restoring property is the modulus of elasticity, and the speed of sound is $v=\sqrt{B/\rho}$ with $B$ the bulk modulus and $\rho$ the density. 🔉⇢

The speed of sound in a gas has a famous history. Newton assumed the compressions and rarefactions occur isothermally, taking $B=P$ and predicting a speed of about $280\ \text{m s}^{-1}$ in air — some 15% below the measured value. Laplace resolved the discrepancy by recognising that the rapid compressions are adiabatic, not isothermal, because heat has no time to flow; then $B=\gamma P$ and the Newton-Laplace formula $v=\sqrt{\gamma P/\rho}$ gives about $331\ \text{m s}^{-1}$ at $0^\circ\text{C}$, in fine agreement with experiment. This result also shows why the speed of sound in an ideal gas depends on temperature but not on pressure, since $v=\sqrt{\gamma RT/M}$. 🔉⇢

When two or more waves travel through the same region at the same time, the principle of superposition states that the resultant displacement at each point is the algebraic sum of the displacements that each wave would produce there on its own. This one principle unlocks the rest of the chapter. Two waves of the same frequency travelling in the same direction combine to give interference, constructive where they arrive in phase and destructive where they arrive out of phase; the resultant amplitude depends on the phase difference between them. Superposition is the reason a chord sounds full, a concert hall has dead spots, and noise-cancelling headphones work. 🔉⇢

A particularly important superposition occurs when a wave meets a boundary and is reflected. At a rigid boundary the reflected wave suffers a phase reversal of $\pi$ — a crest returns as a trough — while at a free (open) boundary it is reflected without any phase change. When a wave and its reflection, of equal amplitude and frequency but travelling in opposite directions, superpose, they form a standing wave: $y(x,t)=2A\sin kx\cos\omega t$. Unlike a travelling wave, a standing wave does not move; it has fixed nodes, where the medium is permanently at rest, and antinodes, where it oscillates with maximum amplitude, spaced half a wavelength apart. No net energy is transported along a standing wave. 🔉⇢

A medium of finite length can support standing waves only at certain special frequencies, its normal modes, fixed by the boundary conditions. A string clamped at both ends must have a node at each end, so its allowed wavelengths are $\lambda_n=2L/n$ and its natural frequencies are $\nu_n=nv/2L$ for $n=1,2,3,\dots$ — the fundamental and its overtones, forming a complete harmonic series. An air column open at both ends behaves similarly, with antinodes at both open ends, but a pipe closed at one end must have a node at the closed end and an antinode at the open end, so it supports only the odd harmonics $\nu_n=(2n-1)v/4L$. These normal modes are why a guitar string and an organ pipe of a given length sound a definite pitch, and they are the single most heavily examined idea in the chapter. 🔉⇢

Superposition also explains beats, the slow throbbing heard when two notes of slightly different frequencies sound together. The resultant amplitude waxes and wanes at a beat frequency equal to the difference of the two frequencies, $\nu_{beat}=|\nu_1-\nu_2|$, a fact musicians exploit to tune instruments by ear: as two strings are brought into tune the beats slow and finally vanish. Beats are a superposition of two waves in time, just as a standing wave pattern is a superposition in space. 🔉⇢

The chapter ends with the Doppler effect: the change in the observed frequency of a wave when the source, the observer, or both are in motion relative to the medium. An approaching source raises the pitch and a receding one lowers it, as anyone who has heard a siren pass will recognise. The general result for sound is $\nu'=\nu\,(v\pm v_o)/(v\mp v_s)$, where $v$ is the speed of sound, $v_o$ the observer's speed and $v_s$ the source's speed, with the signs chosen so that motion which reduces the source-observer separation raises the frequency. Unlike the Doppler effect for light, the effect for sound is not symmetric in the motions of source and observer, because the medium provides a preferred frame. 🔉⇢

For the JEE, waves is a reliable source of one or two marks every year in Main, and it pairs naturally with the preceding chapter on oscillations to feed multi-concept Advanced problems. The examiners return again and again to a small set of ideas: the progressive-wave relation and the difference between particle and wave velocity; the speed of a wave on a string and the speed of sound; the normal modes of strings and of open and closed pipes, including end corrections and resonance-column experiments; beats; and the Doppler effect with its treacherous sign convention. Master the harmonic series for strings and both kinds of pipe, keep the two velocities distinct, and reason out the Doppler signs physically rather than memorising them, and the chapter is largely won. Work the interactive scenes actively — predict a mode's frequency or the shift in pitch before you check it — and the formulas will attach themselves to a physical picture rather than floating free as symbols to be memorised. 🔉⇢

What you will master 🔉⇢

🧠 Concepts Map of the deep dives

This chapter breaks into the concepts below. Each has its own tab with interactive 3D, a full derivation, and a JEE worked example. The Contents tab has the complete prose.

Transverse and LongiProgressive Wave andy(x,t)=a(kx- t+)The Speed of a Travev=/k=The Principle of Supy(x,t)=_i y_i(x,t)▶Reflection of Waves ▶Beats_beat=|_1-_2|The Doppler Effect'=,v v_ov v_s▶
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What you are looking at

A map of the whole chapter. Each box is one concept with its governing relation, and the arrows are the recommended order to learn them in.

Click any box to jump straight to that concept’s tab.

🗺️ Concept map — click any concept to open its tab. Each box shows the concept and its governing formula; the path is the recommended learning order.

Transverse and Longitudinal Waves 🔉⇢

A mechanical wave is a moving pattern of disturbance that transports energy and information through an elastic medium without any bulk transport of matter; it is transverse when the constituents of the medium oscillate perpendicular to the direction of propagation and longitudinal when they oscillate along it.

Progressive Wave and Its Description 🔉⇢

A sinusoidal (harmonic) progressive wave travelling along the positive x-axis is described by $y(x,t)=a\sin(kx-\omega t+\phi)$, in which $a$ is the amplitude, $k=2\pi/\lambda$ is the angular wave number, $\omega=2\pi\nu=2\pi/T$ is the angular frequency, and $(kx-\omega t+\phi)$ is the phase, with $\phi$ the initial phase angle.

The Speed of a Travelling Wave 🔉⇢

The speed of a mechanical wave, $v=\omega/k=\nu\lambda$, is fixed by the medium's elastic and inertial properties: on a stretched string $v=\sqrt{T/\mu}$; for longitudinal waves $v=\sqrt{B/\rho}$, so in a gas Newton's isothermal estimate $v=\sqrt{P/\rho}$ is corrected by Laplace to the adiabatic $v=\sqrt{\gamma P/\rho}=\sqrt{\gamma RT/M}$.

The Principle of Superposition of Waves 🔉⇢

When two or more waves overlap in the same region of a medium, the net displacement of each element is the algebraic sum of the displacements the individual waves would produce separately, $y(x,t)=\sum_i y_i(x,t)$; for two equal-amplitude, same-frequency waves travelling the same way this yields interference with resultant amplitude $2a\cos(\phi/2)$.

Reflection of Waves and Standing Waves 🔉⇢

A wave reflected at a rigid boundary undergoes a phase reversal of $\pi$ (a crest returns as a trough) while reflection at a free/open boundary occurs with no phase change; the superposition of a wave and its oppositely travelling reflection produces a standing wave $y=2a\sin kx\cos\omega t$ whose nodes and antinodes fix the system's normal modes.

Beats 🔉⇢

Beats are the periodic waxing and waning of loudness heard when two harmonic sound waves of nearly equal (but unequal) frequencies are superposed; the intensity fluctuates at the beat frequency $\nu_{beat}=|\nu_1-\nu_2|$, equal to the difference of the two frequencies.

The Doppler Effect 🔉⇢

The Doppler effect is the change in the observed frequency of a wave caused by relative motion between the source and the observer through the medium; for sound the observed frequency is $\nu'=\nu\,\dfrac{v\pm v_o}{v\mp v_s}$, higher on approach and lower on recession, and it is not symmetric in source and observer motion because the medium provides a preferred frame.

📖 Contents Full chapter — read straight through

The complete chapter in reading order. Each section is narratable; the concept tabs add interactive 3D and worked examples on top of this text.

Waves
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What you are looking at

The chapter banner — the physics of this chapter in a single moving picture, with live symbols and values.

Every diagram below it is interactive: drag the controls and the numbers move with the drawing.

Transverse and Longitudinal Waves 🔉⇢

🎯 In a transverse wave the constituents of the medium oscillate perpendicular to the direction the wave travels (a bead on the string only moves up and down); in a longitudinal wave they oscillate along the direction of travel, making compressions and rarefactions (sound). In both, the pattern and its energy move on while each particle stays put. Slide the wavelength and watch the crest spacing change.
transverse (string)longitudinal (sound)
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transverse: particles move ⊥ to propagation; longitudinal: particles move ∥ to propagation. Both carry the pattern to the right while each particle only oscillates in place.
What this shows

In a transverse wave the constituents of the medium oscillate perpendicular to the direction the wave travels (a bead on the string only moves up and down); in a longitudinal wave they oscillate along the direction of travel, making compressions and rarefactions (sound). In both, the pattern and its energy move on while each particle stays put. Slide the wavelength and watch the crest spacing change.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: A mechanical wave is a moving pattern of disturbance that transports energy and information through an elastic medium without any bulk transport of matter; it is transverse when the constituents of the medium oscillate perpendicular to the direction of propagation and longitudinal when they oscillate along it. 🔉⇢

A wave is a disturbance that propagates through a medium while the medium itself, as a whole, stays put. Drop a pebble into a still pond and circular ripples spread outward; yet a cork floating on the surface merely bobs up and down about its mean position and is not carried away with the ripples. This simple observation contains the whole idea of a wave: what moves outward is the pattern of disturbance, carrying energy and information, not the water. In exactly the same way, when we speak, sound moves outward through the air without any net flow of air from our mouth to the listener. 🔉⇢

Mechanical waves owe their existence to the elastic binding between neighbouring constituents of the medium. If one element is displaced from its equilibrium position, elastic (restoring) forces from its neighbours pull it back, and in doing so they disturb those neighbours in turn. The disturbance is therefore handed on from one element to the next, while each element itself only executes small oscillations about its own equilibrium mean position. This is why mechanical waves require a material medium and cannot travel through vacuum, unlike electromagnetic waves. 🔉⇢

Waves are classified by the direction in which the constituents of the medium oscillate relative to the direction in which the wave travels. If the constituents oscillate perpendicular (normal) to the direction of wave propagation, the wave is called a transverse wave. If they oscillate along (parallel to) the direction of propagation, the wave is a longitudinal wave. This single criterion — the geometry of particle motion versus propagation — is the definition you must always return to. 🔉⇢

The standard example of a transverse wave is a wave on a stretched string. When one end of a long string is given a continuous up-and-down (sinusoidal) jerk in the y-direction while the wave travels along the x-direction, each element of the string oscillates up and down about its equilibrium position, normal to the direction of wave motion. The shape that travels along the string is sinusoidal, but no piece of string travels along with it; each element simply moves transversely. 🔉⇢

The standard example of a longitudinal wave is a sound wave in air. Imagine a long pipe filled with air with a piston at one end. A single sudden push-and-pull of the piston generates a pulse consisting of a compression (a region of higher density, also called a condensation) followed by a rarefaction (a region of lower density). If the piston is driven continuously and periodically, a sinusoidal longitudinal wave propagates along the pipe. Here a volume element of air oscillates back and forth in a direction parallel to the direction of wave propagation. 🔉⇢

In a longitudinal sound wave the physical quantity that plays the role of the string's transverse displacement is the change in density (and the associated change in pressure) of a small volume element. When a region is compressed, its molecules are packed closer, so they tend to move out into the adjoining region, compressing it in turn while the first region rarefies. The compression and rarefaction thus travel from one region to the next, making propagation possible. The restoring 'spring' here is the medium's resistance to compression. 🔉⇢

It is essential to distinguish a wave from bulk flow. A wind is the motion of air as a whole from one place to another; a sound wave is a propagation of disturbance (in pressure and density) through air with no net motion of the medium as a whole. Likewise a stream is a bulk flow of water, whereas a water wave carries only a disturbance. Both transverse and longitudinal waves considered here are travelling or progressive waves because the disturbance travels from one part of the medium to another. 🔉⇢

Which type of wave a medium can support depends on the kind of elastic stress it can sustain. In a transverse wave, as the wave passes each element of the medium undergoes a shearing strain. A medium can carry a transverse wave only if it can sustain shearing stress. Solids can sustain shear, and so can support transverse waves; ideal fluids (gases and the interior of liquids) cannot sustain shearing stress and therefore cannot transmit transverse waves through their bulk. 🔉⇢

Longitudinal waves, by contrast, involve only compression and rarefaction — a volumetric (compressive) strain. Both fluids and solids can sustain compressive strain, so longitudinal waves can propagate in all elastic media: solids, liquids and gases. This is why sound (a longitudinal wave) travels through air, water and steel alike, whereas a transverse wave through the bulk cannot travel through air. 🔉⇢

A concrete summary: in a medium like steel, both transverse and longitudinal waves can propagate, because a solid resists both shear and compression. Air can sustain only longitudinal waves, because a gas has no shear rigidity. This distinction is a favourite of examiners: 'solids support both, gases support only longitudinal' is a one-line fact that follows directly from which stress each wave requires. 🔉⇢

Waves on the surface of water are a subtle and important special case. Surface waves are of two kinds: capillary waves and gravity waves. Capillary waves are ripples of short wavelength (not more than a few centimetres) whose restoring force is the surface tension of water. Gravity waves have much longer wavelengths (metres to hundreds of metres) and their restoring force is gravity, which tends to keep the water surface at its lowest level. 🔉⇢

For water surface waves the particle motion is neither purely transverse nor purely longitudinal. The particles do not merely move up and down; they also move back and forth, tracing roughly circular or elliptical paths, and the oscillations are not confined to the surface but extend, with diminishing amplitude, down towards the bottom. Ocean waves are therefore correctly described as a combination of both longitudinal and transverse character. 🔉⇢

This is why, in worked problems, waves produced by a motorboat sailing in water, or ocean surface waves, are classified as 'a combination of both'. A kink in a longitudinal spring produced by displacing one end sideways is likewise a combination, whereas waves produced by moving a piston back and forth in a cylinder of liquid, and ultrasonic waves in air from a vibrating quartz crystal, are purely longitudinal. 🔉⇢

The connection between waves and oscillations is fundamental. The physics of waves in elastic media grew out of the physics of oscillations — masses on springs and the simple pendulum. A useful mental model, due to this history, is a chain of masses connected by springs: pull one mass and the disturbance travels down the chain (like bogies of a train coupled by spring couplings receiving a push from the engine), while each mass only oscillates locally. Waves in elastic media are intimately connected with harmonic oscillations. 🔉⇢

Waves transport two things: energy and information. All communication depends on this. Speech is the production of sound waves in air; hearing is their detection. Signals are routinely converted between forms — sound to an electric current, then to an electromagnetic wave carried by an optical cable or satellite — and detection reverses these steps. The wave, in every case, is the carrier that moves energy and pattern from source to receiver without transporting the medium. 🔉⇢

Not all waves need a medium. Mechanical waves — waves on a string, water waves, sound waves, seismic waves — require a medium and cannot pass through vacuum. Electromagnetic waves (light, radio waves, X-rays) do not require a medium and travel through vacuum at the same speed c = 2.99792458 x 10^8 m/s. A third kind, matter waves, are associated with electrons, protons and other constituents of matter and arise in quantum mechanics. This chapter studies only mechanical waves. 🔉⇢

A word on terminology that unifies the two types. In a transverse wave the crest is the point of maximum positive displacement and the trough is the point of maximum negative displacement. In a longitudinal wave the analogues of crest and trough are the compression (maximum density/pressure) and the rarefaction (minimum density/pressure). Recognising these correspondences lets you carry the mathematics of one type over to the other. 🔉⇢

A further quantitative point worth remembering: it is found that, generally, transverse and longitudinal waves travel with different speeds in the same medium. The speed of each is fixed by the appropriate elastic modulus (shear modulus for transverse, bulk modulus for longitudinal) together with the medium's inertia (mass density). Since a solid has different shear and bulk moduli, the two speeds differ. This is exploited by seismologists, who use the different arrival times of transverse (S) and longitudinal (P) seismic waves. 🔉⇢

Finally, keep the physical picture crisp. A wave is a self-sustaining, travelling pattern of disturbance in an elastic medium: energy and phase move forward, the medium's constituents only oscillate in place, and the geometry of that oscillation (perpendicular versus parallel to propagation) is what sorts every mechanical wave into transverse or longitudinal. Every later result in this chapter — the progressive-wave equation, wave speeds, superposition, standing waves, and beats — is built on this foundation. 🔉⇢

Derivation 🔉⇢

  1. Set up the transverse case: take propagation along the x-axis and let the transverse displacement of a string element be $y$. Because the constituents oscillate perpendicular to propagation, $y$ is measured in the y-direction while the wave advances along $x$.
  2. Assert the transverse-medium condition: a passing transverse wave subjects each element to a shearing strain, so a transverse wave in the bulk requires the medium to sustain shear stress. Hence transverse (bulk) waves exist in solids but not in fluids.
  3. Set up the longitudinal case: take propagation along $x$ and let the displacement of a volume element be $s$, measured along $x$ (parallel to propagation). A compression is a region where $\partial s/\partial x<0$ (density $\rho+\delta\rho$ increases) and a rarefaction where density decreases.
  4. Relate density change to a restoring force: a local compression $\delta\rho$ induces a pressure change $\delta P$; since pressure is force per unit area, this provides a restoring force proportional to the disturbance, exactly analogous to a spring's $F=-kx$. This is why only compressibility (bulk elasticity) is needed.
  5. Assert the longitudinal-medium condition: compressive (volumetric) strain can be sustained by solids, liquids and gases alike, so longitudinal waves propagate in all elastic media.
  6. Compare in one medium: in steel (a solid), shear modulus and bulk modulus are both nonzero, so both wave types exist with generally different speeds; in air (a gas), shear modulus is zero, so only the longitudinal wave survives.
  7. Classify surface water waves: with surface tension providing the restoring force one gets short-wavelength capillary waves, with gravity providing it one gets long-wavelength gravity waves; the particle paths are mixed (up-down and back-forth), so ocean waves are a combination of transverse and longitudinal motion.
⚠️ JEE trap: The classic error is to believe that a wave carries the medium forward with it — that the water in a ripple, or the air in a sound wave, streams outward from the source. It does not. Each element of the medium only oscillates about its fixed equilibrium position (a cork bobs up and down but does not drift with the ripples); what advances is the pattern of disturbance, transporting energy and information, not matter. A related error is to say gases can carry transverse waves. Through the bulk of a gas they cannot, because a gas has no shear rigidity — only the surface of a liquid or a solid can support transverse motion, since a transverse wave demands that the medium sustain a shearing strain. 🔉⇢

Progressive Wave and Its Description 🔉⇢

🎯 A progressive harmonic wave is y(x,t)=A sin(kx-wt): the amplitude A is the peak displacement, the wave number k=2pi/lambda fixes the spatial period and omega=2pi*nu the temporal one. A point of constant phase moves at the wave speed v=omega/k=nu*lambda. Watch the brown particle: it executes SHM up and down while the crest (red) races to the right. Slide A and lambda to reshape the wave.
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y(x,t) = A sin(kx − ωt),   k = 2π/λ,   ω = 2πν,   v = ω/k = νλ. The brown dot is one particle: it only oscillates vertically as the wave passes.
What this shows

A progressive harmonic wave is y(x,t)=A sin(kx-wt): the amplitude A is the peak displacement, the wave number k=2pi/lambda fixes the spatial period and omega=2pi*nu the temporal one. A point of constant phase moves at the wave speed v=omega/k=nu*lambda. Watch the brown particle: it executes SHM up and down while the crest (red) races to the right. Slide A and lambda to reshape the wave.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: A sinusoidal (harmonic) progressive wave travelling along the positive x-axis is described by $y(x,t)=a\sin(kx-\omega t+\phi)$, in which $a$ is the amplitude, $k=2\pi/\lambda$ is the angular wave number, $\omega=2\pi\nu=2\pi/T$ is the angular frequency, and $(kx-\omega t+\phi)$ is the phase, with $\phi$ the initial phase angle. 🔉⇢

To describe a travelling wave mathematically we need a function of both position $x$ and time $t$. At every fixed instant this function must give the shape of the wave in space, and at every fixed location it must describe the oscillatory motion of the constituent of the medium there. For a sinusoidal travelling wave the natural choice is a sine function, and taking the wave to be transverse (displacement $y$, position $x$), the wave is described by $y(x,t)=a\sin(kx-\omega t+\phi)$. 🔉⇢

The phase term $\phi$ is equivalent to writing the wave as a linear combination of sine and cosine: $y(x,t)=A\sin(kx-\omega t)+B\cos(kx-\omega t)$, with $a=\sqrt{A^2+B^2}$ and $\phi=\tan^{-1}(B/A)$. This shows that a single sinusoid with an initial phase is completely general for a wave of given amplitude and given $k,\omega$. 🔉⇢

Why does $y=a\sin(kx-\omega t+\phi)$ represent a wave travelling in the $+x$ direction? Fix an instant $t=t_0$: the argument becomes $kx+\text{constant}$, so the shape of the wave in space is a sine curve. Fix a location $x=x_0$: the argument becomes $-\omega t+\text{constant}$, so the displacement at that point varies sinusoidally with time — each constituent executes simple harmonic motion. 🔉⇢

The travelling character emerges from the combination $kx-\omega t$: as $t$ increases, $x$ must increase to keep the phase $(kx-\omega t+\phi)$ constant, so the wave pattern advances in the $+x$ direction. The companion function $y(x,t)=a\sin(kx+\omega t+\phi)$, with the opposite relative sign, represents a wave travelling in the $-x$ direction. This sign rule is worth memorising: $kx-\omega t \Rightarrow +x$ travel, $kx+\omega t \Rightarrow -x$ travel. 🔉⇢

The amplitude $a$ is the maximum displacement of a constituent from its equilibrium position. Because the sine function ranges between $+1$ and $-1$, the displacement $y$ ranges between $+a$ and $-a$. We take $a$ to be a positive constant without loss of generality; the displacement $y$ may be positive or negative, but the amplitude $a$ is always positive. 🔉⇢

The quantity $(kx-\omega t+\phi)$ is the phase of the wave. Given the amplitude, the phase alone determines the displacement at any position and any instant. The constant $\phi$ is the value of the phase at $x=0,\ t=0$, and is therefore called the initial phase angle (or phase constant). By a suitable choice of the origin of $x$ and of the initial time it is always possible to set $\phi=0$, so there is no loss of generality in dropping $\phi$. 🔉⇢

The wavelength $\lambda$ is defined as the minimum distance between two points having the same phase. For convenience these can be taken as two consecutive crests or two consecutive troughs; the crest is the point of maximum positive displacement and the trough the point of maximum negative displacement. The wavelength is thus the spatial period of the wave. 🔉⇢

To connect $\lambda$ with $k$, take $\phi=0$ and $t=0$, giving $y(x,0)=a\sin kx$. Since the sine function repeats after every change of $2\pi$ in its argument, the displacement repeats when $kx$ increases by $2\pi$, i.e. when $x$ increases by $2\pi/k$. Hence $\lambda=2\pi/k$, or equivalently $k=2\pi/\lambda$. 🔉⇢

The constant $k$ is called the angular wave number or propagation constant. Its SI unit is radian per metre ($\text{rad m}^{-1}$). Physically, $k$ is $2\pi$ times the number of complete waves that fit into unit length; it measures how rapidly the phase advances with distance at a fixed time. 🔉⇢

Now fix a location, say $x=0$, to study time-dependence: $y(0,t)=a\sin(-\omega t)=-a\sin\omega t$. The period $T$ of the wave is the time an element takes to complete one full oscillation, so the displacement must repeat when $\omega t$ increases by $2\pi$: $\omega T=2\pi$, giving $\omega=2\pi/T$. The constant $\omega$ is the angular frequency of the wave, with SI unit $\text{rad s}^{-1}$. 🔉⇢

The frequency $\nu$ is the number of oscillations per second, $\nu=1/T$, so $\nu=\omega/2\pi$ and $\omega=2\pi\nu$. Frequency is measured in hertz (Hz). The frequency (and hence $\omega$) is set by the source of the disturbance, whereas the wavelength (and hence $k$) then follows from the wave speed in the medium. 🔉⇢

The same equation describes a longitudinal wave with one relabelling: the displacement of an element, now parallel to the direction of propagation, is written $s(x,t)=a\sin(kx-\omega t+\phi)$. Here $a$ is the displacement amplitude and all other symbols keep their meaning; only the transverse displacement $y$ is replaced by the longitudinal displacement $s$. Thus one mathematical framework covers both wave types. 🔉⇢

It is vital to distinguish two very different velocities associated with a wave. The wave velocity $v$ is the speed at which the pattern (a point of constant phase, such as a crest) advances; it is a property of the medium. The particle velocity is the velocity of an individual constituent of the medium as it oscillates in place. They are conceptually and numerically distinct, and confusing them is a classic error. 🔉⇢

The particle velocity is obtained by differentiating $y$ with respect to time at fixed $x$: $\partial y/\partial t=-a\omega\cos(kx-\omega t+\phi)$. Its maximum magnitude is $a\omega$, and it oscillates in time exactly as the displacement does, but a quarter cycle ahead. This is the speed of the string element itself, not the speed of the wave. 🔉⇢

The slope of the string at a given instant is $\partial y/\partial x=ak\cos(kx-\omega t+\phi)$. Comparing the two derivatives gives the elegant relation $\dfrac{\partial y}{\partial t}=-v\,\dfrac{\partial y}{\partial x}$, i.e. the particle velocity equals $-v$ times the slope of the waveform. Where the waveform is steep, the element moves fast; at a crest or trough the slope is zero, so the particle is momentarily at rest even though the wave keeps advancing. 🔉⇢

The particle acceleration follows from a second time-derivative: $\partial^2 y/\partial t^2=-\omega^2 y$. This is the defining equation of simple harmonic motion, confirming that every constituent of the medium performs SHM about its equilibrium position with angular frequency $\omega$. A progressive wave is, in this sense, an organised parade of identical simple harmonic oscillators, each lagging its neighbour in phase. 🔉⇢

At a fixed instant, two points a distance $\Delta x$ apart differ in phase by $\Delta\phi=k\,\Delta x=(2\pi/\lambda)\Delta x$. Points one wavelength apart ($\Delta x=\lambda$) are in phase (phase difference $2\pi$); points half a wavelength apart are exactly out of phase (phase difference $\pi$). This phase-difference relation is the workhorse for wave problems and for understanding interference in the later sections. 🔉⇢

In a harmonic progressive wave of a given frequency, all particles vibrate with the same amplitude and the same frequency, but with different phases at a given instant — each point reaches its crest a little later than the one before it. This contrasts sharply with a stationary wave (studied later), in which all particles between two nodes share the same phase but have different amplitudes. 🔉⇢

A worked comparison illustrates the mathematics: a wave $y(x,t)=0.005\sin(80.0x-3.0t)$ (SI units) has amplitude $a=0.005\ \text{m}=5\ \text{mm}$, angular wave number $k=80.0\ \text{m}^{-1}$ and angular frequency $\omega=3.0\ \text{s}^{-1}$. From these, $\lambda=2\pi/k\approx7.85\ \text{cm}$, $T=2\pi/\omega\approx2.09\ \text{s}$, and $\nu=1/T\approx0.48\ \text{Hz}$. Reading off $a,k,\omega$ by comparison with the standard form is the first move in almost every progressive-wave problem. 🔉⇢

In summary, the progressive-wave equation $y=a\sin(kx-\omega t+\phi)$ packages the entire kinematics of a travelling harmonic wave: amplitude and phase, the spatial period through $k=2\pi/\lambda$, the temporal period through $\omega=2\pi/T=2\pi\nu$, and — crucially — the distinction between the wave velocity that carries the pattern and the particle velocity that describes each oscillating element. Mastery of reading and manipulating this single relation underpins the rest of the chapter. 🔉⇢

Derivation 🔉⇢

  1. Start from the general sinusoidal travelling wave $y(x,t)=a\sin(kx-\omega t+\phi)$, with $a$ the amplitude and $(kx-\omega t+\phi)$ the phase.
  2. Wavelength from $k$: set $t=0,\ \phi=0$ to get $y(x,0)=a\sin kx$; the sine repeats when its argument grows by $2\pi$, so displacement repeats over $\Delta x=2\pi/k$. Hence $\lambda=2\pi/k$, i.e. $k=2\pi/\lambda$.
  3. Angular frequency from $T$: fix $x=0$ to get $y(0,t)=-a\sin\omega t$; the motion repeats when $\omega t$ grows by $2\pi$, so $\omega T=2\pi$, giving $\omega=2\pi/T=2\pi\nu$ with $\nu=1/T$.
  4. Particle velocity: differentiate at fixed $x$: $v_{p}=\dfrac{\partial y}{\partial t}=-a\omega\cos(kx-\omega t+\phi)$, with maximum magnitude $a\omega$.
  5. Waveform slope: differentiate at fixed $t$: $\dfrac{\partial y}{\partial x}=a k\cos(kx-\omega t+\phi)$.
  6. Link the two: dividing gives $\dfrac{\partial y}{\partial t}=-\dfrac{\omega}{k}\dfrac{\partial y}{\partial x}=-v\,\dfrac{\partial y}{\partial x}$, since the wave speed is $v=\omega/k$; thus particle velocity $=-v\times(\text{slope})$, distinct from the wave velocity $v$.
  7. Particle acceleration: differentiate the velocity, $\dfrac{\partial^2 y}{\partial t^2}=-\omega^2 y$, the SHM equation, confirming every element performs simple harmonic motion of angular frequency $\omega$.
⚠️ JEE trap: The most common error is to confuse the particle velocity with the wave velocity. The wave velocity $v=\omega/k=\nu\lambda$ is the constant speed at which the pattern (a point of fixed phase, e.g. a crest) travels and is a property of the medium; the particle velocity $\partial y/\partial t=-a\omega\cos(kx-\omega t+\phi)$ is the speed of an individual element oscillating in place, and it changes continuously, reaching zero at the crest and trough and a maximum $a\omega$ as the element crosses its mean position. A second error is to think a crest moves because the material there moves forward; in fact the crest is just the location where the phase currently equals $\pi/2$, and it advances as neighbouring elements successively reach maximum displacement while each element stays on its own line. 🔉⇢

The Speed of a Travelling Wave 🔉⇢

🎯 A transverse pulse on a stretched string travels at v = sqrt(T/mu), a restoring-force factor (tension T) over an inertia factor (linear mass density mu) under a root. Stretch the string tighter and it snaps back faster, so the wave speeds up; load it with mass and it slows. The same restoring/inertia form gives the speed of sound, v = sqrt(gamma*P/rho). Slide T and mu and read v.
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v = √(T/μ)   on a stretched string: T = tension, μ = mass per unit length. Tighten the string (T↑) or lighten it (μ↓) and the pulse travels faster.
What this shows

A transverse pulse on a stretched string travels at v = sqrt(T/mu), a restoring-force factor (tension T) over an inertia factor (linear mass density mu) under a root. Stretch the string tighter and it snaps back faster, so the wave speeds up; load it with mass and it slows. The same restoring/inertia form gives the speed of sound, v = sqrt(gamma*P/rho). Slide T and mu and read v.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: The speed of a mechanical wave, $v=\omega/k=\nu\lambda$, is fixed by the medium's elastic and inertial properties: on a stretched string $v=\sqrt{T/\mu}$; for longitudinal waves $v=\sqrt{B/\rho}$, so in a gas Newton's isothermal estimate $v=\sqrt{P/\rho}$ is corrected by Laplace to the adiabatic $v=\sqrt{\gamma P/\rho}=\sqrt{\gamma RT/M}$. 🔉⇢

To find how fast a travelling wave moves, fix attention on any point of constant phase — conveniently a crest — and follow it in time. Over a small interval $\Delta t$ the whole wave pattern shifts to the right by $\Delta x$, and the crest moves the same $\Delta x$. The speed of the wave is $v=\Delta x/\Delta t$. A point of any other phase moves with the same speed; otherwise the pattern would not stay rigid. 🔉⇢

The condition that a chosen phase point moves at constant speed is that its phase stays fixed: $kx-\omega t=\text{constant}$. Requiring $k\,\Delta x-\omega\,\Delta t=0$ and letting the intervals shrink gives $\dfrac{dx}{dt}=\dfrac{\omega}{k}=v$. Thus the wave speed is the ratio of angular frequency to angular wave number. 🔉⇢

Expressing $\omega$ and $k$ in terms of period and wavelength, $\omega=2\pi/T$ and $k=2\pi/\lambda$, gives $v=\dfrac{\omega}{k}=\dfrac{\lambda}{T}=\lambda\nu$. This general relation $v=\nu\lambda$ holds for all progressive waves: in the time of one full oscillation of any constituent, the wave pattern advances by exactly one wavelength. 🔉⇢

A crucial point of physics: the speed of a mechanical wave is determined by the medium, not by the source. It is set by the inertial property (linear mass density for strings, mass density in general) and the elastic property (an appropriate modulus — tension for a string, Young's modulus for a solid bar, bulk modulus for a fluid). Given the speed, $v=\nu\lambda$ then relates wavelength to frequency; the source chooses the frequency, and the medium fixes the wavelength via $\lambda=v/\nu$. 🔉⇢

For a transverse wave on a stretched string, the restoring force is the tension $T$ and the inertial property is the linear mass density $\mu=m/L$ (mass per unit length). We expect the speed to increase with the restoring force and decrease with the inertia. Dimensional analysis confirms this: $[T]=\text{MLT}^{-2}$ and $[\mu]=\text{ML}^{-1}$, and the only combination with dimension of speed $\text{LT}^{-1}$ is $\sqrt{T/\mu}$. 🔉⇢

Dimensional analysis alone cannot fix the dimensionless constant, so we write $v=C\sqrt{T/\mu}$. An exact derivation from Newton's laws (beyond the present scope) shows $C=1$, so the speed of transverse waves on a stretched string is $v=\sqrt{T/\mu}$. Note that $v$ depends only on the properties of the medium ($T$ and $\mu$) and not on the wavelength or frequency of the wave. 🔉⇢

A numerical illustration: a steel wire $0.72\ \text{m}$ long with mass $5.0\times10^{-3}\ \text{kg}$ has $\mu=6.9\times10^{-3}\ \text{kg m}^{-1}$. Under a tension of $60\ \text{N}$ the transverse-wave speed is $v=\sqrt{60/(6.9\times10^{-3})}\approx93\ \text{m s}^{-1}$. The tension here is a property of the stretched string (arising from the external stretching force), so $v$ is genuinely a medium property. 🔉⇢

For a longitudinal wave the constituents oscillate forward and backward, so the relevant elasticity is resistance to compression — the bulk modulus $B=-\Delta P/(\Delta V/V)$, which measures the pressure change needed to produce a fractional volume change. The inertial property is again the mass density $\rho$. Dimensional analysis on $B$ ($[B]=\text{ML}^{-1}\text{T}^{-2}$, like pressure) and $\rho$ ($[\rho]=\text{ML}^{-3}$) yields $\sqrt{B/\rho}$ as the only speed. 🔉⇢

Writing $v=C\sqrt{B/\rho}$ and using the exact result $C=1$, the general formula for longitudinal waves in a fluid medium is $v=\sqrt{B/\rho}$. For a solid bar, where lateral expansion is negligible and the strain is essentially longitudinal, the relevant modulus is Young's modulus $Y$, giving $v=\sqrt{Y/\rho}$ for longitudinal waves in a bar. 🔉⇢

Liquids and solids generally transmit sound faster than gases. At first this seems paradoxical because solids and liquids have far higher densities $\rho$, which by $v=\sqrt{B/\rho}$ would slow the wave. The resolution is that their bulk moduli $B$ are enormously larger — they are much harder to compress — and this increase in $B$ outweighs the increase in $\rho$, so the net speed is higher. 🔉⇢

To estimate the speed of sound in a gas, treat it as an ideal gas obeying $PV=Nk_{B}T$. Newton assumed the compressions and rarefactions occur isothermally. For an isothermal change $V\Delta P+P\Delta V=0$, so $-\Delta P/(\Delta V/V)=P$; hence the effective bulk modulus is $B=P$, and the speed of sound is $v=\sqrt{P/\rho}$. This is Newton's formula. 🔉⇢

Newton's formula gives about $280\ \text{m s}^{-1}$ for air at STP (using $\rho_0\approx1.29\ \text{kg m}^{-3}$), roughly $15\%$ below the measured value of $331\ \text{m s}^{-1}$. The discrepancy is too large to ignore, and it signals that the isothermal assumption is wrong. 🔉⇢

Laplace corrected the error. The pressure variations in a sound wave are so rapid that heat has no time to flow between compressions and rarefactions; the process is adiabatic, not isothermal. For an adiabatic change of an ideal gas $PV^{\gamma}=\text{constant}$, from which the adiabatic bulk modulus is $B_{ad}=\gamma P$, where $\gamma=C_p/C_v$ is the ratio of specific heats. 🔉⇢

Substituting the adiabatic bulk modulus gives the Laplace-corrected speed of sound in a gas: $v=\sqrt{\gamma P/\rho}$. For air $\gamma=7/5$, and this yields $v\approx331.3\ \text{m s}^{-1}$ at STP, in excellent agreement with the measured value. This modification of Newton's formula is called the Laplace correction. 🔉⇢

The adiabatic formula can be rewritten using the ideal-gas law. Writing $\rho=PM/(RT)$ for a gas of molar mass $M$, the pressure cancels: $v=\sqrt{\gamma P/\rho}=\sqrt{\gamma RT/M}$. This compact form exposes the true dependences of the speed of sound in a gas. 🔉⇢

Three consequences follow immediately and are frequently examined. First, the speed of sound in an ideal gas is independent of pressure at fixed temperature: increasing $P$ increases $\rho$ in the same proportion, so $P/\rho$ (and hence $v$) is unchanged. Second, the speed increases with temperature as $v\propto\sqrt{T}$ (absolute temperature). Third, the speed increases with humidity, because moist air (containing lighter water molecules) has a lower effective molar mass $M$ than dry air. 🔉⇢

The temperature dependence is worth quantifying: because $v\propto\sqrt{T}$, sound travels faster on a hot day than on a cold one. Near room temperature a useful rule is that the speed of sound in air rises by roughly $0.6\ \text{m s}^{-1}$ for each degree Celsius, following directly from $v=\sqrt{\gamma RT/M}$ with $T$ in kelvin. 🔉⇢

Note that these wave speeds are independent of amplitude, wavelength and frequency (for a non-dispersive medium like ideal air). In higher studies one meets dispersive media whose wave speed does depend on frequency, causing a pulse to change shape as it travels; but the ordinary formulas $\sqrt{T/\mu}$, $\sqrt{B/\rho}$ and $\sqrt{\gamma P/\rho}$ describe non-dispersive propagation. 🔉⇢

It is also worth recalling that in the same medium transverse and longitudinal waves generally travel at different speeds, because they sample different moduli: a transverse wave on/in a solid responds to shear, while the longitudinal wave responds to compression (bulk or Young's modulus). Seismic P-waves (longitudinal) and S-waves (transverse) illustrate this — the P-wave arrives first, and the time gap between them is used to locate earthquakes. 🔉⇢

In summary, the speed of any progressive wave is $v=\omega/k=\nu\lambda$, but its value is fixed by the medium: $\sqrt{T/\mu}$ for a string, $\sqrt{B/\rho}$ for a fluid, $\sqrt{Y/\rho}$ for a solid bar, and $\sqrt{\gamma P/\rho}=\sqrt{\gamma RT/M}$ for sound in a gas after the Laplace correction. Newton's isothermal $\sqrt{P/\rho}$ is the instructive wrong answer that motivates the whole correction. 🔉⇢

Derivation 🔉⇢

  1. Wave speed from phase: require the phase of $y=a\sin(kx-\omega t)$ to be constant, $kx-\omega t=\text{const}$; differentiating, $k\,dx-\omega\,dt=0$, so $v=\dfrac{dx}{dt}=\dfrac{\omega}{k}$, and with $\omega=2\pi/T,\ k=2\pi/\lambda$ this is $v=\dfrac{\lambda}{T}=\nu\lambda$.
  2. String speed by dimensions: with $[T]=\text{MLT}^{-2}$ and $[\mu]=\text{ML}^{-1}$, the combination $T/\mu$ has dimension $\text{L}^2\text{T}^{-2}$; hence $v=C\sqrt{T/\mu}$, and the exact theory gives $C=1$, so $v=\sqrt{T/\mu}$.
  3. Longitudinal speed by dimensions: with $[B]=\text{ML}^{-1}\text{T}^{-2}$ and $[\rho]=\text{ML}^{-3}$, $B/\rho$ has dimension $\text{L}^2\text{T}^{-2}$; hence $v=C\sqrt{B/\rho}$ with $C=1$, giving $v=\sqrt{B/\rho}$ (and $v=\sqrt{Y/\rho}$ for a solid bar).
  4. Newton (isothermal): for $PV=Nk_BT$ at constant $T$, $V\Delta P+P\Delta V=0\Rightarrow -\Delta P/(\Delta V/V)=P$, so the effective bulk modulus is $B=P$ and $v=\sqrt{P/\rho}$.
  5. Numerical check of Newton: at STP $\rho_0=(29.0\times10^{-3})/(22.4\times10^{-3})\approx1.29\ \text{kg m}^{-3}$ and $P\approx1.01\times10^5\ \text{Pa}$, giving $v\approx280\ \text{m s}^{-1}$ — about $15\%$ below the measured $331\ \text{m s}^{-1}$.
  6. Laplace correction: sound compressions are adiabatic, $PV^{\gamma}=\text{const}$, so differentiating gives adiabatic bulk modulus $B_{ad}=\gamma P$; hence $v=\sqrt{\gamma P/\rho}$, and with $\gamma=7/5$ for air this yields $\approx331\ \text{m s}^{-1}$.
  7. Reveal the dependences: substitute $\rho=PM/(RT)$ into $v=\sqrt{\gamma P/\rho}$ to obtain $v=\sqrt{\gamma RT/M}$, showing $v$ is independent of $P$, increases as $\sqrt{T}$, and increases as $M$ decreases (humid air is faster).
⚠️ JEE trap: Two errors dominate here. The first is Newton's own: assuming the pressure variations in a sound wave are isothermal, which gives $v=\sqrt{P/\rho}\approx280\ \text{m s}^{-1}$, about $15\%$ too low. The variations are actually adiabatic (heat cannot flow fast enough), so the correct bulk modulus is $\gamma P$, not $P$, and $v=\sqrt{\gamma P/\rho}\approx331\ \text{m s}^{-1}$. The second error is to think the speed of sound depends on pressure: since $v=\sqrt{\gamma P/\rho}$ and $\rho\propto P$ at fixed temperature, the pressure cancels and $v=\sqrt{\gamma RT/M}$ depends only on temperature and the gas's molar mass — raising the pressure of a gas at fixed temperature does not change the speed of sound in it. 🔉⇢

The Principle of Superposition of Waves 🔉⇢

🎯 When two waves cross the same region, the principle of superposition says the resultant displacement is the algebraic sum y = y1 + y2. Two equal waves in phase (phi=0) reinforce into an amplitude 2A (constructive interference); exactly out of phase (phi=pi) they cancel (destructive). Slide the phase difference and watch the red resultant grow and shrink between these limits.
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y = y1 + y2   (principle of superposition). In phase (φ=0) → constructive (amplitude adds); out of phase (φ=π) → destructive (they cancel).
What this shows

When two waves cross the same region, the principle of superposition says the resultant displacement is the algebraic sum y = y1 + y2. Two equal waves in phase (phi=0) reinforce into an amplitude 2A (constructive interference); exactly out of phase (phi=pi) they cancel (destructive). Slide the phase difference and watch the red resultant grow and shrink between these limits.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: When two or more waves overlap in the same region of a medium, the net displacement of each element is the algebraic sum of the displacements the individual waves would produce separately, $y(x,t)=\sum_i y_i(x,t)$; for two equal-amplitude, same-frequency waves travelling the same way this yields interference with resultant amplitude $2a\cos(\phi/2)$. 🔉⇢

Consider what happens when two wave pulses travelling in opposite directions cross each other on a string. Remarkably, after they have crossed, each pulse continues on its way with its original shape and speed, completely unaffected — the pulses retain their identities. It is only during the interval of overlap that the wave pattern differs from either pulse alone. 🔉⇢

Full derivation, worked example and interactive 3D on the The Principle of Superposition of Waves tab →

Reflection of Waves and Standing Waves 🔉⇢

🎯 A wave and its reflection superpose into a standing wave y = 2A sin(kx) cos(omega*t): the pattern does not travel. Points where sin(kx)=0 are nodes (permanently at rest, red dots); halfway between are antinodes that swing with maximum amplitude. A string fixed at both ends must have a node at each end, so only wavelengths 2L/n fit, giving natural frequencies nu_n = n v/2L. Step the harmonic number n.
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y = 2A sin(kx) cos(ωt),   νn = n v / 2L   (string, both ends fixed). Nodes stay at rest; antinodes swing hardest. Mode n has n antinodes.
What this shows

A wave and its reflection superpose into a standing wave y = 2A sin(kx) cos(omega*t): the pattern does not travel. Points where sin(kx)=0 are nodes (permanently at rest, red dots); halfway between are antinodes that swing with maximum amplitude. A string fixed at both ends must have a node at each end, so only wavelengths 2L/n fit, giving natural frequencies nu_n = n v/2L. Step the harmonic number n.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: A wave reflected at a rigid boundary undergoes a phase reversal of $\pi$ (a crest returns as a trough) while reflection at a free/open boundary occurs with no phase change; the superposition of a wave and its oppositely travelling reflection produces a standing wave $y=2a\sin kx\cos\omega t$ whose nodes and antinodes fix the system's normal modes. 🔉⇢

So far we have considered waves in an unbounded medium. What happens when a pulse or wave meets a boundary? If the boundary is rigid, the wave is reflected — the phenomenon of an echo is reflection of sound by a rigid boundary. If the boundary is not completely rigid, or is an interface between two different media, part of the incident wave is reflected and part is transmitted (refracted) into the second medium. 🔉⇢

Full derivation, worked example and interactive 3D on the Reflection of Waves and Standing Waves tab →

Beats 🔉⇢

🎯 Two waves of nearly equal frequency superpose to give a resultant whose amplitude waxes and wanes: beats. The red envelope traces this swelling and fading, and the number of loudness maxima per second is the beat frequency nu_beat = |nu1 - nu2|. Musicians tune by ear this way: as two strings approach the same pitch the beats slow and vanish. Slide the frequency difference and count the throbs.
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νbeat = |ν1 − ν2|. Two near-equal tones add to a sound whose loudness swells and fades; the throbs per second equal the frequency difference.
What this shows

Two waves of nearly equal frequency superpose to give a resultant whose amplitude waxes and wanes: beats. The red envelope traces this swelling and fading, and the number of loudness maxima per second is the beat frequency nu_beat = |nu1 - nu2|. Musicians tune by ear this way: as two strings approach the same pitch the beats slow and vanish. Slide the frequency difference and count the throbs.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: Beats are the periodic waxing and waning of loudness heard when two harmonic sound waves of nearly equal (but unequal) frequencies are superposed; the intensity fluctuates at the beat frequency $\nu_{beat}=|\nu_1-\nu_2|$, equal to the difference of the two frequencies. 🔉⇢

Beats are an interesting phenomenon that arises from the interference of waves — but interference in time rather than in space. When two harmonic sound waves of close (but not equal) frequencies are heard at the same time, we hear a tone of intermediate frequency (the average of the two) whose loudness rises and falls periodically. These audibly distinct swells and fades of intensity are the beats. 🔉⇢

The rate at which the loudness waxes and wanes equals the difference between the two close frequencies. Musicians exploit this constantly: when tuning one instrument to another, they listen for beats and adjust until the beats slow down and finally vanish, at which point the two frequencies are equal and the instruments are in tune. 🔉⇢

To analyse beats mathematically, consider two harmonic sound waves of nearly equal angular frequencies $\omega_1$ and $\omega_2$, and fix attention on a single location, $x=0$, for convenience. Choosing a phase of $\pi/2$ for each (so the sines become cosines) and assuming equal amplitudes, the two displacements are $s_1=a\cos\omega_1 t$ and $s_2=a\cos\omega_2 t$. 🔉⇢

We write $s$ rather than $y$ because sound is a longitudinal wave, so the displacement is along the direction of propagation. Let $\omega_1$ be the slightly greater of the two frequencies. By the principle of superposition the resultant displacement at that point is the algebraic sum $s=s_1+s_2=a(\cos\omega_1 t+\cos\omega_2 t)$. 🔉⇢

Using the identity $\cos A+\cos B=2\cos\!\big(\tfrac{A-B}{2}\big)\cos\!\big(\tfrac{A+B}{2}\big)$, the resultant becomes $s=\Big[2a\cos\!\big(\tfrac{\omega_1-\omega_2}{2}\big)t\Big]\cos\!\big(\tfrac{\omega_1+\omega_2}{2}\big)t$. This is naturally read as a fast oscillation modulated by a slowly varying amplitude. 🔉⇢

Define $\omega_a=\tfrac{\omega_1+\omega_2}{2}$ (the average) and $\omega_b=\tfrac{\omega_1-\omega_2}{2}$ (half the difference). Then $s=[2a\cos\omega_b t]\cos\omega_a t$. Because the two frequencies are close, $|\omega_1-\omega_2|\ll\omega_1,\omega_2$, so $\omega_a\gg\omega_b$: the term $\cos\omega_a t$ is a rapid oscillation and $2a\cos\omega_b t$ is a slowly varying amplitude envelope. 🔉⇢

The resultant is therefore a wave oscillating at the average angular frequency $\omega_a$, but with an amplitude that is no longer constant — unlike a pure harmonic wave. The amplitude $|2a\cos\omega_b t|$ swells to its maximum $2a$ whenever $\cos\omega_b t=\pm1$ and falls to zero whenever $\cos\omega_b t=0$. The ear perceives this changing amplitude as changing loudness. 🔉⇢

Now count how often the loudness peaks. Intensity depends on the square of the amplitude, and $\cos\omega_b t$ reaches its extreme value $+1$ or $-1$ twice per cycle of $\cos\omega_b t$. Hence the intensity waxes and wanes at angular frequency $2\omega_b=\omega_1-\omega_2$. Since $\omega=2\pi\nu$, the beat frequency in hertz is $\nu_{beat}=\nu_1-\nu_2$. 🔉⇢

More precisely, because loudness responds to the magnitude of the amplitude (a maximum occurs for both $+2a$ and $-2a$), the audible beat frequency is the full difference $|\nu_1-\nu_2|$, not half of it. So $\nu_{beat}=|\nu_1-\nu_2|$: the number of intensity maxima heard per second equals the difference of the two source frequencies. 🔉⇢

A concrete example: two waves of $11\ \text{Hz}$ and $9\ \text{Hz}$ produce a resultant that oscillates at the average $10\ \text{Hz}$ but whose amplitude waxes and wanes $2$ times each second — the beats occur at $2\ \text{Hz}=|11-9|\ \text{Hz}$. You would hear a $10\ \text{Hz}$-like tone throbbing twice per second. 🔉⇢

The condition for the analysis to hold — and for beats to be perceived as distinct throbs — is that the two frequencies be close. If they differ widely, the difference frequency is too high for the ear to follow as separate swells (beats above roughly $10\ \text{Hz}$ merge into a rough or dissonant tone). Beats are thus a small-difference phenomenon, which is exactly why they are so useful for fine tuning. 🔉⇢

The tuning application deserves emphasis because it is the standard exam scenario. If a string produces beats with a standard fork, changing the string's tension changes its frequency; watching whether the beat frequency increases or decreases tells you which side of the standard your string is on, and hence which way to adjust. The beats vanish precisely at a match. 🔉⇢

Consider the classic problem: two sitar strings A and B, both playing 'Dha', are slightly out of tune and produce $5$ beats per second, so $|\nu_A-\nu_B|=5\ \text{Hz}$. The tension in B is increased slightly — which raises B's frequency — and the beat frequency falls to $3\ \text{Hz}$. Because raising $\nu_B$ reduced the beat rate, $\nu_B$ must have been below $\nu_A$ (it moved toward A). With $\nu_A=427\ \text{Hz}$ and $\nu_A-\nu_B=5\ \text{Hz}$, we get $\nu_B=422\ \text{Hz}$. 🔉⇢

The sign reasoning in such problems is the crux and repays care. The beat frequency only gives the magnitude $|\nu_1-\nu_2|$, so on its own it cannot say which frequency is larger. The extra piece of information — how the beat rate responds to a known change (increasing tension raises frequency) — resolves the ambiguity: if a change that raises $\nu_B$ lowers the beat frequency, then $\nu_B<\nu_A$; if it raises the beat frequency, then $\nu_B>\nu_A$. 🔉⇢

Physically, beats are just the principle of superposition applied to two nearly equal frequencies: at some instants the two waves arrive in phase and reinforce (loud), a moment later they have drifted to being out of phase and cancel (soft), because the slightly faster wave gains a full cycle on the slower one once every $1/|\nu_1-\nu_2|$ seconds. The 'gaining a full cycle' picture gives the beat period directly as $T_{beat}=1/|\nu_1-\nu_2|$. 🔉⇢

Beats also connect to broader technology and biology. The same mathematics underlies amplitude modulation in radio and the heterodyne technique in electronics, where two frequencies are mixed to produce a difference frequency. In each case a slowly varying envelope carries usable information at the difference frequency $|\nu_1-\nu_2|$. 🔉⇢

It is worth contrasting beats with the spatial interference of the superposition section. There, two waves of the same frequency with a fixed phase difference gave a steady pattern with a position-dependent amplitude $2a\cos(\phi/2)$. Here, two waves of slightly different frequencies at a single point give a time-dependent amplitude $2a\cos\omega_b t$: the roles of space and time are, in a sense, exchanged, but the underlying rule is the same superposition. 🔉⇢

In summary, superposing two harmonic sounds of nearly equal frequencies $\nu_1$ and $\nu_2$ yields a tone at the average frequency whose amplitude is modulated by $2a\cos\omega_b t$, producing audible beats at frequency $\nu_{beat}=|\nu_1-\nu_2|$. Beats are the ear's direct experience of the superposition principle in time, and they provide a sensitive, practical method for detecting tiny frequency differences and for tuning instruments. 🔉⇢

Derivation 🔉⇢

  1. Set two nearly-equal-frequency sounds at a fixed point $x=0$ with equal amplitude: $s_1=a\cos\omega_1 t$ and $s_2=a\cos\omega_2 t$, with $\omega_1$ slightly greater than $\omega_2$.
  2. Superpose: $s=s_1+s_2=a(\cos\omega_1 t+\cos\omega_2 t)$.
  3. Apply $\cos A+\cos B=2\cos\!\big(\tfrac{A-B}{2}\big)\cos\!\big(\tfrac{A+B}{2}\big)$: $s=2a\cos\!\big(\tfrac{\omega_1-\omega_2}{2}\big)t\;\cos\!\big(\tfrac{\omega_1+\omega_2}{2}\big)t$.
  4. Define $\omega_a=\tfrac{\omega_1+\omega_2}{2}$ and $\omega_b=\tfrac{\omega_1-\omega_2}{2}$, so $s=[2a\cos\omega_b t]\cos\omega_a t$; since $|\omega_1-\omega_2|\ll\omega_{1,2}$, we have $\omega_a\gg\omega_b$.
  5. Interpret: a fast oscillation at $\omega_a$ with a slowly varying amplitude envelope $2a\cos\omega_b t$ that reaches magnitude $2a$ when $\cos\omega_b t=\pm1$ and $0$ when $\cos\omega_b t=0$.
  6. Count intensity maxima: loudness peaks each time $\cos\omega_b t=\pm1$, i.e. twice per cycle of $\omega_b$, so the intensity waxes and wanes at $2\omega_b=\omega_1-\omega_2$.
  7. Convert to frequency with $\omega=2\pi\nu$: the beat frequency is $\nu_{beat}=\nu_1-\nu_2$ (magnitude $|\nu_1-\nu_2|$), with beat period $T_{beat}=1/|\nu_1-\nu_2|$.
⚠️ JEE trap: A widespread error is to take the beat frequency as the average, or as half the difference $\tfrac12|\nu_1-\nu_2|$, of the two frequencies. It is neither: the tone you hear is at the average frequency, but the throbbing (the rate at which loudness swells and fades) is the full difference $\nu_{beat}=|\nu_1-\nu_2|$, because intensity peaks each time the amplitude envelope reaches either $+2a$ or $-2a$. A second, subtler error is to think the beat frequency alone tells you which source is higher in pitch. It gives only the magnitude $|\nu_1-\nu_2|$; to decide the sign you need extra information, such as how the beat rate changes when you deliberately alter one frequency (increasing a string's tension raises its frequency) — if that change lowers the beat rate, the altered string was the lower of the two. 🔉⇢

The Doppler Effect 🔉⇢

🎯 The Doppler effect is the change in observed frequency when the source or observer moves relative to the medium. A source moving towards observer B (right) crowds the wavefronts ahead of it, so B hears a higher frequency; the fronts behind spread out, so observer A (left) hears a lower one: nu' = nu (v +/- v_o)/(v -/+ v_s). Slide the source speed v_s and read the shifted frequencies. The signs are fixed by whether the motion is closing the gap (pitch up) or opening it (pitch down).
observer Aobserver B
🔉⇢
ν′ = ν (v ± vo)/(v ∓ vs). A source moving towards an observer crowds its wavefronts (pitch ↑); moving away spreads them (pitch ↓).
What this shows

The Doppler effect is the change in observed frequency when the source or observer moves relative to the medium. A source moving towards observer B (right) crowds the wavefronts ahead of it, so B hears a higher frequency; the fronts behind spread out, so observer A (left) hears a lower one: nu' = nu (v +/- v_o)/(v -/+ v_s). Slide the source speed v_s and read the shifted frequencies. The signs are fixed by whether the motion is closing the gap (pitch up) or opening it (pitch down).

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: The Doppler effect is the change in the observed frequency of a wave caused by relative motion between the source and the observer through the medium; for sound the observed frequency is $\nu'=\nu\,\dfrac{v\pm v_o}{v\mp v_s}$, higher on approach and lower on recession, and it is not symmetric in source and observer motion because the medium provides a preferred frame. 🔉⇢

You have surely noticed that the pitch of a train whistle or an ambulance siren sounds higher as the vehicle approaches you and abruptly lower as it recedes. This change in the observed frequency of a wave due to relative motion between the source and the observer (through the medium) is called the Doppler effect, after Christian Doppler who first described it. The emitted frequency of the source never changes; only the frequency received by the observer does. 🔉⇢

Full derivation, worked example and interactive 3D on the The Doppler Effect tab →

The Principle of Superposition of Waves 🔉⇢deep concept

Definition: When two or more waves overlap in the same region of a medium, the net displacement of each element is the algebraic sum of the displacements the individual waves would produce separately, $y(x,t)=\sum_i y_i(x,t)$; for two equal-amplitude, same-frequency waves travelling the same way this yields interference with resultant amplitude $2a\cos(\phi/2)$. 🔉⇢

🔬 Interactive 3D · Two equal harmonic waves combine; a phase-difference slider morphs the resultant amplitude between $2a$ (in phase, constructive) and $0$ ($\pi$ out of phase, destructive).

Consider what happens when two wave pulses travelling in opposite directions cross each other on a string. Remarkably, after they have crossed, each pulse continues on its way with its original shape and speed, completely unaffected — the pulses retain their identities. It is only during the interval of overlap that the wave pattern differs from either pulse alone. 🔉⇢

During the overlap, the net displacement of any element of the medium is the algebraic sum of the displacements that each pulse would give it separately. This is the principle of superposition of waves. Because displacements can be positive or negative, they add with sign: two crests reinforce, while a crest meeting an equal trough cancels. 🔉⇢

A dramatic illustration is two pulses of equal and opposite shape approaching each other. At the instant they exactly overlap, the algebraic sum of their displacements is zero everywhere, so the string is momentarily flat with zero displacement throughout. Yet the energy is not lost: it resides in the transverse motion (kinetic energy) of the string elements at that instant, and the pulses re-emerge unchanged immediately afterwards. 🔉⇢

Stated mathematically, if $y_1(x,t)$ and $y_2(x,t)$ are the displacements due to two disturbances that overlap in a region, the net displacement is $y(x,t)=y_1(x,t)+y_2(x,t)$. The principle says each wave propagates as if the other were not present; the constituents of the medium simply respond to the sum of the two demands placed on them. 🔉⇢

The principle generalises to any number of waves. If waves with wave functions $y_1=f_1(x-vt),\ y_2=f_2(x-vt),\ \dots,\ y_n=f_n(x-vt)$ move in the medium, the resultant disturbance is $y=\sum_{i=1}^{n}f_i(x-vt)$. This additive rule is the mathematical statement that the governing wave equation is linear, so sums of solutions are again solutions. 🔉⇢

The principle of superposition is basic to the phenomenon of interference — the systematic reinforcement or cancellation that results when waves of the same frequency overlap with a definite phase relationship. To see interference emerge, we specialise to two harmonic travelling waves on a stretched string. 🔉⇢

Take two harmonic waves with the same angular frequency $\omega$ and the same angular wave number $k$ (hence the same wavelength $\lambda$ and the same wave speed), equal amplitudes $a$, and both travelling in the $+x$ direction. Let them differ only in initial phase: $y_1(x,t)=a\sin(kx-\omega t)$ and $y_2(x,t)=a\sin(kx-\omega t+\phi)$, where $\phi$ is the constant phase difference. 🔉⇢

By superposition the resultant is $y(x,t)=a\sin(kx-\omega t)+a\sin(kx-\omega t+\phi)$. Applying the identity $\sin A+\sin B=2\sin\!\big(\tfrac{A+B}{2}\big)\cos\!\big(\tfrac{A-B}{2}\big)$ gives $y(x,t)=\big[2a\cos\tfrac{\phi}{2}\big]\sin\!\big(kx-\omega t+\tfrac{\phi}{2}\big)$. 🔉⇢

This resultant is itself a harmonic travelling wave in the $+x$ direction, with the same frequency $\omega$ and the same wavelength $\lambda$. Its initial phase is $\phi/2$, midway between the two waves. The important new feature is that its amplitude, $A(\phi)=2a\cos(\phi/2)$, depends on the phase difference $\phi$ between the two constituent waves. 🔉⇢

The amplitude relation $A(\phi)=2a\cos(\phi/2)$ is the heart of two-wave interference. It shows that by simply changing the relative phase of two identical waves, the resultant amplitude can be tuned continuously from a maximum of $2a$ down to zero, without changing the amplitude of either individual wave. 🔉⇢

When $\phi=0$ the two waves are exactly in phase. Then $A=2a\cos0=2a$ and $y(x,t)=2a\sin(kx-\omega t)$, the largest possible resultant. This is constructive interference: the amplitudes add up in the resultant wave. Crest falls on crest and trough on trough everywhere. 🔉⇢

When $\phi=\pi$ the two waves are exactly out of phase. Then $A=2a\cos(\pi/2)=0$ and $y(x,t)=0$ at all points and all times. This is destructive interference: the amplitudes subtract out and the string remains undisturbed. Crest falls on trough everywhere, so they cancel completely. 🔉⇢

More generally, constructive interference occurs whenever $\phi$ is an even multiple of $\pi$ (i.e. $\phi=2n\pi$, an integral multiple of $2\pi$), because then $\cos(\phi/2)=\pm1$ and the amplitude is maximal. Destructive interference occurs whenever $\phi$ is an odd multiple of $\pi$ (i.e. $\phi=(2n+1)\pi$), because then $\cos(\phi/2)=0$. 🔉⇢

A subtle but important consequence concerns energy. In destructive interference the energy is not destroyed; it is redistributed. In a full interference pattern (as in Young's double slit, studied later), the energy missing from the dark regions reappears in the bright regions, so the total energy is conserved. The superposition principle rearranges energy in space; it never creates or annihilates it. 🔉⇢

The phase difference $\phi$ is often produced by a path difference. Two waves that set out in phase but travel different distances to a point arrive with a phase difference $\phi=\dfrac{2\pi}{\lambda}\times(\text{path difference})$. Constructive interference then corresponds to a path difference of a whole number of wavelengths, $n\lambda$, and destructive interference to an odd number of half-wavelengths, $(n+\tfrac{1}{2})\lambda$. 🔉⇢

The linearity that underpins superposition is an approximation that holds for waves of small amplitude, which is the regime of this chapter. For very large-amplitude disturbances the medium can respond non-linearly and pulses may distort one another; but for the ordinary sound, string and water waves treated here, the linear superposition principle is accurate and extraordinarily useful. 🔉⇢

Superposition is the single idea from which the rest of the chapter is built. Reflection combined with superposition produces standing waves; superposition of two nearly-equal frequencies produces beats; and superposition of many harmonics is how a plucked string or a musical instrument produces its characteristic tone. Recognising a problem as 'add the waves algebraically, then simplify with a trigonometric identity' is the key skill. 🔉⇢

It is instructive to contrast the two ways two identical waves can combine. Two same-frequency waves travelling in the same direction (this section) produce a travelling resultant whose amplitude $2a\cos(\phi/2)$ is set by their phase difference. Two same-frequency waves travelling in opposite directions (next section) instead produce a standing wave in which the position, not merely the amplitude, decides the motion. The difference lies entirely in the relative direction of travel. 🔉⇢

In summary, the principle of superposition states that overlapping waves add algebraically, $y=\sum_i y_i$; for two equal, same-direction harmonic waves this gives a resultant of the same frequency and wavelength but amplitude $2a\cos(\phi/2)$, which is $2a$ (constructive) when they are in phase and $0$ (destructive) when they are $\pi$ out of phase. This principle, and the interference it explains, is the conceptual engine of wave physics. 🔉⇢

It is convenient to state the principle of superposition as a precise statement about displacements. When two or more waves travel through the same region of a medium at the same instant, the resultant displacement of any particle at any moment is the algebraic sum of the displacements that each wave would have produced there on its own. The waves pass through one another without being changed; after they have crossed, each wave continues to propagate with its original amplitude, wavelength and speed, exactly as though the other wave had never been present. This is a remarkable and convenient property of the linear waves studied in this chapter. 🔉⇢

The physical reason the principle holds is that the equation governing small disturbances in an elastic medium is linear. Because the restoring force on a displaced element of the medium is proportional to the displacement, the disturbances add without one distorting the other. If the amplitude of a wave is very large, the medium no longer responds in this simple proportional manner and the superposition principle fails; but for the ordinary waves of sound and for waves on a stretched string the amplitudes are small and the principle is very accurate. 🔉⇢

Consider two sinusoidal waves of equal frequency and equal amplitude travelling in the same direction along a stretched string, differing only by a constant phase. Writing each as a sine function of position and time and adding them, the trigonometric identity for the sum of two sine functions gives a single travelling wave of the same frequency and the same wavelength, whose amplitude depends on the phase difference between the two waves. The resultant is still a sinusoidal wave, and its amplitude is largest when the waves arrive in phase and smallest when they arrive exactly out of phase. 🔉⇢

When the phase difference is zero the two crests coincide and the two troughs coincide; the displacements add to give a resultant of double the amplitude. This is constructive interference, and the intensity, which is proportional to the square of the amplitude, is four times that of a single wave. When the phase difference is half a period, a crest of one wave meets a trough of the other; the displacements cancel and the resultant amplitude is reduced to zero. This is destructive interference, and no energy appears at that location. 🔉⇢

It is important to understand where the energy goes in destructive interference. Energy is conserved overall: the total energy carried by the combined waves is simply redistributed in space. Where the waves interfere destructively the intensity falls to a minimum; where they interfere constructively the intensity rises to a maximum. Averaged over the whole pattern the energy is exactly what the two waves brought with them. Interference does not create or destroy energy; it only rearranges how the energy is distributed over the region where the waves overlap. 🔉⇢

The same additive relation applies to a short pulse meeting another short pulse. Two pulses travelling in opposite directions on a string approach, overlap, and for a moment produce a resultant shape that is the sum of the two separate shapes; then they continue and emerge with their original forms unchanged. If the pulses are on the same side of the string they reinforce while overlapping; if they are on opposite sides they partially or completely cancel at the instant of meeting, after which each reappears undisturbed on the far side. This is a direct and simple illustration of the principle. 🔉⇢

The principle of superposition is the foundation for several later results in the study of waves. The standing wave produced by two identical waves travelling in opposite directions, the beats produced by two sound waves of slightly different frequency, and the interference pattern produced by two coherent sources are all direct consequences of adding displacements. In every case one does not solve a new wave problem; one simply superposes known travelling waves and interprets the resultant. Recognising this economy of ideas is an important part of understanding the chapter. 🔉⇢

A careful point about interference concerns the conditions under which a steady pattern is seen. A fixed pattern of maxima and minima appears only when the two waves maintain a constant phase difference over time; such sources are described as coherent. If the phase difference changes rapidly and at random, the positions of the maxima and minima shift continuously and no steady interference pattern is observed. This is why a stable interference pattern requires sources whose relative phase remains constant, a requirement that becomes central when interference is studied in detail. 🔉⇢

To apply the principle in a problem, first identify each individual wave, writing its displacement as a function of position and time with its own amplitude, wavelength and phase. Then add these displacement functions algebraically at the point and instant of interest. Finally interpret the resultant: read off its amplitude, decide whether the interference is constructive or destructive, and relate the amplitude to the intensity through the square relation. This ordered procedure reduces every superposition question to simple addition followed by careful interpretation. 🔉⇢

It is worth stressing that superposition is a statement about the medium as much as about the waves. At each point of the medium the particle can have only one displacement at a given instant, and that single displacement is whatever the separate waves demand added together. The medium does not keep the waves apart in separate compartments; it responds to the total disturbance. This is why two sound waves crossing a room, or two ripples spreading on a water surface, produce a single combined pattern of motion rather than two independent patterns that ignore each other. 🔉⇢

A useful consequence of the principle is that any complicated periodic wave can be regarded as a sum of simple sinusoidal waves of suitable amplitudes and phases. A plucked string or a musical instrument produces a resultant waveform that is the superposition of a fundamental and its harmonics, and the particular combination fixes the quality of the sound. Because the medium adds these components without distorting them, the separate sinusoidal waves travel together and can be treated one at a time and then added, which is a powerful method of analysis throughout the study of waves. 🔉⇢

Finally, superposition explains why waves are so different from particles. Two particles cannot occupy the same position, but two waves can occupy the same region at the same time and simply add. After they separate, each carries on with its original amplitude, wavelength, speed and direction, showing no memory of the meeting. This ability of waves to pass through one another and to add their displacements, and then continue unchanged, is one of the most important and characteristic properties of wave motion, and the principle of superposition is its exact and general statement. 🔉⇢

It is helpful to distinguish clearly between the displacement and the intensity when applying the principle. The superposition principle adds displacements, which are signed quantities and can therefore cancel; the intensity, being proportional to the square of the resultant amplitude, is always positive and is not simply the sum of the separate intensities. Two waves that each carry a certain intensity can combine to give four times that intensity at points of constructive interference and zero at points of destructive interference. Adding intensities directly is a common mistake; one must first add the displacements and only then square the resultant amplitude. 🔉⇢

The dependence of the resultant amplitude on the phase difference can be summarised in a single relation. For two waves of equal amplitude the resultant amplitude is largest when the phase difference is zero or a whole number of complete cycles, and smallest when the phase difference is an odd number of half cycles. Between these extremes the resultant amplitude varies smoothly, so a gradual change in the path difference between two sources produces a gradual change from a bright maximum to a dark minimum and back again. This smooth variation is exactly the pattern of maxima and minima observed in an interference experiment. 🔉⇢

A closing remark places superposition in the wider structure of the chapter. Every later phenomenon in the study of waves is built on this one principle: reflection and the standing waves it produces, the beats heard when two close frequencies sound together, and the interference of coherent sources all follow by adding travelling waves that are already understood. Because the medium responds linearly, these combinations require no new physics; they need only the careful addition of known waves and the correct interpretation of the resultant amplitude and intensity. The principle of superposition is therefore among the most useful single ideas in the entire chapter. 🔉⇢

Derivation from first principles 🔉⇢

  1. State the principle: for overlapping disturbances $y_1,\ y_2,\dots,y_n$, the net displacement is the algebraic (signed) sum $y(x,t)=\sum_{i=1}^{n} y_i(x,t)$; each wave propagates as if the others were absent.
  2. Choose two identical harmonic waves in the $+x$ direction differing only in phase: $y_1=a\sin(kx-\omega t)$ and $y_2=a\sin(kx-\omega t+\phi)$.
  3. Superpose: $y=y_1+y_2=a\sin(kx-\omega t)+a\sin(kx-\omega t+\phi)$.
  4. Apply $\sin A+\sin B=2\sin\!\big(\tfrac{A+B}{2}\big)\cos\!\big(\tfrac{A-B}{2}\big)$ with $A=kx-\omega t$ and $B=kx-\omega t+\phi$: $y=2a\cos\!\big(\tfrac{\phi}{2}\big)\sin\!\big(kx-\omega t+\tfrac{\phi}{2}\big)$.
  5. Identify the resultant: a travelling wave of the same $\omega$ and $k$ (same frequency and wavelength), initial phase $\phi/2$, and amplitude $A(\phi)=2a\cos(\phi/2)$.
  6. Constructive case: for $\phi=0$ (or $\phi=2n\pi$), $\cos(\phi/2)=\pm1$, so $A=2a$ and $y=2a\sin(kx-\omega t)$ — amplitudes add.
  7. Destructive case: for $\phi=\pi$ (or $\phi=(2n+1)\pi$), $\cos(\phi/2)=0$, so $A=0$ and $y=0$ everywhere — amplitudes cancel.
  8. Translate to path difference: since $\phi=\dfrac{2\pi}{\lambda}\Delta x$, constructive interference needs $\Delta x=n\lambda$ and destructive interference needs $\Delta x=(n+\tfrac{1}{2})\lambda$.
⚠️ JEE trap: A frequent misconception is that when two waves interfere destructively the energy is destroyed, or that the two waves permanently annihilate each other. Neither is true. When two equal-and-opposite pulses overlap and the string is momentarily flat, the energy is stored in the transverse kinetic energy of the moving string elements, and the pulses emerge unchanged an instant later — superposition never alters a wave's identity, only the pattern during overlap. In a sustained interference pattern the energy removed from the destructive (dark) regions is exactly what appears in the constructive (bright) regions, so total energy is conserved. A second error is to think the resultant amplitude is always $2a$ or always $a+a$; in fact it is $2a\cos(\phi/2)$ and depends entirely on the phase difference $\phi$. 🔉⇢

Worked example · JEE Advanced 🔉⇢

SITUATION Two harmonic waves of equal amplitude $a$ and equal frequency travel in the same direction along a string, with a constant phase difference $\phi$ between them.
TARGET Find the amplitude of the resultant wave, and evaluate it for $\phi=0$, $\phi=\pi/2$, $\phi=2\pi/3$ and $\phi=\pi$; identify which cases are constructive and which destructive.
STRATEGY Add the two waves by superposition and simplify with $\sin A+\sin B=2\sin(\tfrac{A+B}{2})\cos(\tfrac{A-B}{2})$ to read off the resultant amplitude $A(\phi)=2a\cos(\phi/2)$, then substitute each phase.
EXECUTE $A(\phi)=2a\cos(\phi/2)$. For $\phi=0$: $A=2a\cos0=2a$ (fully constructive). For $\phi=\pi/2$: $A=2a\cos(\pi/4)=\sqrt{2}\,a\approx1.41a$ (partial). For $\phi=2\pi/3$: $A=2a\cos(\pi/3)=a$ (partial). For $\phi=\pi$: $A=2a\cos(\pi/2)=0$ (fully destructive).
REFLECT The resultant amplitude varies continuously from $2a$ (in phase) to $0$ ($\pi$ out of phase) as $\phi$ increases, without either individual amplitude changing. This single expression, $2a\cos(\phi/2)$, captures the whole of two-wave interference and is the algebraic content of the superposition principle.

Source: NCERT Class XI Physics, Chapter 14 (Waves)

Reflection of Waves and Standing Waves 🔉⇢deep concept

Definition: A wave reflected at a rigid boundary undergoes a phase reversal of $\pi$ (a crest returns as a trough) while reflection at a free/open boundary occurs with no phase change; the superposition of a wave and its oppositely travelling reflection produces a standing wave $y=2a\sin kx\cos\omega t$ whose nodes and antinodes fix the system's normal modes. 🔉⇢

🔬 Interactive 3D · A string fixed at both ends is driven through its harmonics; nodes stay fixed while antinodes oscillate, showing $\nu_n=nv/2L$ and node-to-node spacing $\lambda/2$.

So far we have considered waves in an unbounded medium. What happens when a pulse or wave meets a boundary? If the boundary is rigid, the wave is reflected — the phenomenon of an echo is reflection of sound by a rigid boundary. If the boundary is not completely rigid, or is an interface between two different media, part of the incident wave is reflected and part is transmitted (refracted) into the second medium. 🔉⇢

Consider a pulse travelling along a stretched string toward a rigid boundary (the string tied to a fixed wall). Assuming no energy is absorbed, the reflected pulse has the same shape and amplitude as the incident pulse, but it suffers a phase change of $\pi$ (that is, $180^{\circ}$) on reflection. A crest returns as a trough and vice versa; the pulse is inverted. 🔉⇢

The reason for the inversion is a boundary condition: the fixed end can never move, so the displacement must be zero there at all times. By superposition, the only way the incident and reflected waves can always sum to zero at the wall is for them to differ in phase by $\pi$, so that they cancel there. This is the boundary condition on a rigid wall. 🔉⇢

The same conclusion follows dynamically. As the pulse reaches the wall it exerts an upward force on it; by Newton's third law the wall exerts an equal and opposite (downward) force back on the string, and this generates a reflected pulse inverted relative to the incident one — a phase change of $\pi$. 🔉⇢

If instead the boundary point is not rigid but completely free to move — for example a string tied to a light ring that slides frictionlessly on a rod — the reflected pulse has the same phase and amplitude as the incident pulse; there is no inversion. At such a free end the incident and reflected pulses add, so the net maximum displacement at the boundary is twice the amplitude of each pulse. The open end of an organ pipe is a physical example of a (nearly) free boundary. 🔉⇢

To summarise the reflection rule: a travelling wave or pulse suffers a phase change of $\pi$ on reflection at a rigid boundary, and no phase change on reflection at an open (free) boundary. Mathematically, for an incident wave $y_i=a\sin(kx-\omega t)$, the reflected wave at a rigid boundary is $y_r=a\sin(kx-\omega t+\pi)=-a\sin(kx-\omega t)$, while at an open boundary $y_r=a\sin(kx-\omega t)$. At the rigid boundary the sum $y_i+y_r$ is zero at all times, as required. 🔉⇢

Now consider reflection at two boundaries — a string fixed at both ends, or an air column in a pipe. A wave travelling one way reflects at one end, travels back, reflects at the other, and so on, until a steady pattern is established. Such steady patterns are called standing waves or stationary waves. They arise from the superposition of two identical waves travelling in opposite directions. 🔉⇢

To obtain the standing wave, superpose $y_1(x,t)=a\sin(kx-\omega t)$ (travelling in $+x$) and $y_2(x,t)=a\sin(kx+\omega t)$ (travelling in $-x$). Using $\sin(A-B)+\sin(A+B)=2\sin A\cos B$ with $A=kx,\ B=\omega t$, the resultant is $y(x,t)=2a\sin kx\cos\omega t$. 🔉⇢

This wave pattern is fundamentally different from a travelling wave. The variables $x$ and $t$ appear separately, not in the travelling combination $kx-\omega t$. Each element oscillates in time as $\cos\omega t$ with the same angular frequency, but its amplitude, $2a\sin kx$, is fixed by its location. The pattern neither moves left nor right — hence 'standing' or 'stationary'. 🔉⇢

Because the amplitude depends on position, some points never move while others swing with maximum amplitude. The points at which the amplitude is zero (no motion at all) are called nodes; the points at which the amplitude is the largest are called antinodes. In a standing wave all particles between two adjacent nodes move in phase (they reach their extremes together) but with different amplitudes — the exact opposite of a progressive wave, where amplitudes are equal but phases differ. 🔉⇢

The nodes occur where $\sin kx=0$, i.e. $kx=n\pi$ with $n=0,1,2,\dots$; since $k=2\pi/\lambda$, this gives node positions $x=n\lambda/2$. The antinodes occur where $\sin kx=\pm1$, i.e. $kx=(n+\tfrac12)\pi$, giving $x=(n+\tfrac12)\lambda/2$. Consequently the distance between two consecutive nodes (or two consecutive antinodes) is $\lambda/2$, and a node and its adjacent antinode are separated by $\lambda/4$. 🔉⇢

The most significant feature of standing waves is that the boundary conditions constrain the allowed wavelengths and frequencies. Unlike a travelling wave, which can have any frequency, a bounded system can only oscillate at a discrete set of natural frequencies called its normal modes. Each end condition (node at a fixed end, antinode at a free/open end) selects which standing waves fit. 🔉⇢

For a string of length $L$ fixed at both ends, both ends must be nodes. Taking one end at $x=0$ (already a node), the condition that $x=L$ is also a node requires $L=n\lambda/2$, i.e. the allowed wavelengths are $\lambda=2L/n$ for $n=1,2,3,\dots$. The corresponding frequencies are $\nu_n=v/\lambda=nv/2L$. 🔉⇢

The lowest frequency, $\nu_1=v/2L$ (with $n=1$), is the fundamental mode or first harmonic; $n=2$ gives the second harmonic, $n=3$ the third, and so on. A string fixed at both ends therefore supports all integer harmonics, $\nu_n=nv/2L$. A real string generally vibrates in a superposition of several modes; which modes are prominent depends on where it is plucked or bowed, and this is the basis of instruments like the sitar and violin. 🔉⇢

An air column in a pipe open at both ends behaves like the string: each open end is a pressure node but a displacement antinode, and fitting antinodes at both ends again gives $\lambda=2L/n$ and $\nu_n=nv/2L$ for $n=1,2,3,\dots$. So an open-open pipe, like the fixed-fixed string, sounds all harmonics. 🔉⇢

A pipe closed at one end and open at the other is different. The closed end is a displacement node (and a pressure antinode), while the open end is a displacement antinode. Fitting a node at one end and an antinode at the other requires $L=(n+\tfrac12)\lambda/2$, giving $\lambda=\dfrac{2L}{n+\tfrac12}$ and frequencies $\nu=(n+\tfrac12)\dfrac{v}{2L}$ for $n=0,1,2,\dots$. 🔉⇢

Rewriting the closed-open frequencies makes the pattern transparent: the fundamental is $\nu_1=v/4L$, and the higher modes are $3v/4L,\ 5v/4L,\ 7v/4L,\dots$. A closed-open pipe therefore produces only the odd harmonics (odd multiples of the fundamental $v/4L$); the even harmonics are absent. This is a favourite examination contrast: open-open gives all harmonics, closed-open gives only odd harmonics. 🔉⇢

These bounded systems can be driven into forced oscillation, and when the driving frequency matches one of the natural frequencies the system responds strongly — resonance. This is how a tuning fork excites an air column of the right length, and how the length of an organ pipe or the tension of a string is tuned to a desired pitch. Each normal-mode frequency is a resonant frequency of the system. 🔉⇢

A note on pressure versus displacement in sound: at a closed end the air cannot move, so the displacement is minimum (a node) while the pressure change is largest (a pressure antinode); at an open end it is the reverse — maximum displacement (antinode) and least pressure change. This complementarity — a displacement node is a pressure antinode and vice versa — is essential when analysing pipes and is frequently tested. 🔉⇢

In summary, reflection inverts a wave at a rigid boundary ($\pi$ phase change) but not at a free/open one; the superposition of a wave and its opposite-going reflection gives the standing wave $y=2a\sin kx\cos\omega t$, with nodes and antinodes spaced $\lambda/2$ apart and separated from each other by $\lambda/4$. Boundary conditions then quantise the modes: $\nu_n=nv/2L$ (all harmonics) for a string fixed at both ends and for an open-open pipe, and $\nu_n=(2n-1)v/4L$ (odd harmonics only) for a closed-open pipe. 🔉⇢

A standing wave, also called a stationary wave, is produced when two travelling waves of the same frequency and amplitude move through a medium in opposite directions and superpose. The most common way to set this up is to send a wave along a stretched string and let it reflect from a boundary, so that the incident wave and the reflected wave travel in opposite directions in the same region. Adding the two waves gives a resultant in which the pattern of crests and troughs no longer progresses along the medium but instead oscillates in a fixed position. 🔉⇢

The behaviour of a wave at a boundary depends on the nature of that boundary. When a pulse on a string reaches a rigid, fixed end, the reflected pulse returns inverted, with a phase change of half a period; the fixed point cannot move, so the incident and reflected displacements must always cancel there. When the pulse reaches a free end that can move without restraint, the reflected pulse returns with the same phase and is not inverted. These two rules for reflection are the basis for locating the nodes and antinodes when a standing wave is established. 🔉⇢

Adding an incident sinusoidal wave and its reflection gives a resultant displacement that is a product of two factors: one factor depends only on position and the other only on time. The position factor fixes the amplitude of oscillation at each point of the medium; the time factor makes every point oscillate together with the same frequency. Because the pattern is a product rather than a travelling function of position minus speed times time, the disturbance does not move along the medium. Each particle simply oscillates in place with an amplitude set by where it sits. 🔉⇢

The points where the position factor is zero never move at all; these are the nodes. The points where the position factor reaches its maximum oscillate with the greatest amplitude; these are the antinodes. Adjacent nodes are separated by half a wavelength, and adjacent antinodes are likewise separated by half a wavelength, so that a node and the neighbouring antinode are a quarter of a wavelength apart. This regular spacing of nodes and antinodes is the clearest visible signature that a standing wave, rather than a travelling wave, has been produced. 🔉⇢

A standing wave on a string of fixed length can exist only for certain wavelengths, because the boundary conditions must be satisfied at both ends. A string clamped rigidly at both ends must have a node at each end, so the length of the string must equal a whole number of half wavelengths. This condition selects a discrete set of allowed wavelengths and therefore a discrete set of allowed frequencies. These special frequencies are the natural frequencies of the string, and the standing wave patterns that correspond to them are its normal modes of vibration. 🔉⇢

The lowest allowed frequency is the fundamental, for which the string vibrates in a single loop with a node at each end and one antinode in the middle. The higher allowed frequencies are the harmonics, which are whole number multiples of the fundamental frequency; the string then vibrates in two, three or more loops with additional nodes in between. A real string plucked or bowed vibrates in several of these modes at once, and the particular mixture of the fundamental and its harmonics gives a musical instrument its characteristic quality of sound. 🔉⇢

The same ideas apply to sound waves in an air column, such as the air inside an organ pipe. A closed end of a pipe forces a displacement node there, because the air cannot move along the pipe at a rigid wall, while an open end is approximately a displacement antinode, because the air is free to move. A pipe closed at one end and open at the other therefore supports a different set of allowed frequencies from a pipe open at both ends, and comparing the two cases is a standard and instructive exercise. 🔉⇢

Resonance is closely connected to these natural frequencies. If a medium that can support standing waves is driven by an external periodic force whose frequency matches one of the natural frequencies, the amplitude of the resulting standing wave grows large and the system is said to resonate. This is why an air column resonates strongly only at particular lengths for a given frequency, and why musical instruments are built to favour a chosen set of natural frequencies. The standing wave picture and the idea of resonance together explain how instruments produce definite musical notes. 🔉⇢

To analyse a standing wave problem, first decide the nature of each boundary and hence whether a node or an antinode must sit there. Next fit a whole number of half wavelengths, or the appropriate quarter wavelength pattern for a closed end, into the length of the medium to find the allowed wavelengths. Then use the relation between speed, frequency and wavelength to convert these allowed wavelengths into the natural frequencies. This ordered method turns every standing wave question into a matter of fitting the pattern to the boundaries and applying the basic wave relation. 🔉⇢

It is important to see that a standing wave does not transport energy along the medium in the way a travelling wave does. In a progressive wave energy is carried steadily from one region to another, but in a standing wave the nodes never move and no energy crosses them; the energy merely oscillates back and forth between the kinetic form, greatest as the string sweeps through its central position, and the potential form, greatest when the string is momentarily at rest in its most distorted shape. The pattern stores energy and exchanges it internally rather than passing it along. 🔉⇢

A further point concerns the reflection that sets up the standing wave. At a fixed boundary the medium beyond cannot be displaced, so the reflected wave is inverted and the boundary is forced to be a node; at a free boundary the medium is not constrained, so the reflected wave keeps its phase and the boundary becomes an antinode. Because the standing wave is simply the superposition of the incident and reflected waves, the type of boundary directly decides the pattern of nodes and antinodes, and therefore which wavelengths and frequencies the medium can support. 🔉⇢

The idea of natural frequencies and normal modes reaches far beyond strings and pipes. Any bounded elastic system, whether a stretched membrane, a column of air, or a solid rod, has its own set of standing wave patterns and its own natural frequencies fixed by its size and by the speed of waves within it. When such a system is disturbed it tends to vibrate in these natural modes, and when it is driven at one of these frequencies it responds with a large amplitude. This is the general reason that the study of standing waves is so important throughout physics. 🔉⇢

Derivation from first principles 🔉⇢

  1. Reflection rule: at a rigid end the displacement must vanish for all $t$; writing incident $y_i=a\sin(kx-\omega t)$ and reflected $y_r=a\sin(kx-\omega t+\pi)=-a\sin(kx-\omega t)$ ensures $y_i+y_r=0$ at the wall. At a free end there is no phase change: $y_r=a\sin(kx-\omega t)$.
  2. Form the standing wave: superpose oppositely travelling waves $y_1=a\sin(kx-\omega t)$ and $y_2=a\sin(kx+\omega t)$, so $y=a[\sin(kx-\omega t)+\sin(kx+\omega t)]$.
  3. Apply $\sin(A-B)+\sin(A+B)=2\sin A\cos B$ with $A=kx,\ B=\omega t$: $y(x,t)=2a\sin kx\,\cos\omega t$. The variables separate, so the pattern does not travel; its position-dependent amplitude is $2a\sin kx$.
  4. Locate nodes and antinodes: nodes where $\sin kx=0\Rightarrow kx=n\pi\Rightarrow x=n\lambda/2$; antinodes where $\sin kx=\pm1\Rightarrow x=(n+\tfrac12)\lambda/2$. Adjacent nodes (or antinodes) are $\lambda/2$ apart; a node and next antinode are $\lambda/4$ apart.
  5. String fixed at both ends: require nodes at $x=0$ and $x=L$, so $L=n\lambda/2\Rightarrow\lambda=2L/n$; hence $\nu_n=v/\lambda=nv/2L,\ n=1,2,3,\dots$ — all harmonics, fundamental $v/2L$.
  6. Pipe open at both ends: antinodes at both ends give the same condition $\lambda=2L/n$ and $\nu_n=nv/2L$ — all harmonics.
  7. Pipe closed at one end, open at the other: node at the closed end, antinode at the open end give $L=(n+\tfrac12)\lambda/2\Rightarrow\lambda=2L/(n+\tfrac12)$; hence $\nu=(n+\tfrac12)v/2L$, i.e. $\nu_1=v/4L$ and modes $3v/4L,5v/4L,\dots$ — only odd harmonics, $\nu_n=(2n-1)v/4L$.
⚠️ JEE trap: Students often believe every pipe and string produces the same set of harmonics, or that a closed-open pipe merely has a lower fundamental but the same overtone series. In fact the boundary conditions matter decisively: a string fixed at both ends and a pipe open at both ends support all integer harmonics ($\nu_n=nv/2L$), whereas a pipe closed at one end supports only the odd harmonics ($\nu_1=v/4L,\ 3v/4L,\ 5v/4L,\dots$, i.e. $(2n-1)v/4L$) — its even harmonics are simply not allowed. A second common error is to forget that reflection at a rigid boundary inverts the wave (a $\pi$ phase reversal, crest returns as trough), while reflection at a free/open boundary does not; getting this sign wrong ruins the node/antinode assignment. Finally, do not confuse the standing-wave picture with a travelling one: in a standing wave the nodes stay fixed and particles between adjacent nodes share the same phase but differ in amplitude. 🔉⇢

Worked example · JEE Advanced 🔉⇢

SITUATION A pipe $30.0\ \text{cm}$ long is open at both ends. Take the speed of sound in air as $330\ \text{m s}^{-1}$.
TARGET Determine which harmonic mode of the pipe resonates with a $1.1\ \text{kHz}$ source, and state whether resonance with the same source persists if one end of the pipe is then closed.
STRATEGY For the open-open pipe use $\nu_n=nv/2L$ and solve for the integer $n$ that gives $1.1\ \text{kHz}$. For the closed-open pipe use $\nu_n=(2n-1)v/4L$ and check whether $1.1\ \text{kHz}$ corresponds to an allowed (odd) harmonic.
EXECUTE Open-open: $\nu_n=n(330)/(2\times0.30)=550\,n\ \text{Hz}$. Setting $550n=1100$ gives $n=2$, so the source resonates with the second harmonic. Closed-open: fundamental $\nu_1=v/4L=330/(4\times0.30)=275\ \text{Hz}$, and allowed modes are odd multiples $275,825,1375,\dots$. Now $1100/275=4$, an even multiple, which is not an allowed mode, so no resonance occurs once one end is closed.
REFLECT The open pipe resonates at $1.1\ \text{kHz}$ as its second harmonic, but closing one end removes all even harmonics and shifts the fundamental to $v/4L=275\ \text{Hz}$; since $1.1\ \text{kHz}$ would be the fourth harmonic (even), it is forbidden and resonance disappears. This is the textbook demonstration that a closed-open pipe sounds only odd harmonics.

Source: NCERT Class XI Physics, Chapter 14 (Waves)

The Doppler Effect 🔉⇢deep concept

Definition: The Doppler effect is the change in the observed frequency of a wave caused by relative motion between the source and the observer through the medium; for sound the observed frequency is $\nu'=\nu\,\dfrac{v\pm v_o}{v\mp v_s}$, higher on approach and lower on recession, and it is not symmetric in source and observer motion because the medium provides a preferred frame. 🔉⇢

🔬 Interactive 3D · A moving source emits spherical wavefronts that bunch ahead (higher pitch) and spread behind (lower pitch); sliders for $v_s$ and $v_o$ update $\nu'=\nu(v\pm v_o)/(v\mp v_s)$.

You have surely noticed that the pitch of a train whistle or an ambulance siren sounds higher as the vehicle approaches you and abruptly lower as it recedes. This change in the observed frequency of a wave due to relative motion between the source and the observer (through the medium) is called the Doppler effect, after Christian Doppler who first described it. The emitted frequency of the source never changes; only the frequency received by the observer does. 🔉⇢

The key to the sound Doppler effect is that a mechanical wave travels at a fixed speed set by the medium alone. Relative to an observer at rest in the medium, the speed of a mechanical wave in that medium depends only on the elastic and inertial (mass-density) properties of the medium; it does not depend on the motion of the source. This medium-fixed speed provides a preferred frame of reference, and it is what makes the sound Doppler formulas asymmetric between source motion and observer motion. 🔉⇢

Begin with a stationary source and a moving observer. The source at rest emits wavefronts (compressions) separated by one wavelength $\lambda=v/\nu$, all travelling outward at speed $v$ relative to the medium. If the observer moves toward the source with speed $v_o$, they sweep through the oncoming wavefronts faster: relative to the observer the waves approach at speed $v+v_o$, while the wavelength is unchanged at $\lambda=v/\nu$. 🔉⇢

The frequency the moving observer records is (relative wave speed)/(wavelength): $\nu'=\dfrac{v+v_o}{\lambda}=\dfrac{v+v_o}{v/\nu}=\nu\,\dfrac{v+v_o}{v}=\nu\Big(1+\dfrac{v_o}{v}\Big)$. Moving toward the source raises the frequency; moving away replaces $+v_o$ by $-v_o$ and lowers it. Note that here the wavelength in the medium is untouched — only the rate of encountering wavefronts changes. 🔉⇢

Now take a moving source and a stationary observer, which is physically different. In one period $T=1/\nu$ the source advances a distance $v_s T$ toward the observer, so it emits each successive wavefront from a point closer to the observer than the last. The wavefronts ahead of the source are therefore crowded together: the effective wavelength ahead is shortened to $\lambda'=(v-v_s)T=\dfrac{v-v_s}{\nu}$. 🔉⇢

The observer, at rest in the medium, still receives these waves travelling at speed $v$, but now with the compressed wavelength $\lambda'$. The observed frequency is $\nu'=\dfrac{v}{\lambda'}=\dfrac{v}{(v-v_s)/\nu}=\nu\,\dfrac{v}{v-v_s}$. An approaching source thus raises the pitch by shortening the wavelength; a receding source ($v_s\to-v_s$) lengthens the wavelength and lowers the pitch. 🔉⇢

Compare the two single-motion results carefully, because their difference is the conceptual heart of the topic. A moving observer changes the effective wave speed relative to themselves but leaves the wavelength alone, giving $\nu'=\nu(1+v_o/v)$. A moving source changes the wavelength in the medium but leaves the wave speed alone, giving $\nu'=\nu\,v/(v-v_s)$. These two expressions are not the same even for equal speeds — the Doppler effect for sound is not symmetric in source and observer motion. 🔉⇢

The asymmetry has a clean physical origin: the medium supplies a preferred frame. Sound travels at speed $v$ with respect to the medium, not with respect to the source or the observer, so it matters physically whether it is the source or the observer that moves relative to that medium. (For light in vacuum there is no medium and no preferred frame, so the light Doppler effect depends only on the relative velocity — but that is a Class XII / relativity topic.) 🔉⇢

Combining both motions gives the general formula for sound. With the observer moving at $v_o$ and the source at $v_s$ (both measured relative to the medium), the observed frequency is $\nu'=\nu\,\dfrac{v\pm v_o}{v\mp v_s}$. The signs are chosen by the physical rule that any motion tending to reduce the source-observer separation raises the pitch, and any motion increasing the separation lowers it. 🔉⇢

A reliable sign convention avoids errors. In the numerator use $v+v_o$ when the observer moves toward the source and $v-v_o$ when moving away. In the denominator use $v-v_s$ when the source moves toward the observer and $v+v_s$ when moving away. The compact rule 'approach raises pitch, recession lowers pitch' is the check: whatever combination of signs you pick, verify that approach gives $\nu'\gt \nu$ and recession gives $\nu'\lt \nu$. 🔉⇢

Consider approach as the leading case: observer moving toward a source that is moving toward the observer gives $\nu'=\nu\,\dfrac{v+v_o}{v-v_s}$, clearly greater than $\nu$. As a source approaches, $v-v_s$ shrinks and $\nu'$ climbs; in the limit $v_s\to v$ the wavelength ahead collapses toward zero and $\nu'\to\infty$ — the onset of the shock wave (sonic boom) when the source reaches the speed of sound. 🔉⇢

For recession, both effects lower the pitch: observer receding from a receding source gives $\nu'=\nu\,\dfrac{v-v_o}{v+v_s}\lt \nu$. This is exactly why the siren's pitch drops the moment the ambulance passes and begins to move away: the sign of the source's motion relative to you flips from 'toward' to 'away'. 🔉⇢

When the speeds are small compared with the wave speed, $v_o,v_s\ll v$, both formulas reduce to the same first-order result $\nu'\approx\nu\Big(1+\dfrac{v_{rel}}{v}\Big)$, where $v_{rel}$ is the speed of approach (positive) or recession (negative) of source relative to observer. In this low-speed limit the asymmetry between source and observer motion becomes negligible and only the relative velocity matters — which is why the effect looks 'symmetric' in everyday, slow situations. 🔉⇢

The fractional change in frequency in the low-speed limit is $\dfrac{\Delta\nu}{\nu}\approx\dfrac{v_{rel}}{v}$. This handy relation lets you estimate speeds from measured frequency shifts and is the working basis of Doppler speed measurement — a small, easily measured fractional shift corresponds to a speed that is that same fraction of the wave speed. 🔉⇢

The Doppler effect is not a curiosity; it is a workhorse of technology and science. Doppler radar and police speed guns bounce waves off a moving target and read the shift to infer its speed; Doppler ultrasound (echocardiography) measures blood-flow velocities in the body; and in astronomy the redshift of light from distant galaxies (their spectral lines shifted toward longer wavelengths) reveals that the universe is expanding. In each case a frequency shift is decoded into a velocity. 🔉⇢

A careful reasoning point: for sound, all velocities in the formula must be measured relative to the medium (the air). If there is a wind, the medium itself is moving, and one must add the wind velocity to the wave speed appropriately — a common source of exam traps. This again reflects the fact that the medium defines the frame in which the wave speed is the fixed value $v$. 🔉⇢

It is instructive to place the Doppler effect within the logic of this chapter. It rests directly on two earlier ideas: that a mechanical wave's speed is set by the medium alone (the speed section) and that the observed frequency is (wave speed relative to observer)/(wavelength). Doppler does not introduce any new wave physics; it carefully applies these two facts to the case where the source and/or observer move relative to the medium. 🔉⇢

In summary, relative motion between a sound source and an observer changes the observed frequency to $\nu'=\nu\,\dfrac{v\pm v_o}{v\mp v_s}$: approach raises the pitch, recession lowers it, and the effect is genuinely asymmetric in source versus observer motion because the medium provides a preferred frame in which the wave speed is fixed. Choosing signs so that 'approach raises pitch' is respected, and remembering that all speeds are measured relative to the medium, will handle every standard problem. 🔉⇢

It is convenient to state the central idea in physical terms before using any equation. A source of sound produces a steady stream of compressions and rarefactions that travel outward through the medium at the fixed speed set by the medium. The number of these compressions that pass an observer in one second is the frequency the observer detects. Anything that changes how often the compressions arrive changes the observed frequency, even when the source itself emits at a steady rate. The whole effect is about the rate at which the travelling compressions reach the observer. 🔉⇢

Consider again a stationary source and think about the pattern of compressions it produces in the medium. Each period the source produces one compression, and that compression then travels outward at the medium speed. The distance in the medium between one compression and the next is the wavelength, equal to the speed divided by the frequency of the source. When neither the source nor the observer moves relative to the medium, the observer meets these compressions at exactly the rate the source produces them, so the observed frequency equals the frequency of the source. 🔉⇢

Now let the observer move through the medium while the source is at rest. The compressions still sit in the medium separated by the same wavelength, because the source has not moved and the medium speed has not changed. But the observer, by moving through this fixed pattern, meets the compressions at a different rate. Moving in the direction from which the waves come, the observer meets them more often and the observed frequency increases; moving in the same direction as the waves travel, the observer meets them less often and the observed frequency is lower. 🔉⇢

The amount of this change follows directly from the speed of the waves relative to the observer. If the observer moves through the medium at a given speed, the compressions pass at the medium speed increased or reduced by the observer speed, while the distance between compressions in the medium remains equal to the original wavelength. Dividing this relative speed by the original wavelength gives the observed frequency. The important feature of a moving observer is therefore that the wavelength in the medium is not changed and only the rate of meeting the compressions is altered. 🔉⇢

Contrast this with a source that moves through the medium while the observer is at rest. Now the medium speed of the waves is still fixed, but during each period the source itself moves forward a small distance, so it produces each new compression from a point a little further along than the last one. In the direction of the motion the compressions are therefore packed close together and the wavelength in the medium is reduced. In the opposite direction the compressions are separated by a greater distance and the wavelength in the medium is greater. 🔉⇢

The observer at rest still meets these compressions travelling at the medium speed, but now they are separated by the altered wavelength. Where the compressions have been packed close together the observer meets them more often, so the observed frequency is higher; where they have been separated by a greater distance the observer meets them less often, so the observed frequency is lower. The important feature of a moving source is exactly the reverse of the moving observer: here the medium speed relative to the observer is not changed and it is the wavelength in the medium that is altered. 🔉⇢

This difference between the two cases is the important physical point of the whole discussion, and it should be stated clearly. A moving observer changes the speed of the waves relative to the observer but leaves the wavelength in the medium equal to its original value. A moving source changes the wavelength in the medium but leaves the medium speed of the waves equal to its fixed value. Because one case changes the relative speed and the other changes the wavelength, the two situations produce different results even when the speeds involved are equal. 🔉⇢

The deep reason for this difference is that the medium provides the reference in which the wave speed has its fixed value. A mechanical wave such as sound travels at a definite speed relative to the medium, set only by the elastic and inertial properties of that medium, and not relative to the source or the observer. Because the speed is fixed relative to the medium, it physically matters which of the two, the source or the observer, is the one that moves relative to the medium. The two kinds of motion are simply not equal. 🔉⇢

A convenient way to keep the physics correct is to require that every speed be measured relative to the medium. The speed of the source and the speed of the observer are both measured relative to the still medium, and the wave speed is the fixed value that also belongs to the medium. Once all three speeds are referred to this single common reference, the number of compressions reaching the observer is found without difficulty, and the observed frequency follows without confusion. Using the wrong reference is the most common difficulty in these exercises. 🔉⇢

It is instructive to see how both expressions agree when the speeds are small compared with the wave speed. When the source speed and the observer speed are both much smaller than the medium speed of the waves, the change in observed frequency depends, to a very good approximation, only on the relative speed of approach or of separation of the source and the observer. In this slow situation the difference between a moving source and a moving observer becomes very small and can be neglected, which is why the effect appears the same for both in ordinary situations. 🔉⇢

The fractional change in the observed frequency, in this slow situation, equals the relative speed of approach divided by the wave speed. This simple relation is of great practical value, because it lets one estimate a speed that is not known from a measured change in frequency. A small measured change in frequency corresponds to a speed that is the same small fraction of the wave speed. The relation shows clearly that the frequency change and the speed are directly proportional when the speeds are small, and it is the basis of the everyday reasoning about the effect. 🔉⇢

A wind in the medium is a fine point that exercises often test. All the speeds in the reasoning are measured relative to the medium, so if the whole body of air is itself moving as a wind, the medium is no longer at rest and one must add the speed of the wind to the wave speed in the correct direction. The wave still travels at its fixed speed relative to the air, but the air itself is now in motion, and only by referring everything to the moving medium does the counting of arriving compressions remain correct. 🔉⇢

It is worth relating this effect to the rest of the chapter, because it requires no new wave physics at all. It depends on just two earlier points already established. The first is that a mechanical wave travels at a fixed speed set by the medium alone, independent of the motion of the source. The second is that the frequency an observer detects equals the speed of the waves relative to the observer divided by the wavelength that reaches the observer. The whole discussion simply applies these two facts to the case where the source or the observer moves. 🔉⇢

The behaviour also relates naturally to the earlier idea of compressions and rarefactions in a sound wave. When the source moves forward, the compressions ahead of it are produced from points that move forward and so lie close together in the medium, giving a smaller wavelength and a higher observed frequency. Behind the moving source the compressions are produced from points that move backward and lie separated by a greater distance, giving a greater wavelength and a lower observed frequency. The change in the observed sound is thus a direct change in the separation of the compressions in the medium. 🔉⇢

A short numerical example makes the reasoning simple. Suppose the source produces a definite number of compressions each second and the observer records how many reach a fixed position each second. When the observer is at rest and the source is at rest, the two numbers are equal. When either the source or the observer moves, the number at the observer differs from the number at the source, and the ratio of the two numbers is exactly the ratio of the observed frequency to the frequency of the source. The effect is, at bottom, a difference between two numbers of arriving compressions. 🔉⇢

Finally, it is convenient to summarise the physical relations that never change, whatever the situation. Motion that reduces the distance between the source and the observer always makes the observed frequency higher, and motion that increases that distance always makes the observed frequency lower. The wave speed itself, being a property of the medium, is not altered by any of this motion. And because the medium provides the fixed reference for the wave speed, a moving source and a moving observer are genuinely different, agreeing only in the slow situation where the relative speed alone matters. 🔉⇢

Derivation from first principles 🔉⇢

  1. Foundational fact: a sound wave travels at a fixed speed $v$ set by the medium's elastic and inertial properties, independent of the source's motion; this makes the medium a preferred frame and all speeds below are measured relative to it.
  2. Moving observer, stationary source: the wavelength in the medium is unchanged, $\lambda=v/\nu$, but the observer approaching at $v_o$ meets the waves at relative speed $v+v_o$, so $\nu'=\dfrac{v+v_o}{\lambda}=\nu\dfrac{v+v_o}{v}=\nu\big(1+\tfrac{v_o}{v}\big)$ (use $-v_o$ for recession).
  3. Moving source, stationary observer: in one period $T=1/\nu$ the source advances $v_s T$, compressing the wavefronts ahead to wavelength $\lambda'=(v-v_s)/\nu$; the observer receives them at speed $v$, so $\nu'=\dfrac{v}{\lambda'}=\nu\dfrac{v}{v-v_s}$ (use $v+v_s$ for recession).
  4. Note the asymmetry: the moving-observer result changes the relative wave speed but not $\lambda$, whereas the moving-source result changes $\lambda$ but not the wave speed; the two are unequal even for equal speeds, because the medium is the preferred frame.
  5. Combine both motions: $\nu'=\nu\,\dfrac{v\pm v_o}{v\mp v_s}$, with the upper signs for motions of approach (observer toward source: $+v_o$; source toward observer: $-v_s$ in the denominator).
  6. Fix the sign convention by physics: choose signs so that approach gives $\nu'\gt \nu$ and recession gives $\nu'\lt \nu$; e.g. mutual approach gives $\nu'=\nu\dfrac{v+v_o}{v-v_s}$ and mutual recession gives $\nu'=\nu\dfrac{v-v_o}{v+v_s}$.
  7. Low-speed limit: for $v_o,v_s\ll v$, expand to first order to get $\nu'\approx\nu\big(1+\tfrac{v_{rel}}{v}\big)$ and $\dfrac{\Delta\nu}{\nu}\approx\dfrac{v_{rel}}{v}$, where $v_{rel}$ is the speed of approach; the source/observer asymmetry then becomes negligible.
  8. Stated in physical terms: the observed frequency equals the number of compressions that reach the observer each second, while the source frequency equals the number the source produces each second; the two numbers are equal only when neither the source nor the observer moves relative to the medium.
  9. Moving observer through a fixed pattern: the wavelength in the medium equals the original value while the compressions pass at the medium speed increased by the observer speed for approach, so the observed frequency increases in the ratio of that relative speed to the medium speed.
  10. Moving source through the medium: during one period the source moves forward a small distance, so the compressions ahead lie close together and the wavelength in the medium is reduced, while the medium speed is not changed and the observer meets the closer compressions more often.
  11. General reasoning: because a moving observer changes the relative speed while a moving source changes the wavelength, the two results differ even for equal speeds; the medium is the fixed reference in which the wave speed keeps its value, so the source motion and the observer motion are not equal.
⚠️ JEE trap: The most common and important misconception is that the Doppler effect for sound depends only on the relative velocity of source and observer, so that a source approaching a fixed observer at speed $u$ gives the same shift as an observer approaching a fixed source at the same speed $u$. It does not: $\nu'=\nu\,v/(v-u)$ for the moving source but $\nu'=\nu\,(v+u)/v$ for the moving observer, and these are unequal. The reason is that sound travels at a fixed speed relative to the medium, which therefore singles out a preferred frame — it physically matters whether it is the source or the observer that moves relative to the air. Only in the low-speed limit $u\ll v$, or for light in vacuum (which has no medium), does the effect reduce to depending on relative velocity alone. A second error is sign confusion; always check that your chosen signs make an approaching configuration raise the pitch and a receding one lower it. 🔉⇢

Worked example · JEE Advanced 🔉⇢

SITUATION A source of sound emits a note of frequency $\nu=400\ \text{Hz}$. Take the speed of sound in air as $v=340\ \text{m s}^{-1}$. Consider two separate cases at the same speed $u=34\ \text{m s}^{-1}$: (i) the source moves toward a stationary observer, and (ii) the observer moves toward a stationary source.
TARGET Find the observed frequency in each case and show explicitly that the two results differ, illustrating the source-observer asymmetry.
STRATEGY Apply the moving-source formula $\nu'=\nu\,v/(v-v_s)$ for case (i) and the moving-observer formula $\nu'=\nu(v+v_o)/v$ for case (ii), both configured for approach so that the pitch rises, then compare.
EXECUTE Case (i) moving source: $\nu'=400\times\dfrac{340}{340-34}=400\times\dfrac{340}{306}\approx444.4\ \text{Hz}$. Case (ii) moving observer: $\nu'=400\times\dfrac{340+34}{340}=400\times\dfrac{374}{340}=440.0\ \text{Hz}$. Both exceed $400\ \text{Hz}$ (approach raises pitch), but they are not equal: the moving source gives $\approx444\ \text{Hz}$ while the moving observer gives $440\ \text{Hz}$.
REFLECT Even though the relative speed of approach is $34\ \text{m s}^{-1}$ in both cases, the observed frequencies differ ($\approx444\ \text{Hz}$ vs $440\ \text{Hz}$) because sound has a fixed speed relative to the air: a moving source compresses the wavelength while a moving observer changes only the encounter rate. The discrepancy shrinks as $u/v\to0$, recovering the naive relative-velocity result in the low-speed limit.

Source: NCERT Class XI Physics, Chapter 14 (Waves)

✍️ Worked Examples Polya 5-move · JEE tier

WE1 · Transverse and Longitudinal Waves · JEE Advanced 🔉⇢

SITUATION You are given four wave situations: (a) a kink in a longitudinal spring produced by displacing one end sideways, (b) waves produced in a cylinder of liquid by moving its piston back and forth, (c) waves produced by a motorboat sailing in water, and (d) ultrasonic waves in air produced by a vibrating quartz crystal.
TARGET State in each case whether the wave motion is transverse, longitudinal, or a combination of both.
STRATEGY Apply the single criterion: compare the direction of particle oscillation with the direction of wave propagation. A sideways displacement of a spring introduces transverse motion, while a lengthwise push introduces longitudinal motion. Bulk fluids (gases, and the interior of liquids) can carry only longitudinal waves because they cannot sustain shear; water surfaces carry a mixed motion.
EXECUTE (a) A sideways displacement of the spring's end sets up transverse motion, but along the spring the disturbance also compresses and stretches the coils, so the motion is transverse and longitudinal. (b) A piston pushing a liquid back and forth compresses it along the axis, so the wave is purely longitudinal. (c) A motorboat generates water surface waves whose particles move both up-down and back-forth, so the motion is transverse and longitudinal. (d) Ultrasonic waves in air are sound waves, and air sustains only compressive strain, so the wave is longitudinal.
REFLECT The answers are (a) both, (b) longitudinal, (c) both, (d) longitudinal. Notice the deciding factor is never the source alone but which strains the medium can sustain: air permits only longitudinal waves, a solid spring permits both, and a water surface intrinsically mixes the two.

Source: NCERT Class XI Physics, Chapter 14 (Waves)

WE2 · Progressive Wave and Its Description · JEE Advanced 🔉⇢

SITUATION A wave travelling along a string is described by $y(x,t)=0.005\sin(80.0x-3.0t)$, with all numerical constants in SI units (so $0.005\ \text{m}$, $80.0\ \text{rad m}^{-1}$, $3.0\ \text{rad s}^{-1}$).
TARGET Find (a) the amplitude, (b) the wavelength, and (c) the period and frequency; then evaluate the displacement $y$ at $x=30.0\ \text{cm}$ and $t=20\ \text{s}$.
STRATEGY Compare the given expression term-by-term with the standard progressive-wave form $y=a\sin(kx-\omega t)$ to read off $a$, $k$ and $\omega$; then use $\lambda=2\pi/k$, $T=2\pi/\omega$ and $\nu=1/T$, and finally substitute the numbers into the argument (in radians) to get $y$.
EXECUTE Matching gives $a=0.005\ \text{m}=5\ \text{mm}$, $k=80.0\ \text{m}^{-1}$, $\omega=3.0\ \text{s}^{-1}$. Then $\lambda=2\pi/80.0\approx7.85\ \text{cm}$; $T=2\pi/3.0\approx2.09\ \text{s}$; $\nu=1/T\approx0.48\ \text{Hz}$. At $x=0.30\ \text{m},\ t=20\ \text{s}$: argument $=80.0(0.30)-3.0(20)=24-60=-36\ \text{rad}$, so $y=0.005\sin(-36)$. Reducing $-36\ \text{rad}$ modulo $2\pi$ gives $\sin(-36)\approx1$, hence $y\approx5\ \text{mm}$.
REFLECT Reading off $a,k,\omega$ by comparison is the standard first step, and it immediately yields the wave's spatial and temporal periods. The final substitution reminds us the phase argument is in radians; the displacement at that instant happens to be near its maximum, but it never exceeds the amplitude $5\ \text{mm}$.

Source: NCERT Class XI Physics, Chapter 14 (Waves)

WE3 · The Speed of a Travelling Wave · JEE Advanced 🔉⇢

SITUATION Estimate the speed of sound in air at standard temperature and pressure (STP), given that the mass of one mole of air is $29.0\times10^{-3}\ \text{kg}$, and compare the isothermal (Newton) and adiabatic (Laplace) predictions with the measured value.
TARGET Compute the density of air at STP, apply Newton's isothermal formula $v=\sqrt{P/\rho}$, then apply the Laplace-corrected adiabatic formula $v=\sqrt{\gamma P/\rho}$ with $\gamma=7/5$, and compare both with the experimental $331\ \text{m s}^{-1}$.
STRATEGY One mole of any gas occupies $22.4\ \text{L}$ at STP, so use $\rho_0=M/V_m$ for the density; take $P=1.01\times10^5\ \text{Pa}$. First find $\rho_0$, then substitute into each speed formula.
EXECUTE $\rho_0=\dfrac{29.0\times10^{-3}\ \text{kg}}{22.4\times10^{-3}\ \text{m}^3}=1.29\ \text{kg m}^{-3}$. Newton: $v=\sqrt{P/\rho_0}=\sqrt{1.01\times10^5/1.29}\approx280\ \text{m s}^{-1}$. Laplace: $v=\sqrt{\gamma P/\rho_0}=\sqrt{(7/5)(1.01\times10^5)/1.29}\approx331\ \text{m s}^{-1}$.
REFLECT Newton's isothermal estimate is about $15\%$ below the measured $331\ \text{m s}^{-1}$; multiplying the bulk modulus by $\gamma=7/5$ (the Laplace correction for adiabatic compressions) brings the prediction into agreement. This is the textbook demonstration that sound propagation in a gas is adiabatic, not isothermal.

Source: NCERT Class XI Physics, Chapter 14 (Waves)

WE4 · Beats · JEE Advanced 🔉⇢

SITUATION Two sitar strings A and B playing the note 'Dha' are slightly out of tune and produce beats of frequency $5\ \text{Hz}$. The tension of string B is slightly increased and the beat frequency is found to decrease to $3\ \text{Hz}$. The frequency of A is $427\ \text{Hz}$.
TARGET Find the original frequency of string B.
STRATEGY Use $\nu_{beat}=|\nu_A-\nu_B|=5\ \text{Hz}$, then resolve the sign ambiguity using the fact that increasing a string's tension increases its frequency and observing whether the beat frequency rose or fell.
EXECUTE Since increasing B's tension raises $\nu_B$, if $\nu_B$ had been greater than $\nu_A$ the gap $|\nu_A-\nu_B|$ would have widened and beats would increase; instead the beats decreased (from $5$ to $3\ \text{Hz}$), so raising $\nu_B$ moved it toward $\nu_A$, meaning $\nu_B<\nu_A$. Then $\nu_A-\nu_B=5\ \text{Hz}$ with $\nu_A=427\ \text{Hz}$ gives $\nu_B=422\ \text{Hz}$.
REFLECT The original frequency of B is $422\ \text{Hz}$. The decisive step is not the arithmetic but the sign logic: the beat frequency supplies only the magnitude of the difference, and the response of the beats to a known change in tension tells us which string was flat.

Source: NCERT Class XI Physics, Chapter 14 (Waves)

WE5 · Problem 1 · easy 🔉⇢

SITUATION Three disturbances are listed: (i) sound propagating through air, (ii) a pulse sent down a stretched string by flicking one end sideways, (iii) the P-waves and S-waves generated by an earthquake travelling through rock.
TARGET Classify each as transverse or longitudinal, and explain why S-waves cannot travel through the liquid outer core of the Earth.
STRATEGY A wave is transverse if particle displacement is perpendicular to propagation and longitudinal if parallel. Transverse mechanical waves require a restoring shear stress, which only solids and strings under tension can supply; fluids have zero shear modulus.
EXECUTE (i) Sound in air is $\textbf{longitudinal}$ — air can support only compressions/rarefactions ($\text{bulk modulus}\neq0$, shear modulus $=0$). (ii) The string pulse is $\textbf{transverse}$ — tension provides the transverse restoring force. (iii) P-waves are $\textbf{longitudinal}$ and S-waves are $\textbf{transverse}$; since a liquid ($\mu_{\text{shear}}=0$) cannot sustain shear, S-waves are stopped by the outer core.
REFLECT The disappearance of S-waves beyond a certain angle (the S-wave shadow zone) is the seismological proof that the Earth's outer core is liquid — a wave classification with planetary consequences.

Source: NCERT XI §14.2 (derived)

WE6 · Problem 2 · medium 🔉⇢

SITUATION A pure tone of frequency $340\ \text{Hz}$ travels through air at $v=340\ \text{m s}^{-1}$ as a longitudinal wave.
TARGET Find the wavelength, and the shortest distance between a point of maximum pressure variation (pressure antinode) and the nearest point of maximum displacement (displacement antinode).
STRATEGY Use $\lambda=v/\nu$. In a longitudinal sound wave the pressure variation and the particle displacement are $90^\circ$ out of phase in space, so a pressure antinode coincides with a displacement node; the nearest displacement antinode is a quarter wavelength away.
EXECUTE $\lambda=\dfrac{v}{\nu}=\dfrac{340}{340}=1\ \text{m}$. Pressure and displacement are in space-quadrature, so the required separation is $\dfrac{\lambda}{4}=\dfrac{1}{4}=0.25\ \text{m}$.
REFLECT The pressure description and displacement description of the same sound wave peak at different places — a microphone (pressure sensor) and a hot-wire (velocity sensor) placed together would report signals a quarter wavelength out of step.

Source: NCERT XI §14.2 / §14.3 (derived)

WE7 · Problem 3 · medium 🔉⇢

SITUATION A student claims that by shaking the free surface of water in a tank one can launch a purely transverse wave, and that the same is possible deep inside the bulk of a large body of water.
TARGET Decide whether a purely transverse mechanical wave can propagate through the bulk of a liquid or gas, and describe the actual motion of surface water waves.
STRATEGY A transverse bulk wave needs a shear restoring force ($\text{shear modulus}>0$). Fluids in bulk have no shear rigidity, so bulk waves in them are longitudinal only. Surface waves are a special case where gravity and surface tension provide restoring forces.
EXECUTE In the $\textbf{bulk}$ of a gas or liquid only $\textbf{longitudinal}$ waves propagate (no shear rigidity). At a free $\textbf{surface}$, however, water particles trace nearly circular paths — the disturbance is a combination of transverse and longitudinal motion driven by gravity and surface tension, not a pure transverse elastic wave.
REFLECT This is why 'transverse water waves' are not a counterexample to the shear-modulus rule: they live only on the interface, and their restoring force is gravity, not elastic shear of the bulk.

Source: NCERT XI §14.2 (derived)

WE8 · Problem 4 · easy 🔉⇢

SITUATION A wave on a string is described in SI units by $y(x,t)=0.005\,\sin(80.0\,x-3.0\,t)$.
TARGET Find (a) amplitude, (b) wavelength, (c) period and frequency, (d) wave speed, and (e) the displacement at $x=30.0\ \text{cm}$, $t=20\ \text{s}$.
STRATEGY Compare term by term with the standard progressive wave $y=a\sin(kx-\omega t)$ to read off $a$, $k=80.0\ \text{m}^{-1}$, $\omega=3.0\ \text{s}^{-1}$; then use $\lambda=2\pi/k$, $T=2\pi/\omega$, $\nu=1/T$, $v=\omega/k$.
EXECUTE (a) $a=0.005\ \text{m}=5\ \text{mm}$. (b) $\lambda=\dfrac{2\pi}{80}=7.85\times10^{-2}\ \text{m}$. (c) $T=\dfrac{2\pi}{3}=2.09\ \text{s}$, $\nu=1/T=0.48\ \text{Hz}$. (d) $v=\dfrac{\omega}{k}=\dfrac{3.0}{80}=3.75\times10^{-2}\ \text{m s}^{-1}$. (e) $y=0.005\sin(80{\times}0.30-3{\times}20)=0.005\sin(-36\ \text{rad})\approx 5\ \text{mm}$.
REFLECT The wave speed $0.0375\ \text{m s}^{-1}$ is a property of the medium, whereas the maximum particle speed $a\omega=0.015\ \text{m s}^{-1}$ is smaller here — the two speeds are unrelated in general.

Source: NCERT XI §14.3, Example 14.2 (derived)

WE9 · Problem 5 · medium 🔉⇢

SITUATION A transverse harmonic wave is written as $y(x,t)=0.02\,\sin\!\big(\pi(0.5\,x-200\,t)\big)$ m, with $x$ in metres and $t$ in seconds.
TARGET Determine the direction of propagation, wavelength, frequency, and wave speed, and confirm $v=\nu\lambda$.
STRATEGY Expand to $y=a\sin(kx-\omega t)$: here $k=0.5\pi\ \text{m}^{-1}$ and $\omega=200\pi\ \text{s}^{-1}$. A $(kx-\omega t)$ argument means motion along $+x$. Use $\lambda=2\pi/k$, $\nu=\omega/2\pi$, $v=\omega/k$.
EXECUTE $\lambda=\dfrac{2\pi}{0.5\pi}=4\ \text{m}$; $\nu=\dfrac{200\pi}{2\pi}=100\ \text{Hz}$; $v=\dfrac{\omega}{k}=\dfrac{200\pi}{0.5\pi}=400\ \text{m s}^{-1}$, along $+x$. Check: $\nu\lambda=100\times4=400\ \text{m s}^{-1}=v.$
REFLECT The sign of the coefficient of $t$ relative to $x$ is the whole story about direction: a common minus sign gives $+x$ travel, opposite signs give $-x$ travel.

Source: NCERT XI §14.3 (derived)

WE10 · Problem 6 · medium 🔉⇢

SITUATION A sound wave in air of frequency $500\ \text{Hz}$ and amplitude $2.0\ \text{mm}$ travels at $340\ \text{m s}^{-1}$.
TARGET Compare the maximum speed of an air particle with the speed of the wave itself.
STRATEGY Wave speed is $v$ (property of the medium). The maximum particle speed for $y=a\sin(kx-\omega t)$ is $v_{p,\max}=a\omega=2\pi\nu a$. Compute both and take the ratio.
EXECUTE $v_{p,\max}=2\pi\nu a=2\pi(500)(2.0\times10^{-3})=6.28\ \text{m s}^{-1}$. Wave speed $v=340\ \text{m s}^{-1}$. Ratio $\dfrac{v_{p,\max}}{v}=\dfrac{6.28}{340}\approx0.018$.
REFLECT Even for an audibly loud tone the air molecules jiggle at only a few metres per second, hundreds of times slower than the wave that races through them — energy is transported without any bulk transport of the medium.

Source: NCERT XI §14.3 (derived)

WE11 · Problem 7 · advanced 🔉⇢

SITUATION A progressive wave is $y(x,t)=5.0\times10^{-3}\,\sin(50\,x-1068\,t)$ m (SI units), so $k=50\ \text{m}^{-1}$ and $\omega=1068\ \text{s}^{-1}$ (i.e. $\nu\approx170\ \text{Hz}$).
TARGET Find (a) the phase difference between two particles separated by $\Delta x=\lambda/3$, and (b) the maximum acceleration of a particle.
STRATEGY Two ideas chain here. Spatial phase difference is $\Delta\phi=k\,\Delta x$; for $\Delta x=\lambda/3$ this is simply $2\pi/3$ independent of the numbers. Maximum particle acceleration for SHM is $a_{\max}=\omega^{2}A$.
EXECUTE (a) $\Delta\phi=k\,\Delta x=k\cdot\dfrac{\lambda}{3}=\dfrac{2\pi}{\lambda}\cdot\dfrac{\lambda}{3}=\dfrac{2\pi}{3}\ \text{rad}=120^\circ$. (b) $a_{\max}=\omega^{2}A=(1068)^{2}(5.0\times10^{-3})\approx5.7\times10^{3}\ \text{m s}^{-2}$.
REFLECT A one-third-wavelength gap always means a $120^\circ$ phase lag whatever the frequency, while the huge peak acceleration ($\sim580g$) shows how violently a medium point must accelerate to carry even a modest-amplitude high-frequency wave.

Source: JEE-pattern

WE12 · Problem 8 · easy 🔉⇢

SITUATION A steel wire $0.72\ \text{m}$ long has mass $5.0\times10^{-3}\ \text{kg}$ and is stretched to a tension of $60\ \text{N}$.
TARGET Find the speed of transverse waves on the wire.
STRATEGY Speed on a string is $v=\sqrt{T/\mu}$ with linear mass density $\mu=m/L$. Compute $\mu$ first, then substitute.
EXECUTE $\mu=\dfrac{5.0\times10^{-3}}{0.72}=6.9\times10^{-3}\ \text{kg m}^{-1}$. $v=\sqrt{\dfrac{T}{\mu}}=\sqrt{\dfrac{60}{6.9\times10^{-3}}}=\sqrt{8.64\times10^{3}}\approx93\ \text{m s}^{-1}$.
REFLECT The speed depends only on tension and mass per length, not on frequency or amplitude — the same wire carries a $10\ \text{Hz}$ and a $1000\ \text{Hz}$ wave at the identical $93\ \text{m s}^{-1}$.

Source: NCERT XI §14.4, Example 14.3 (derived)

WE13 · Problem 9 · medium 🔉⇢

SITUATION Air at $0^\circ\text{C}$ has pressure $P=1.01\times10^{5}\ \text{Pa}$, density $\rho=1.29\ \text{kg m}^{-3}$, and ratio of specific heats $\gamma=1.4$.
TARGET Estimate the speed of sound using Laplace's (adiabatic) formula.
STRATEGY Sound compressions/rarefactions are so rapid that they are adiabatic, not isothermal. Laplace's correction gives $v=\sqrt{\gamma P/\rho}$ rather than Newton's $\sqrt{P/\rho}$.
EXECUTE $v=\sqrt{\dfrac{\gamma P}{\rho}}=\sqrt{\dfrac{1.4\times1.01\times10^{5}}{1.29}}=\sqrt{1.096\times10^{5}}\approx331\ \text{m s}^{-1}$.
REFLECT The result $331\ \text{m s}^{-1}$ matches the measured speed of sound at $0^\circ\text{C}$, confirming that the compressions are adiabatic — the factor $\sqrt{\gamma}=1.18$ is exactly what Newton's model was missing.

Source: NCERT XI §14.4, Example 14.4 (derived)

WE14 · Problem 10 · medium 🔉⇢

SITUATION For the same air ($P=1.01\times10^{5}\ \text{Pa}$, $\rho=1.29\ \text{kg m}^{-3}$, $\gamma=1.4$), Newton assumed the compressions in a sound wave are isothermal.
TARGET Compute the speed Newton's formula predicts and quantify how far it falls short of the measured $331\ \text{m s}^{-1}$.
STRATEGY Newton took $v=\sqrt{P/\rho}$ (isothermal bulk modulus $=P$). Compute and compare with the Laplace/measured value; the ratio is exactly $\sqrt{\gamma}$.
EXECUTE $v_{\text{Newton}}=\sqrt{\dfrac{P}{\rho}}=\sqrt{\dfrac{1.01\times10^{5}}{1.29}}=\sqrt{7.83\times10^{4}}\approx280\ \text{m s}^{-1}$. This is about $15\%$ low; the correcting factor is $\dfrac{v_{\text{Laplace}}}{v_{\text{Newton}}}=\sqrt{\gamma}=\sqrt{1.4}\approx1.18$.
REFLECT Newton's $280\ \text{m s}^{-1}$ disagreed with experiment for over a century until Laplace realised heat has no time to flow during the rapid oscillations — a rare case where a famous formula was quantitatively, not just conceptually, wrong.

Source: NCERT XI §14.4 (derived)

WE15 · Problem 11 · advanced 🔉⇢

SITUATION The speed of sound in air is $331\ \text{m s}^{-1}$ at $0^\circ\text{C}$ ($273\ \text{K}$). Using $v=\sqrt{\gamma RT/M}$ so that $v\propto\sqrt{T}$ at fixed composition.
TARGET Find the speed of sound at $27^\circ\text{C}$, and the approximate increase in speed per degree Celsius near room temperature.
STRATEGY Since $v\propto\sqrt{T}$, use $v_2=v_1\sqrt{T_2/T_1}$ with absolute temperatures. For the per-degree rate, differentiate: $\dfrac{dv}{dT}=\dfrac{v}{2T}$.
EXECUTE $v_{300}=331\sqrt{\dfrac{300}{273}}=331\times1.048\approx347\ \text{m s}^{-1}$. Rate near $273\ \text{K}$: $\dfrac{dv}{dT}=\dfrac{331}{2\times273}\approx0.61\ \text{m s}^{-1}$ per kelvin (per $^\circ\text{C}$).
REFLECT The familiar rule 'sound gains about $0.6\ \text{m s}^{-1}$ for every degree of warming' drops straight out of $v\propto\sqrt{T}$; note that pressure cancels (raising $P$ raises $\rho$ in step), so only temperature matters for a given gas.

Source: NCERT XI §14.4 (derived)

WE16 · Problem 12 · medium 🔉⇢

SITUATION Two waves of the same frequency arrive at a point with amplitudes $a_1=3\ \text{units}$ and $a_2=4\ \text{units}$, with a constant phase difference $\phi=90^\circ$ between them.
TARGET Find the amplitude of the resultant wave.
STRATEGY By the principle of superposition the resultant of two same-frequency sinusoids has amplitude $A=\sqrt{a_1^{2}+a_2^{2}+2a_1a_2\cos\phi}$.
EXECUTE $A=\sqrt{3^{2}+4^{2}+2(3)(4)\cos90^\circ}=\sqrt{9+16+0}=\sqrt{25}=5\ \text{units}$.
REFLECT At $\phi=90^\circ$ the cross term vanishes and amplitudes add in quadrature like perpendicular vectors; contrast with $\phi=0$ (giving $7$) and $\phi=180^\circ$ (giving $1$) for the same two waves.

Source: NCERT XI §14.5 (derived)

WE17 · Problem 13 · advanced 🔉⇢

SITUATION Two identical loudspeakers, driven in phase by the same $170\ \text{Hz}$ oscillator ($v=340\ \text{m s}^{-1}$), face a listener. The path lengths from the two speakers to the listener differ by $1.0\ \text{m}$.
TARGET Determine whether the listener hears a loud sound or near-silence at that spot.
STRATEGY Interference is set by the path difference in wavelengths: $\Delta=m\lambda$ gives constructive (loud), $\Delta=(m+\tfrac12)\lambda$ gives destructive (silence). First find $\lambda=v/\nu$, then express $1.0\ \text{m}$ in units of $\lambda$.
EXECUTE $\lambda=\dfrac{v}{\nu}=\dfrac{340}{170}=2.0\ \text{m}$. Path difference $\Delta=1.0\ \text{m}=\dfrac{\lambda}{2}$, i.e. a half-integer number of wavelengths, corresponding to a phase difference of $\pi$. Hence the waves arrive out of phase and the listener hears $\textbf{near-silence}$ (destructive interference).
REFLECT A geometric half-metre becomes acoustically decisive because it equals $\lambda/2$; move the listener another $1\ \text{m}$ ($\Delta=\lambda$) and the sound becomes loud again — the basis of interference-based noise cancellation.

Source: JEE-pattern

WE18 · Problem 14 · medium 🔉⇢

SITUATION Two coherent waves of equal amplitude $a$ superpose at a point with phase difference $\phi=60^\circ$.
TARGET Find the resultant amplitude and the resultant intensity in terms of the single-wave intensity $I_0$.
STRATEGY For equal amplitudes the resultant amplitude simplifies to $A=2a\cos(\phi/2)$, and since intensity $\propto$ amplitude$^2$, $I=4I_0\cos^{2}(\phi/2)$.
EXECUTE $A=2a\cos30^\circ=2a\cdot\dfrac{\sqrt3}{2}=\sqrt3\,a\approx1.73\,a$. Intensity $I=4I_0\cos^{2}30^\circ=4I_0(0.75)=3I_0$.
REFLECT Two equal sources do not simply give $2I_0$: in phase they give $4I_0$ (constructive), out of phase $0$; at $60^\circ$ we land at $3I_0$. Energy is redistributed, never created — averaged over all $\phi$ it returns to $2I_0$.

Source: NCERT XI §14.5 (derived)

WE19 · Problem 15 · easy 🔉⇢

SITUATION A transverse pulse of positive (upward) displacement travels along a string toward a rigidly clamped end, and a second identical pulse travels toward a free (ring-on-frictionless-rod) end.
TARGET State the phase change on reflection at each boundary and the nature of the end point (node or antinode) for a standing wave.
STRATEGY A rigid boundary cannot move, so the reflected wave must cancel the incident displacement there — this forces a $\pi$ phase flip (crest returns as trough) and makes the end a displacement node. A free end has no such constraint: no phase change, and it is a displacement antinode.
EXECUTE At the $\textbf{rigid}$ end: reflection with a $\pi$ ($180^\circ$) phase change; the upward pulse returns inverted; the end is a $\textbf{node}$. At the $\textbf{free}$ end: reflection with $\textbf{no}$ phase change; the pulse returns erect; the end is an $\textbf{antinode}$.
REFLECT The boundary condition, not the wave, decides the inversion: 'hard' reflections flip the wave (like a light wave off a denser medium), 'soft' reflections do not — a single rule that recurs across all of wave physics.

Source: NCERT XI §14.6 (derived)

WE20 · Problem 16 · advanced 🔉⇢

SITUATION A sonometer wire of length $L=0.50\ \text{m}$ and linear mass density $\mu=1.0\times10^{-3}\ \text{kg m}^{-1}$ is clamped at both ends and stretched to a tension $T=40\ \text{N}$.
TARGET Find the wave speed on the wire and the frequencies of the first two harmonics (fundamental and first overtone).
STRATEGY Chain two results. First the wave speed $v=\sqrt{T/\mu}$; then, for a string fixed at both ends, standing waves require $L=n\lambda/2$, giving $\nu_n=\dfrac{nv}{2L}$, $n=1,2,3,\dots$
EXECUTE $v=\sqrt{\dfrac{40}{1.0\times10^{-3}}}=\sqrt{4.0\times10^{4}}=200\ \text{m s}^{-1}$. Fundamental $\nu_1=\dfrac{v}{2L}=\dfrac{200}{2(0.50)}=200\ \text{Hz}$. First overtone $\nu_2=2\nu_1=400\ \text{Hz}$.
REFLECT A string fixed at both ends emits the complete harmonic series $\nu_1,2\nu_1,3\nu_1,\dots$ — this integer-multiple structure is why plucked strings sound musical, and doubling the tension would raise every note by the same factor $\sqrt2$.

Source: NCERT XI §14.4 + §14.6 (derived)

WE21 · Problem 17 · medium 🔉⇢

SITUATION A pipe $30.0\ \text{cm}$ long is open at both ends; the speed of sound is $330\ \text{m s}^{-1}$.
TARGET Which harmonic of the pipe resonates with a $1.1\ \text{kHz}$ source?
STRATEGY An open pipe supports all harmonics $\nu_n=\dfrac{nv}{2L}$, $n=1,2,3,\dots$ Compute $\nu_1$ and see which integer multiple equals $1100\ \text{Hz}$.
EXECUTE $\nu_n=\dfrac{n\,v}{2L}=\dfrac{n(330)}{2(0.30)}=550\,n\ \text{Hz}$. Setting $550n=1100$ gives $n=2$, so the source resonates with the $\textbf{second harmonic}$ (first overtone).
REFLECT An open pipe has an antinode at each end and contains every integer harmonic, so it is richer in overtones than a closed pipe of the same length — one reason open flue organ pipes sound brighter.

Source: NCERT XI §14.6, Example 14.5 (derived)

WE22 · Problem 18 · medium 🔉⇢

SITUATION The same pipe ($L=30.0\ \text{cm}$, $v=330\ \text{m s}^{-1}$) is now closed at one end.
TARGET Determine whether the $1.1\ \text{kHz}$ source can still produce resonance.
STRATEGY A closed (stopped) pipe has a node at the closed end and antinode at the open end, so only odd harmonics exist: $\nu_n=\dfrac{(2n-1)v}{4L}$, $n=1,2,3,\dots$ Find the fundamental and test whether $1100\ \text{Hz}$ is an odd multiple of it.
EXECUTE $\nu_1=\dfrac{v}{4L}=\dfrac{330}{4(0.30)}=275\ \text{Hz}$; allowed frequencies are $275,825,1375,\dots$ ($275\times\{1,3,5,\dots\}$). We need $\dfrac{1100}{275}=4$, but $4$ is even, so $1100\ \text{Hz}$ is $\textbf{not}$ an allowed mode — $\textbf{no resonance}$.
REFLECT Closing one end halves the fundamental (deepens the pitch by an octave) and deletes all even harmonics; a frequency that resonated in the open pipe can fall silent when the pipe is stopped.

Source: NCERT XI §14.6, Example 14.5 (derived)

WE23 · Problem 19 · advanced 🔉⇢

SITUATION In a resonance-tube experiment a tuning fork of frequency $340\ \text{Hz}$ is held over a tube whose air column length is varied. The first resonance occurs at $l_1=0.24\ \text{m}$ and the second at $l_2=0.74\ \text{m}$.
TARGET Determine the speed of sound and the end correction of the tube.
STRATEGY For a closed air column the resonances are at $l_1+e=\lambda/4$ and $l_2+e=3\lambda/4$. Subtracting eliminates the unknown end correction: $l_2-l_1=\lambda/2$, so $v=2\nu(l_2-l_1)$. Then back-substitute for $e=\dfrac{l_2-3l_1}{2}$.
EXECUTE $v=2\nu(l_2-l_1)=2(340)(0.74-0.24)=2(340)(0.50)=340\ \text{m s}^{-1}$. End correction $e=\dfrac{l_2-3l_1}{2}=\dfrac{0.74-3(0.24)}{2}=\dfrac{0.74-0.72}{2}=0.01\ \text{m}=1\ \text{cm}$.
REFLECT Taking the difference $l_2-l_1$ cancels the end correction entirely, which is why the two-resonance method gives an accurate $v$; the leftover $e\approx0.6\times$(tube radius) is the small distance the antinode sits beyond the open mouth.

Source: JEE-pattern

WE24 · Problem 20 · easy 🔉⇢

SITUATION Two tuning forks of frequencies $256\ \text{Hz}$ and $260\ \text{Hz}$ are sounded together.
TARGET Find the beat frequency heard, and the time interval between successive maxima of loudness.
STRATEGY When two close frequencies superpose, the loudness waxes and wanes at the beat frequency $\nu_{\text{beat}}=|\nu_1-\nu_2|$; the period between maxima is $1/\nu_{\text{beat}}$.
EXECUTE $\nu_{\text{beat}}=|260-256|=4\ \text{Hz}$. Time between successive loudness maxima $=\dfrac{1}{4}=0.25\ \text{s}$.
REFLECT The ear hears one tone at the average pitch $258\ \text{Hz}$ swelling four times a second; beats vanish as the forks are brought into tune, which is exactly how musicians tune by ear.

Source: NCERT XI §14.7 (derived)

WE25 · Problem 21 · medium 🔉⇢

SITUATION Two sitar strings A and B playing the same note are slightly out of tune and give $5$ beats per second. When the tension of B is slightly increased, the beat frequency falls to $3\ \text{Hz}$. String A has frequency $427\ \text{Hz}$.
TARGET Find the original frequency of string B.
STRATEGY Increasing tension raises a string's frequency ($\nu\propto\sqrt T$). If raising $\nu_B$ made the beat decrease, then $\nu_B$ was moving toward $\nu_A$, so originally $\nu_B<\nu_A$. Combine with $|\nu_A-\nu_B|=5$.
EXECUTE Since the beat decreased on raising $\nu_B$, we have $\nu_B<\nu_A$, so $\nu_B=\nu_A-5=427-5=422\ \text{Hz}$.
REFLECT The direction the beat frequency moves when you change one source resolves the sign ambiguity in $|\nu_A-\nu_B|$ — without that clue $\nu_B$ could equally have been $432\ \text{Hz}$.

Source: NCERT XI §14.7, Example 14.6 (derived)

WE26 · Problem 22 · advanced 🔉⇢

SITUATION A tuning fork A of frequency $256\ \text{Hz}$ produces $4$ beats per second with an unknown fork B. A little wax is loaded onto B (which lowers B's frequency), and the beat frequency is now observed to increase to $6\ \text{Hz}$.
TARGET Determine the original frequency of fork B.
STRATEGY The unknown is $\nu_B=256\pm4$. Loading wax lowers $\nu_B$. Test each candidate: if lowering $\nu_B$ moves it away from $256$, the beat increases; if toward $256$, the beat decreases. Pick the candidate consistent with the observed increase to $6\ \text{Hz}$.
EXECUTE Candidates: $260$ or $252\ \text{Hz}$. If $\nu_B=260$, waxing lowers it toward $256$, so the beat would $\textbf{decrease}$ — rejected. If $\nu_B=252$, waxing lowers it further below $256$ (e.g. to $250$), so the beat $\textbf{increases}$ to $6\ \text{Hz}$ — consistent. Hence $\nu_B=252\ \text{Hz}$.
REFLECT Loading (or filing) a fork is the standard trick to break the $\pm$ degeneracy of the beat equation: watching whether beats rise or fall pins down which side of A the unknown lies.

Source: JEE-pattern

WE27 · Problem 23 · medium 🔉⇢

SITUATION An ambulance siren emits a steady $400\ \text{Hz}$ tone and moves toward a stationary listener at $34\ \text{m s}^{-1}$. Speed of sound $v=340\ \text{m s}^{-1}$.
TARGET Find the frequency heard by the listener.
STRATEGY For a source approaching a stationary observer, wavefronts bunch up: $\nu'=\nu\,\dfrac{v}{v-v_s}$ (source speed subtracts in the denominator when approaching).
EXECUTE $\nu'=\nu\dfrac{v}{v-v_s}=400\times\dfrac{340}{340-34}=400\times\dfrac{340}{306}=400\times1.111\approx444\ \text{Hz}$.
REFLECT The pitch rises by about $44\ \text{Hz}$ while approaching; after the ambulance passes, $v_s$ flips sign and the heard frequency drops to $400\times340/374\approx364\ \text{Hz}$ — the sudden fall you hear as it goes by.

Source: JEE-pattern

WE28 · Problem 24 · advanced 🔉⇢

SITUATION A listener runs at $34\ \text{m s}^{-1}$ toward a stationary loudspeaker that emits $400\ \text{Hz}$; separately, consider the listener at rest while the speaker moves toward her at $34\ \text{m s}^{-1}$. Take $v=340\ \text{m s}^{-1}$.
TARGET Compute the heard frequency in each case and show that they are not equal.
STRATEGY Doppler formulas are asymmetric because the medium (air) defines the rest frame. Moving observer: $\nu'=\nu\dfrac{v+v_o}{v}$. Moving source: $\nu'=\nu\dfrac{v}{v-v_s}$. Evaluate both for the same $34\ \text{m s}^{-1}$.
EXECUTE Moving observer: $\nu'=400\dfrac{340+34}{340}=400\times1.10=440\ \text{Hz}$. Moving source: $\nu'=400\dfrac{340}{340-34}=400\times1.111\approx444\ \text{Hz}$. The two differ by about $4\ \text{Hz}$.
REFLECT Although both cases 'close the gap' at the same speed, the results differ because only the source's motion actually changes the wavelength in the medium; the asymmetry disappears only in the limit $v_o,v_s\ll v$, where both reduce to $\nu(1+34/340)$.

Source: JEE-pattern

WE29 · Problem 25 · advanced 🔉⇢

SITUATION A car sounds a horn of frequency $400\ \text{Hz}$ while driving straight toward a large vertical wall at $20\ \text{m s}^{-1}$. The driver hears both the horn directly and its echo from the wall. Speed of sound $v=340\ \text{m s}^{-1}$.
TARGET Find the beat frequency the driver hears between the direct horn and the reflected echo.
STRATEGY Chain Doppler twice, then beats. (1) The wall is a stationary observer receiving from an approaching source: $\nu_w=\nu\dfrac{v}{v-v_s}$. (2) The wall re-emits $\nu_w$ as a stationary source; the driver is a moving observer approaching it: $\nu_{\text{echo}}=\nu_w\dfrac{v+v_s}{v}$. The direct horn heard by the co-moving driver stays at $\nu=400$. Beat $=\nu_{\text{echo}}-\nu$.
EXECUTE $\nu_w=400\dfrac{340}{340-20}=400\dfrac{340}{320}=425\ \text{Hz}$. $\nu_{\text{echo}}=425\dfrac{340+20}{340}=425\dfrac{360}{340}=450\ \text{Hz}$. Beat frequency $=450-400=50\ \text{Hz}$.
REFLECT The echo is Doppler-shifted twice (wall receives high, driver receives higher), giving a large $50\ \text{Hz}$ beat; for small speeds the echo-beat is approximately $\nu\cdot\dfrac{2v_s}{v}=400\cdot\dfrac{40}{340}\approx47\ \text{Hz}$, the basis of Doppler speed radar.

Source: JEE-pattern

On the concept tabs

These worked examples are taught in full alongside their interactive scene:

📐 Formula Sheet Printable · every formula cited

The Progressive Wave and its Parameters

QuantityFormulaWhat it means / when to useSource
Displacement relation (progressive wave) 🔉⇢$y(x,t)=A\sin(kx-\omega t+\phi)$A harmonic wave travelling in the $+x$ direction. $A$ = amplitude, $(kx-\omega t+\phi)$ = phase. Use $(kx+\omega t)$ for a wave travelling in the $-x$ direction.NCERT XI Ch 14 (§14.3)
Angular wave number 🔉⇢$k=\dfrac{2\pi}{\lambda}$$\lambda$ = wavelength, the distance between two consecutive points in the same phase. $k$ has units of $\mathrm{rad\,m^{-1}}$.NCERT XI Ch 14 (§14.3.2)
Angular frequency 🔉⇢$\omega=2\pi\nu=\dfrac{2\pi}{T}$$T$ = period, $\nu$ = frequency. A single particle executes SHM of angular frequency $\omega$ as the wave passes.NCERT XI Ch 14 (§14.3.3)
Wave speed 🔉⇢$v=\dfrac{\omega}{k}=\nu\lambda=\dfrac{\lambda}{T}$The speed at which a point of constant phase (a crest) travels. Fixed by the medium, not by the source.NCERT XI Ch 14 (§14.4)
Particle velocity 🔉⇢$v_p=\dfrac{\partial y}{\partial t}=-A\omega\cos(kx-\omega t)$The velocity of an element of the medium — distinct from the wave velocity. Maximum particle speed $=A\omega$. Also $v_p=-v\,(\partial y/\partial x)$.NCERT XI Ch 14 (§14.3)

Speed of a Travelling Wave

QuantityFormulaWhat it means / when to useSource
Wave on a stretched string 🔉⇢$v=\sqrt{\dfrac{T}{\mu}}$$T$ = tension, $\mu$ = linear mass density (mass per unit length). A restoring-force factor over an inertia factor, under a root.NCERT XI Ch 14 (§14.4.1)
Speed of sound (Newton-Laplace) 🔉⇢$v=\sqrt{\dfrac{B}{\rho}}=\sqrt{\dfrac{\gamma P}{\rho}}$$B=\gamma P$ = adiabatic bulk modulus, $\rho$ = density. Laplace's adiabatic correction to Newton's isothermal $v=\sqrt{P/\rho}$.NCERT XI Ch 14 (§14.4.2)
Speed of sound vs temperature 🔉⇢$v=\sqrt{\dfrac{\gamma RT}{M}}\;\Rightarrow\;v\propto\sqrt{T}$For an ideal gas $P/\rho=RT/M$, so the speed of sound depends on temperature but not on pressure. $M$ = molar mass.NCERT XI Ch 14 (§14.4.2)

Superposition, Standing Waves and Normal Modes

QuantityFormulaWhat it means / when to useSource
Principle of superposition 🔉⇢$y=\sum_i y_i$The resultant displacement at a point is the algebraic sum of the displacements due to each wave separately.NCERT XI Ch 14 (§14.5)
Standing wave 🔉⇢$y(x,t)=2A\sin kx\,\cos\omega t$Superposition of two identical waves travelling in opposite directions. Amplitude $2A\sin kx$ varies with position; the pattern does not move.NCERT XI Ch 14 (§14.6)
Nodes and antinodes 🔉⇢$x_{node}=\dfrac{n\lambda}{2},\quad x_{antinode}=\dfrac{(2n+1)\lambda}{4}$Nodes (zero amplitude) and antinodes (maximum amplitude) alternate, each set spaced $\lambda/2$ apart; a node and the next antinode are $\lambda/4$ apart.NCERT XI Ch 14 (§14.6.1)
String fixed at both ends 🔉⇢$\nu_n=\dfrac{nv}{2L}=\dfrac{n}{2L}\sqrt{\dfrac{T}{\mu}},\;n=1,2,3,\dots$Node at each end. All harmonics present; fundamental $\nu_1=v/2L$. The $n$th mode has $n$ antinodes.NCERT XI Ch 14 (§14.6.1)
Pipe open at both ends 🔉⇢$\nu_n=\dfrac{nv}{2L},\;n=1,2,3,\dots$Antinode at each open end. All harmonics present, like a string but with the boundary conditions inverted.NCERT XI Ch 14 (§14.6.2)
Pipe closed at one end 🔉⇢$\nu_n=\dfrac{(2n-1)v}{4L},\;n=1,2,3,\dots$Node at the closed end, antinode at the open end. Only odd harmonics ($v/4L,\,3v/4L,\,5v/4L,\dots$) are present.NCERT XI Ch 14 (§14.6.2)

Beats and the Doppler Effect

QuantityFormulaWhat it means / when to useSource
Beat frequency 🔉⇢$\nu_{beat}=|\nu_1-\nu_2|$Two waves of nearly equal frequency superpose to give an amplitude that waxes and wanes; the number of beats per second equals the frequency difference.NCERT XI Ch 14 (§14.7)
Doppler effect (general) 🔉⇢$\nu'=\nu\,\dfrac{v\pm v_o}{v\mp v_s}$$v$ = speed of sound, $v_o$ = observer speed, $v_s$ = source speed. Choose signs so that motion reducing the source-observer distance raises $\nu'$.NCERT XI Ch 14 (§14.8)
Doppler — source moving only 🔉⇢$\nu'=\nu\,\dfrac{v}{v\mp v_s}$Observer at rest. Upper sign (denominator $v-v_s$) for a source approaching, raising the pitch; lower sign for a receding source.NCERT XI Ch 14 (§14.8)
Doppler — observer moving only 🔉⇢$\nu'=\nu\,\dfrac{v\pm v_o}{v}$Source at rest. Upper sign (numerator $v+v_o$) for an observer approaching the source, raising the pitch; lower sign for a receding observer.NCERT XI Ch 14 (§14.8)

📜 Previous-Year Questions Authentic NTA · 60 questions

Every PYQ traces to an official NTA/JAB paper (year + shift + paper). Answer keys match the official key. OCR-extracted questions are flagged for SME verification.

JEE Main 2023 · Paper 1 · February 1 Shift 1 · Q10 (official key (printed in paper)) Answer: (C) 100 m/s

A steel wire with mass per unit length $7.0 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}$ is under tension of $70 \mathrm{~N}$. The speed of transverse waves in the wire will be:

  • (A) $10 \mathrm{~m} / \mathrm{s}$
  • (B) $50 \mathrm{~m} / \mathrm{s}$
  • (C) $100 \mathrm{~m} / \mathrm{s}$
  • (D) $200 \pi\mathrm{~m} / \mathrm{s}$
Solution + reasoning
The speed of a transverse wave on a wire is v = √(T/μ). Here T = 70 N and μ = 7.0×10⁻³ kg/m, so v = √(70 / 7.0×10⁻³) = √(1.0×10⁴) = 100 m/s. Hence option (C).
JEE Main 2023 · Paper 1 · January 24 Shift 1 · Q12 (official key (printed in paper)) Answer: (D) 0.5 m/s

A travelling wave is described by the equation $y(x,t) = [0.05\sin (8x - 4t)]$ m The velocity of the wave is : [all the quantities are in SI unit]

  • (A) $\mathrm{4~ms^{-1}}$
  • (B) $\mathrm{2~ms^{-1}}$
  • (C) $\mathrm{8~ms^{-1}}$
  • (D) $\mathrm{0.5~ms^{-1}}$
Solution + reasoning
For y = 0.05 sin(8x − 4t), compare with A sin(kx − ωt): k = 8 rad/m and ω = 4 rad/s. The wave speed is v = ω/k = 4/8 = 0.5 m/s. Hence option (D).
JEE Main 2023 · Paper 1 · January 29 Shift 1 · Q16 (official key (printed in paper)) Answer: (B) 55 Hz

A person observes two moving trains, 'A' reaching the station and 'B' leaving the station with equal speed of $30 \mathrm{~m} / \mathrm{s}$. If both trains emit sounds with frequency $300 \mathrm{~Hz}$, (Speed of sound: $330 \mathrm{~m} / \mathrm{s}$) approximate difference of frequencies heard by the person will be:

  • (A) 10 Hz
  • (B) 55 Hz
  • (C) 80 Hz
  • (D) 33 Hz
Solution + reasoning
The observer is at rest at the station. Train A approaches: f_A = f·v/(v − v_s) = 300·330/(330 − 30) = 330 Hz. Train B recedes: f_B = f·v/(v + v_s) = 300·330/(330 + 30) = 275 Hz. The difference heard is 330 − 275 = 55 Hz. Hence option (B).
IIT-JEE 2008 · Paper 2 · Q29 (official key) Answer: A

A vibrating string of certain length $\ell$ under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency $n$. Now when the tension of the string is slightly increased the number of beats reduces to 2 per second. Assuming the velocity of sound in air to be 340 m/s, the frequency $n$ of the tuning fork in Hz is

  • (A) 344
  • (B) 336
  • (C) 117.3
  • (D) 109.3
IIT-JEE 2009 · Paper 2 · Q44 (official key) Answer: A,D

A student performed the experiment to measure the speed of sound in air using resonance air-column method. Two resonances in the air-column were obtained by lowering the water level. The resonance with the shorter air-column is the first resonance and that with the longer air-column is the second resonance. Then,

  • (A) the intensity of the sound heard at the first resonance was more than that at the second resonance
  • (B) the prongs of the tuning fork were kept in a horizontal plane above the resonance tube
  • (C) the amplitude of vibration of the ends of the prongs is typically around 1 cm
  • (D) the length of the air-column at the first resonance was somewhat shorter than 1/4th of the wavelength of the sound in air
IIT-JEE 2009 · Paper 2 · Q57 (official key) Answer: 5

A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.

IIT-JEE 2010 · Paper 2 · Q40 (official key) Answer: B

A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 m s$^{-1}$, the mass of the string is

  • (A) 5 grams
  • (B) 10 grams
  • (C) 20 grams
  • (D) 40 grams
IIT-JEE 2010 · Paper 1 · Q79 (official key) Answer: 5

When two progressive waves $y_1=4\sin(2x-6t)$ and $y_2=3\sin\left(2x-6t-\dfrac{\pi}{2}\right)$ are superimposed, the amplitude of the resultant wave is

IIT-JEE 2010 · Paper 1 · Q84 (official key) Answer: 7

A stationary source is emitting sound at a fixed frequency $f_0$, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of $f_0$. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is 330 m s$^{-1}$.

IIT-JEE 2011 · Paper 1 · Q24 (official key) Answer: A

A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/hr towards a tall building which reflects the sound waves. The speed of sound in air is 320 m/s. The frequency of the siren heard by the car driver is

  • (A) 8.50 kHz
  • (B) 8.25 kHz
  • (C) 7.75 kHz
  • (D) 7.50 kHz
IIT-JEE 2012 · Paper 1 · Q15 (official key) Answer: BD

A person blows into open-end of a long pipe. As a result, a high-pressure pulse of air travels down the pipe. When this pulse reaches the other end of the pipe,

  • (A) a high-pressure pulse starts traveling up the pipe, if the other end of the pipe is open.
  • (B) a low-pressure pulse starts traveling up the pipe, if the other end of the pipe is open.
  • (C) a low-pressure pulse starts traveling up the pipe, if the other end of the pipe is closed.
  • (D) a high-pressure pulse starts traveling up the pipe, if the other end of the pipe is closed.
IIT-JEE 2012 · Paper 2 · Q7 (official key) Answer: B

A student is performing the experiment of Resonance Column. The diameter of the column tube is $4$ cm. The frequency of the tuning fork is $512$ Hz. The air temperature is $38^\circ$C in which the speed of sound is $336$ m/s. The zero of the meter scale coincides with the top end of the Resonance Column tube. When the first resonance occurs, the reading of the water level in the column is

  • (A) $14.0$ cm
  • (B) $15.2$ cm
  • (C) $16.4$ cm
  • (D) $17.6$ cm
JEE Advanced 2013 · Paper 1 · Q11 (official key) Answer: BC

A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation $y(x,\,t)=(0.01\ \text{m})\sin\left[(62.8\ \text{m}^{-1})x\right]\cos\left[(628\ \text{s}^{-1})t\right]$. Assuming $\pi=3.14$, the correct statement(s) is (are)

  • (A) The number of nodes is $5$.
  • (B) The length of the string is $0.25$ m.
  • (C) The maximum displacement of the midpoint of the string, from its equilibrium position is $0.01$ m.
  • (D) The fundamental frequency is $100$ Hz.
JEE Advanced 2013 · Paper 2 · Q8 (official key) Answer: AB

Two vehicles, each moving with speed $u$ on the same horizontal straight road, are approaching each other. Wind blows along the road with velocity $w$. One of these vehicles blows a whistle of frequency $f_1$. An observer in the other vehicle hears the frequency of the whistle to be $f_2$. The speed of sound in still air is $V$. The correct statement(s) is (are)

  • (A) If the wind blows from the observer to the source, $f_2>f_1$.
  • (B) If the wind blows from the source to the observer, $f_2>f_1$.
  • (C) If the wind blows from observer to the source, $f_2<f_1$.
  • (D) If the wind blows from the source to the observer, $f_2<f_1$.
JEE Advanced 2014 · Paper 1 · Q3 (official key) Answer: A, C, D

One end of a taut string of length $3\ \text{m}$ along the $x$ axis is fixed at $x = 0$. The speed of the waves in the string is $100\ \text{m s}^{-1}$. The other end of the string is vibrating in the $y$ direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary waves is(are)

  • (A) $y(t) = A\sin\dfrac{\pi x}{6}\cos\dfrac{50\pi t}{3}$
  • (B) $y(t) = A\sin\dfrac{\pi x}{3}\cos\dfrac{100\pi t}{3}$
  • (C) $y(t) = A\sin\dfrac{5\pi x}{6}\cos\dfrac{250\pi t}{3}$
  • (D) $y(t) = A\sin\dfrac{5\pi x}{2}\cos 250\pi t$
JEE Advanced 2014 · Paper 1 · Q6 (official key) Answer: D

A student is performing an experiment using a resonance column and a tuning fork of frequency $244\ \text{s}^{-1}$. He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is $(0.350 \pm 0.005)\ \text{m}$, the gas in the tube is (Useful information: $\sqrt{167RT} = 640\ \text{J}^{1/2}\ \text{mole}^{-1/2}$; $\sqrt{140RT} = 590\ \text{J}^{1/2}\ \text{mole}^{-1/2}$. The molar masses $M$ in grams are given in the options. Take the values of $\sqrt{\dfrac{10}{M}}$ for each gas as given there.)

  • (A) Neon $\left(M = 20,\ \sqrt{\dfrac{10}{20}} = \dfrac{7}{10}\right)$
  • (B) Nitrogen $\left(M = 28,\ \sqrt{\dfrac{10}{28}} = \dfrac{3}{5}\right)$
  • (C) Oxygen $\left(M = 32,\ \sqrt{\dfrac{10}{32}} = \dfrac{9}{16}\right)$
  • (D) Argon $\left(M = 36,\ \sqrt{\dfrac{10}{36}} = \dfrac{17}{32}\right)$
JEE Advanced 2015 · Paper 2 · Q4 (official key) Answer: 3

Four harmonic waves of equal frequencies and equal intensities $I_0$ have phase angles $0$, $\pi/3$, $2\pi/3$ and $\pi$. When they are superposed, the intensity of the resulting wave is $nI_0$. The value of $n$ is

Solution + reasoning
Add the four equal-amplitude waves as phasors of unit length at angles 0, π/3, 2π/3 and π. The x-components 1, ½, −½, −1 cancel to 0; the y-components 0, √3/2, √3/2, 0 sum to √3. The resultant amplitude is therefore √3·a, so the resultant intensity is (√3)² I₀ = 3 I₀. Hence n = 3.
JEE Advanced 2017 · Paper 1 · Q11 (official key) Answer: 6

A stationary source emits sound of frequency $f_0 = 492\ \text{Hz}$. The sound is reflected by a large car approaching the source with a speed of $2\ \text{m s}^{-1}$. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz? (Given that the speed of sound in air is $330\ \text{m s}^{-1}$ and the car reflects the sound at the frequency it has received).

JEE Advanced 2017 · Paper 1 · Q3 (official key) Answer: A, D

A block $M$ hangs vertically at the bottom end of a uniform rope of constant mass per unit length. The top end of the rope is attached to a fixed rigid support at $O$. A transverse wave pulse (Pulse 1) of wavelength $\lambda_0$ is produced at point $O$ on the rope. The pulse takes time $T_{OA}$ to reach point $A$ (the bottom end of the rope, where $M$ is attached). If the wave pulse of wavelength $\lambda_0$ is produced at point $A$ (Pulse 2) without disturbing the position of $M$ it takes time $T_{AO}$ to reach point $O$. Which of the following options is/are correct?

  • (A) The time $T_{AO} = T_{OA}$
  • (B) The velocities of the two pulses (Pulse 1 and Pulse 2) are the same at the midpoint of rope
  • (C) The wavelength of Pulse 1 becomes longer when it reaches point $A$
  • (D) The velocity of any pulse along the rope is independent of its frequency and wavelength
JEE Advanced 2018 · Paper 2 · Q6 (official key) Answer: A, B, C

In an experiment to measure the speed of sound by a resonating air column, a tuning fork of frequency $500\ \mathrm{Hz}$ is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonances are heard at air columns of length $50.7\ \mathrm{cm}$ and $83.9\ \mathrm{cm}$. Which of the following statements is (are) true?

  • (A) The speed of sound determined from this experiment is $332\ \mathrm{m\,s^{-1}}$
  • (B) The end correction in this experiment is $0.9\ \mathrm{cm}$
  • (C) The wavelength of the sound wave is $66.4\ \mathrm{cm}$
  • (D) The resonance at $50.7\ \mathrm{cm}$ corresponds to the fundamental harmonic
JEE Advanced 2018 · Paper 1 · Q8 (official key) Answer: 5.00

Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed $1.0\ \mathrm{m\,s^{-1}}$ and the man behind walks at a speed $2.0\ \mathrm{m\,s^{-1}}$. A third man is standing at a height $12\ \mathrm{m}$ above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency $1430\ \mathrm{Hz}$. The speed of sound in air is $330\ \mathrm{m\,s^{-1}}$. At the instant, when the moving men are $10\ \mathrm{m}$ apart, the stationary man is equidistant from them. The frequency of beats in $Hz$, heard by the stationary man at this instant, is __________.

JEE Advanced 2019 · Paper 1 · Q15 (official key) Answer: 8.13

A train S1, moving with a uniform velocity of 108 km/h, approaches another train S2 standing on a platform. An observer O moves with a uniform velocity of 36 km/h towards S2, as shown in figure. Both the trains are blowing whistles of same frequency 120 Hz. When O is 600 m away from S2 and distance between S1 and S2 is 800 m, the number of beats heard by O is ____. [Speed of the sound $= 330$ m/s] [Figure: S2 and S1 lie on a horizontal line 800 m apart, with S1 moving along that line towards S2 at 108 km/h; O lies 600 m from S2 on the line through S2 perpendicular to S1S2, and moves along that perpendicular towards S2 at 36 km/h.]

JEE Advanced 2019 · Paper 2 · Q15 (official key) Answer: B

A musical instrument is made using four different metal strings, 1, 2, 3 and 4 with mass per unit length $\mu$, $2\mu$, $3\mu$ and $4\mu$ respectively. The instrument is played by vibrating the strings by varying the free length in between the range $L_0$ and $2L_0$. It is found that in string-1 ($\mu$) at free length $L_0$ and tension $T_0$ the fundamental mode frequency is $f_0$. List-I gives the above four strings while list-II lists the magnitude of some quantity. List-I: (I) String-1 ($\mu$); (II) String-2 ($2\mu$); (III) String-3 ($3\mu$); (IV) String-4 ($4\mu$). List-II: (P) 1; (Q) $1/2$; (R) $1/\sqrt{2}$; (S) $1/\sqrt{3}$; (T) $3/16$; (U) $1/16$. If the tension in each string is $T_0$, the correct match for the highest fundamental frequency in $f_0$ units will be,

  • (A) I $\to$ P, II $\to$ Q, III $\to$ T, IV $\to$ S
  • (B) I $\to$ P, II $\to$ R, III $\to$ S, IV $\to$ Q
  • (C) I $\to$ Q, II $\to$ S, III $\to$ R, IV $\to$ P
  • (D) I $\to$ Q, II $\to$ P, III $\to$ R, IV $\to$ T
JEE Advanced 2019 · Paper 2 · Q16 (official key) Answer: B

A musical instrument is made using four different metal strings, 1, 2, 3 and 4 with mass per unit length $\mu$, $2\mu$, $3\mu$ and $4\mu$ respectively. The instrument is played by vibrating the strings by varying the free length in between the range $L_0$ and $2L_0$. It is found that in string-1 ($\mu$) at free length $L_0$ and tension $T_0$ the fundamental mode frequency is $f_0$. List-I gives the above four strings while list-II lists the magnitude of some quantity. List-I: (I) String-1 ($\mu$); (II) String-2 ($2\mu$); (III) String-3 ($3\mu$); (IV) String-4 ($4\mu$). List-II: (P) 1; (Q) $1/2$; (R) $1/\sqrt{2}$; (S) $1/\sqrt{3}$; (T) $3/16$; (U) $1/16$. The length of the strings 1, 2, 3 and 4 are kept fixed at $L_0$, $\frac{3L_0}{2}$, $\frac{5L_0}{4}$, and $\frac{7L_0}{4}$, respectively. Strings 1, 2, 3, and 4 are vibrated at their $1^{st}$, $3^{rd}$, $5^{th}$, and $14^{th}$ harmonics, respectively such that all the strings have same frequency. The correct match for the tension in the four strings in the units of $T_0$ will be,

  • (A) I $\to$ P, II $\to$ R, III $\to$ T, IV $\to$ U
  • (B) I $\to$ P, II $\to$ Q, III $\to$ T, IV $\to$ U
  • (C) I $\to$ P, II $\to$ Q, III $\to$ R, IV $\to$ T
  • (D) I $\to$ T, II $\to$ Q, III $\to$ R, IV $\to$ U
JEE Advanced 2020 · Paper 1 · Q17 (official key) Answer: 0.62

A stationary tuning fork is in resonance with an air column in a pipe. If the tuning fork is moved with a speed of $2\ \mathrm{ms^{-1}}$ in front of the open end of the pipe and parallel to it, the length of the pipe should be changed for the resonance to occur with the moving tuning fork. If the speed of sound in air is $320\ \mathrm{ms^{-1}}$, the smallest value of the percentage change required in the length of the pipe is ______.

JEE Advanced 2021 · Paper 2 · Q2 (official key) Answer: A, D

A source, approaching with speed $u$ towards the open end of a stationary pipe of length $L$, is emitting a sound of frequency $f_s$. The farther end of the pipe is closed. The speed of sound in air is $v$ and $f_0$ is the fundamental frequency of the pipe. For which of the following combination(s) of $u$ and $f_s$, will the sound reaching the pipe lead to a resonance?

  • (A) $u = 0.8v$ and $f_s = f_0$
  • (B) $u = 0.8v$ and $f_s = 2f_0$
  • (C) $u = 0.8v$ and $f_s = 0.5f_0$
  • (D) $u = 0.5v$ and $f_s = 1.5f_0$
JEE Advanced 2023 · Paper 2 · Q10 (official key) Answer: 5

A string of length $1$ m and mass $2 \times 10^{-5}$ kg is under tension $T$. When the string vibrates, two successive harmonics are found to occur at frequencies $750$ Hz and $1000$ Hz. The value of tension $T$ is _____ Newton.

JEE Advanced 2023 · Paper 2 · Q14 (official key) Answer: 648.00

$S_1$ and $S_2$ are two identical sound sources of frequency $656$ Hz. The source $S_1$ is located at $O$ and $S_2$ moves anti-clockwise with a uniform speed $4\sqrt{2}\ \text{m}\,\text{s}^{-1}$ on a circular path around $O$. There are three points $P$, $Q$ and $R$ on this path such that $P$ and $R$ are diametrically opposite while $Q$ is equidistant from them. A sound detector is placed at point $P$. The source $S_1$ can move along direction $OP$. [Given: The speed of sound in air is $324\ \text{m}\,\text{s}^{-1}$] When only $S_2$ is emitting sound and it is at $Q$, the frequency of sound measured by the detector in Hz is _________.

JEE Advanced 2023 · Paper 2 · Q15 (official key) Answer: 8.20

$S_1$ and $S_2$ are two identical sound sources of frequency $656$ Hz. The source $S_1$ is located at $O$ and $S_2$ moves anti-clockwise with a uniform speed $4\sqrt{2}\ \text{m}\,\text{s}^{-1}$ on a circular path around $O$. There are three points $P$, $Q$ and $R$ on this path such that $P$ and $R$ are diametrically opposite while $Q$ is equidistant from them. A sound detector is placed at point $P$. The source $S_1$ can move along direction $OP$. [Given: The speed of sound in air is $324\ \text{m}\,\text{s}^{-1}$] Consider both sources emitting sound. When $S_2$ is at $R$ and $S_1$ approaches the detector with a speed $4\ \text{m}\,\text{s}^{-1}$, the beat frequency measured by the detector is _______ Hz.

JEE Advanced 2024 · Paper 1 · Q11 (official key) Answer: 200

A source (S) of sound has frequency $240$ Hz. When the observer (O) and the source move towards each other at a speed $v$ with respect to the ground (as shown in Case 1 in the figure), the observer measures the frequency of the sound to be $288$ Hz. However, when the observer and the source move away from each other at the same speed $v$ with respect to the ground (as shown in Case 2 in the figure), the observer measures the frequency of sound to be $n$ Hz. The value of $n$ is _____.

JEE Advanced 2024 · Paper 1 · Q6 (official key) Answer: A, C, D

Two uniform strings of mass per unit length $\mu$ and $4\mu$, and length $L$ and $2L$, respectively, are joined at point O, and tied at two fixed ends P and Q, as shown in the figure. The strings are under a uniform tension $T$. If we define the frequency $\nu_0 = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$, which of the following statement(s) is(are) correct?

  • (A) With a node at O, the minimum frequency of vibration of the composite string is $\nu_0$.
  • (B) With an antinode at O, the minimum frequency of vibration of the composite string is $2\nu_0$.
  • (C) When the composite string vibrates at the minimum frequency with a node at O, it has 6 nodes, including the end nodes.
  • (D) No vibrational mode with an antinode at O is possible for the composite string.
JEE Advanced 2025 · Paper 2 · Q16 (official key) Answer: 32

An audio transmitter (T) and a receiver (R) are hung vertically from two identical massless strings of length $8$ m with their pivots well separated along the $X$ axis. They are pulled from the equilibrium position in opposite directions along the $X$ axis by a small angular amplitude $\theta_0 = \cos^{-1}(0.9)$ and released simultaneously. If the natural frequency of the transmitter is $660$ Hz and the speed of sound in air is $330$ m/s, the maximum variation in the frequency (in Hz) as measured by the receiver (Take the acceleration due to gravity $g = 10\ \text{m/s}^2$) is ___

JEE Advanced 2025 · Paper 1 · Q7 (official key) Answer: A, D

Consider a system of three connected strings, $S_1$, $S_2$ and $S_3$ with uniform linear mass densities $\mu$ kg/m, $4\mu$ kg/m and $16\mu$ kg/m, respectively. $S_1$ and $S_2$ are connected at the point $P$, whereas $S_2$ and $S_3$ are connected at the point $Q$, and the other end of $S_3$ is connected to a wall. A wave generator O is connected to the free end of $S_1$. The wave from the generator is represented by $y = y_0\cos(\omega t - kx)$ cm, where $y_0$, $\omega$ and $k$ are constants of appropriate dimensions. Which of the following statements is/are correct:

  • (A) When the wave reflects from $P$ for the first time, the reflected wave is represented by $y = \alpha_1 y_0 \cos(\omega t + kx + \pi)$ cm, where $\alpha_1$ is a positive constant.
  • (B) When the wave transmits through $P$ for the first time, the transmitted wave is represented by $y = \alpha_2 y_0 \cos(\omega t - kx)$ cm, where $\alpha_2$ is a positive constant.
  • (C) When the wave reflects from $Q$ for the first time, the reflected wave is represented by $y = \alpha_3 y_0 \cos(\omega t - kx + \pi)$ cm, where $\alpha_3$ is a positive constant.
  • (D) When the wave transmits through $Q$ for the first time, the transmitted wave is represented by $y = \alpha_4 y_0 \cos(\omega t - 4kx)$ cm, where $\alpha_4$ is a positive constant.
JEE Main 2019 · Paper 1 · January 11 Shift 1 · Q13 (published compilation) Answer: D⚑ verify

Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin(450 t – 9x) where distance and time are measured in SI units. The tension in the string is :

  • (A) 10 N
  • (B) 7.5 N
  • (C) 5 N
  • (D) 12.5 N
JEE Main 2020 · Paper 1 · January 7 Shift 2 · Q12 (published compilation) Answer: C⚑ verify

A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is $v_{0}$ = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to :

  • (A) 1 m/s
  • (B) ${1 \over 8}$ m/s
  • (C) ${1 \over 4}$ m/s
  • (D) ${1 \over 2}$ m/s
JEE Main 2020 · Paper 1 · September 5 Shift 1 · Q14 (published compilation) Answer: A⚑ verify

In a resonance tube experiment when the tube is filled with water up to a height of 17.0 cm from bottom, it resonates with a given tuning fork. When the water level is raised the next resonance with the same tuning fork occurs at a height of 24.5 cm. If the velocity of sound in air is 330 m/s, the tuning fork frequency is :

  • (A) 2200 Hz
  • (B) 3300 Hz
  • (C) 1100 Hz
  • (D) 550 Hz
JEE Main 2020 · Paper 1 · September 5 Shift 2 · Q16 (published compilation) Answer: B⚑ verify

A driver in a car, approaching a vertical wall notices that the frequency of his car horn, has changed from 440 Hz to 480 Hz, when it gets reflected from the wall. If the speed of sound in air is 345 m/s, then the speed of the car is :

  • (A) 36 km/hr
  • (B) 54 km/hr
  • (C) 24 km/hr
  • (D) 18 km/hr
JEE Main 2020 · Paper 1 · January 7 Shift 1 · Q17 (published compilation) Answer: A⚑ verify

Speed of a transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross-section 1.0 $mm^{2}$) is 90 $ms^{-1}$. If the Young's modulus of wire is 16 $\times 10^{11} Nm^{-2}$, the extension of wire over its natural length is :

  • (A) 0.03 mm
  • (B) 0.04 mm
  • (C) 0.02 mm
  • (D) 0.01 mm
JEE Main 2020 · Paper 1 · September 3 Shift 1 · Q2 (published compilation) Answer: A⚑ verify

A uniform thin rope of length 12 m and mass 6 kg hangs vertically from a rigid support and a block of mass 2 kg is attached to its free end. A transverse short wavetrain of wavelength 6 cm is produced at the lower end of the rope. What is the wavelength of the wavetrain (in cm) when it reaches the top of the rope ?

  • (A) 12
  • (B) 3
  • (C) 9
  • (D) 6
JEE Main 2020 · Paper 1 · January 8 Shift 2 · Q4 (published compilation) Answer: D⚑ verify

A transverse wave travels on a taut steel wire with a velocity of v when tension in it is 2.06 × $10^{4}$ N. When the tension is changed to T, the velocity changed to v/2. The value of T is close to :

  • (A) 30.5 × $10^{4}$ N
  • (B) 2.50 × $10^{4}$ N
  • (C) 10.2 × $10^{2}$ N
  • (D) 5.15 × $10^{3}$ N
JEE Main 2020 · Paper 1 · September 5 Shift 1 · Q4 (published compilation) Answer: B⚑ verify

Assume that the displacement(s) of air is proportional to the pressure difference ($\Delta$p) created by a sound wave. Displacement (s) further depends on the speed of sound (v), density of air ($\rho$) and the frequency (f). If $\Delta$p ~ 10 Pa, v ~ 300 m/s, $\rho$ ~ 1 kg/$m^{3}$ and f ~ 1000 Hz, then s will be of the order of (take the multiplicative constant to be 1) :

  • (A) 1 mm
  • (B) ${3 \over {100}}$ mm
  • (C) 10 mm
  • (D) ${1 \over {10}}$ mm
JEE Main 2022 · Paper 1 · June 24 Shift 1 · Q12 (published compilation) Answer: A⚑ verify

The equations of two waves are given by : $y_{1}$ = 5 sin 2$\pi$(x - vt) cm $y_{2}$ = 3 sin 2$\pi$(x $-$ vt + 1.5) cm These waves are simultaneously passing through a string. The amplitude of the resulting wave is :

  • (A) 2 cm
  • (B) 4 cm
  • (C) 5.8 cm
  • (D) 8 cm
JEE Main 2022 · Paper 1 · July 26 Shift 2 · Q13 (published compilation) Answer: A⚑ verify

A transverse wave is represented by $y=2 \sin (\omega t-k x)\, \mathrm{cm}$. The value of wavelength (in $\mathrm{cm}$) for which the wave velocity becomes equal to the maximum particle velocity, will be :

  • (A) 4$\pi$
  • (B) 2$\pi$
  • (C) $\pi$
  • (D) 2
JEE Main 2022 · Paper 1 · June 27 Shift 1 · Q16 (published compilation) Answer: A⚑ verify

An observer moves towards a stationary source of sound with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be :

  • (A) 20%
  • (B) 10%
  • (C) 5%
  • (D) 0%
JEE Main 2022 · Paper 1 · June 27 Shift 2 · Q17 (published compilation) Answer: C⚑ verify

If a wave gets refracted into a denser medium, then which of the following is true?

  • (A) wavelength, speed and frequency decreases.
  • (B) wavelength increases, sped decreases and frequency remains constant.
  • (C) wavelength and speed decreases but frequency remains constant.
  • (D) wavelength, speed and frequency increases.
JEE Main 2023 · Paper 1 · January 25 Shift 2 · Q22 (published compilation) Answer: 400⚑ verify

A train blowing a whistle of frequency 320 Hz approaches an observer standing on the platform at a speed of 66 m/s. The frequency observed by the observer will be (given speed of sound = 330 ms$^{-1}$) __________ Hz.

JEE Main 2023 · Paper 1 · January 29 Shift 1 · Q22 (published compilation) Answer: 120⚑ verify

Two simple harmonic waves having equal amplitudes of 8 cm and equal frequency of 10 Hz are moving along the same direction. The resultant amplitude is also 8 cm. The phase difference between the individual waves is _________ degree.

JEE Main 2023 · Paper 1 · January 25 Shift 1 · Q27 (published compilation) Answer: 18⚑ verify

The distance between two consecutive points with phase difference of 60$^\circ$ in a wave of frequency 500 Hz is 6.0 m. The velocity with which wave is travelling is __________ km/s

JEE Main 2023 · Paper 1 · April 8 Shift 1 · Q32 (published compilation) Answer: C⚑ verify

The engine of a train moving with speed $10 \mathrm{~ms}^{-1}$ towards a platform sounds a whistle at frequency $400 \mathrm{~Hz}$. The frequency heard by a passenger inside the train is: (neglect air speed. Speed of sound in air $=330 \mathrm{~ms}^{-1}$ )

  • (A) 200 Hz
  • (B) 412 Hz
  • (C) 400 Hz
  • (D) 388 Hz
JEE Main 2023 · Paper 1 · April 11 Shift 2 · Q36 (published compilation) Answer: C⚑ verify

A car P travelling at $20 \mathrm{~ms}^{-1}$ sounds its horn at a frequency of $400 \mathrm{~Hz}$. Another car $\mathrm{Q}$ is travelling behind the first car in the same direction with a velocity $40 \mathrm{~ms}^{-1}$. The frequency heard by the passenger of the car $\mathrm{Q}$ is approximately [Take, velocity of sound $=360 \mathrm{~ms}^{-1}$ ]

  • (A) 485 Hz
  • (B) 514 Hz
  • (C) 421 Hz
  • (D) 471 Hz
JEE Main 2023 · Paper 1 · April 13 Shift 2 · Q51 (published compilation) Answer: 500⚑ verify

In an experiment with sonometer when a mass of $180 \mathrm{~g}$ is attached to the string, it vibrates with fundamental frequency of $30 \mathrm{~Hz}$. When a mass $\mathrm{m}$ is attached, the string vibrates with fundamental frequency of $50 \mathrm{~Hz}$. The value of $\mathrm{m}$ is ___________ g.

JEE Main 2023 · Paper 1 · April 10 Shift 1 · Q54 (published compilation) Answer: 20⚑ verify

A transverse harmonic wave on a string is given by $y(x,t) = 5\sin (6t + 0.003x)$ where x and y are in cm and t in sec. The wave velocity is _______________ ms$^{-1}$.

JEE Main 2023 · Paper 1 · April 8 Shift 1 · Q54 (published compilation) Answer: 900⚑ verify

An organ pipe $40 \mathrm{~cm}$ long is open at both ends. The speed of sound in air is $360 \mathrm{~ms}^{-1}$. The frequency of the second harmonic is ___________ $\mathrm{Hz}$.

JEE Main 2023 · Paper 1 · April 15 Shift 1 · Q56 (published compilation) Answer: 90⚑ verify

The fundamental frequency of vibration of a string stretched between two rigid support is $50 \mathrm{~Hz}$. The mass of the string is $18 \mathrm{~g}$ and its linear mass density is $20 \mathrm{~g} / \mathrm{m}$. The speed of the transverse waves so produced in the string is ___________ $\mathrm{ms}^{-1}$

JEE Main 2023 · Paper 1 · April 11 Shift 1 · Q58 (published compilation) Answer: 1152⚑ verify

The equation of wave is given by $\mathrm{Y}=10^{-2} \sin 2 \pi(160 t-0.5 x+\pi / 4)$ where $x$ and $Y$ are in $\mathrm{m}$ and $\mathrm{t}$ in $s$. The speed of the wave is ________ $\mathrm{km} ~\mathrm{h}^{-1}$.

JEE Main 2023 · Paper 1 · April 6 Shift 1 · Q59 (published compilation) Answer: 420⚑ verify

A person driving car at a constant speed of $15 \mathrm{~m} / \mathrm{s}$ is approaching a vertical wall. The person notices a change of $40 \mathrm{~Hz}$ in the frequency of his car's horn upon reflection from the wall. The frequency of horn is _______________ $\mathrm{Hz}$. (Given: Speed of sound : $330 \mathrm{~m} / \mathrm{s}$ )

JEE Main 2023 · Paper 1 · April 8 Shift 2 · Q59 (published compilation) Answer: 60⚑ verify

A guitar string of length 90 cm vibrates with a fundamental frequency of 120 Hz. The length of the string producing a fundamental frequency of 180 Hz will be _________ cm.

JEE Main 2024 · Paper 1 · January 31 Shift 1 · Q9 (published compilation) Answer: A⚑ verify

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is $60 \mathrm{~cm}$, the length of the closed pipe will be:

  • (A) 15 cm
  • (B) 60 cm
  • (C) 45 cm
  • (D) 30 cm
JEE Main 2026 · Paper 1 · April 2 Shift 1 · Q35 (published compilation) Answer: D⚑ verify

The equation of a plane progressive wave is given by $y = 5 \cos \pi \left( 200t - \frac{x}{150} \right)$ where $x$ and $y$ are in cm and $t$ is in second. The velocity of the wave is ________ m/s.

  • (A) 120
  • (B) 150
  • (C) 200
  • (D) 300
JEE Main 2026 · Paper 1 · April 5 Shift 1 · Q48 (published compilation) Answer: 349 cm⚑ verify

A transverse wave on a string is described by $y=3 \sin (36 t+0.018 x+\pi / 4)$. where $x, y$ are in cm and $t$ in seconds. The least distance between the two successive crests in the wave is $\_\_\_\_$ cm . (Nearest integer) $(\pi=3.14)$

🎯 Question Bank 100 MCQs · graded

Distribution — advanced: 13 · easy: 29 · hard: 24 · medium: 34. Every question carries a source trace; each ends in an SME-verify solution.

Q1 In a transverse wave, the particles of the medium vibrate easy
Step solution + source
In a transverse wave the particles of the medium oscillate perpendicular to the direction in which the wave energy travels, as seen on a plucked string. Motion along the propagation direction describes a longitudinal wave instead, while circular or random particle motion is not how a simple transverse wave carries its disturbance forward through the medium. 🔉⇢

Source: NCERT-derived

Q2 Sound waves travelling through air are easy
Step solution + source
A gas has no shear elasticity, so it cannot provide a restoring force against sideways displacement and therefore cannot carry transverse waves. It can only be compressed and expanded, so a disturbance travels as alternate compressions and rarefactions along the propagation direction, which is exactly a longitudinal wave. Electromagnetic waves need no medium, so that option is wrong. 🔉⇢

Source: NCERT-derived

Q3 Which of the following waves requires a material medium to propagate? easy
Step solution + source
Sound is a mechanical wave produced by vibrating particles, so it must have a material medium such as air, water, or a solid to propagate and cannot travel through vacuum. Light, radio waves, and X-rays are all electromagnetic waves that travel freely through empty space, so only the sound wave truly requires a medium to reach a listener. 🔉⇢

Source: NCERT-derived

Q4 Ripples on the surface of water are best described as medium
Step solution + source
Particles on a water surface move in nearly circular orbits, combining an up-and-down (transverse) motion with a back-and-forth (longitudinal) motion as the wave passes. Because both components are present, a surface water wave is neither purely transverse nor purely longitudinal, and it is mechanical rather than electromagnetic, so the combined-motion option correctly describes it. 🔉⇢

Source: NCERT-derived

Q5 In a longitudinal sound wave, a region of compression is a region of easy
Step solution + source
A compression is the region where the medium's particles are pushed closest together, so both the density and the pressure rise to their maximum values there. Rarefactions, by contrast, are regions of spreading where pressure and density fall below normal. Maximum particle displacement occurs elsewhere, so the correct answer is maximum density and pressure at a compression. 🔉⇢

Source: NCERT-derived

Q6 Transverse mechanical waves cannot travel through the bulk of a fluid because a fluid medium
Step solution + source
Transverse waves rely on a restoring shear force that resists the sideways sliding of adjacent layers, which requires shear (rigidity) elasticity. Fluids such as gases and liquids have no shear modulus in bulk, so they cannot sustain a shear stress and cannot support bulk transverse waves. The gas is not too dense or massless; it simply lacks shear rigidity. 🔉⇢

Source: NCERT-derived

Q7 A wave transports easy
Step solution + source
As a wave passes, the particles of the medium merely oscillate about fixed equilibrium positions and return there; they are not carried along with the wave. What actually moves forward through the medium is energy, and with it information, transferred from particle to particle. Hence a wave transports energy without any net bulk transport of matter. 🔉⇢

Source: NCERT-derived

Q8 In a sound wave, a rarefaction is a region of medium
Step solution + source
A rarefaction is the region of a longitudinal wave where the particles are spread farthest apart, so both the pressure and the density there drop below their normal undisturbed values. This is the opposite of a compression, where particles crowd together and pressure and density peak. Therefore a rarefaction corresponds to reduced pressure and reduced density. 🔉⇢

Source: NCERT-derived

Q9 For a longitudinal sound wave, the pressure variation is hard
Step solution + source
Since pressure change $\propto -\partial y/\partial x$, it leads/lags the displacement by a quarter cycle. 🔉⇢

Source: NCERT-derived

Q10 Which of the following is a transverse wave? easy
Step solution + source
On a stretched string the particles vibrate up and down, perpendicular to the direction in which the wave travels along the string, which is the defining feature of a transverse wave. Sound in air, a compressional wave in a spring pushed lengthwise, and a pressure wave in a gas all involve motion along the propagation direction, so those are longitudinal. 🔉⇢

Source: NCERT-derived

Q11 A slinky is pushed and pulled repeatedly along its length. The wave produced is medium
Step solution + source
When a slinky is pushed and pulled along its length, regions of compression and rarefaction travel down the coil in the same direction the coils momentarily move. Since the particle motion is parallel to the direction of propagation, the wave is longitudinal. It is not transverse, not electromagnetic, and it travels rather than forming a fixed standing pattern. 🔉⇢

Source: NCERT-derived

Q12 Seismic S-waves (transverse) cannot pass through the Earth's liquid outer core because hard
Step solution + source
Seismic S-waves are transverse, so they need a medium with shear rigidity to provide the sideways restoring force. The Earth's outer core is liquid, and liquids cannot sustain a shear stress because they have no shear modulus. Consequently S-waves are blocked by the liquid core, whereas longitudinal P-waves, needing only compressibility, pass straight through it. 🔉⇢

Source: NCERT-derived

Q13 At the instant a particle of a transverse wave is at its maximum displacement, its medium
Step solution + source
A particle in a wave executes simple harmonic motion. At the extreme (maximum displacement) position it momentarily stops, so its velocity is zero, while the restoring force and hence the acceleration are largest there, directed back toward equilibrium. At the mean position the reverse holds, with maximum speed and zero acceleration. So at the extreme, velocity is zero and acceleration is maximum. 🔉⇢

Source: NCERT-derived

Q14 A symmetric pulse moves in the $+x$ direction. A particle on the string is momentarily moving upward. This particle lies on the advanced
Step solution + source
Transverse velocity $v_y=-v\,(\partial y/\partial x)$; the front of a right-moving hump has negative slope, giving upward velocity. 🔉⇢

Source: JEE-pattern

Q15 The angular wave number $k$ is related to the wavelength $\lambda$ by easy
Step solution + source
$k$ is the phase change per unit length, $k=2\pi/\lambda$. 🔉⇢

Source: NCERT-derived

Q16 The angular frequency $\omega$ in terms of the frequency $\nu$ is easy
Step solution + source
One cycle corresponds to a phase of $2\pi$, so $\omega=2\pi\nu$. 🔉⇢

Source: NCERT-derived

Q17 For the wave $y=0.02\sin(3x-60t)$ (SI units), the wavelength is medium
Step solution + source
$k=3\ \mathrm{rad/m}$, so $\lambda=2\pi/k=2\pi/3\approx2.09\ \mathrm{m}$. 🔉⇢

Source: JEE-pattern

Q18 For the wave $y=0.02\sin(3x-60t)$ (SI units), the wave speed is medium
Step solution + source
$v=\omega/k=60/3=20\ \mathrm{m/s}$. 🔉⇢

Source: JEE-pattern

Q19 For the wave $y=0.02\sin(3x-60t)$ (SI units), the frequency is medium
Step solution + source
$\nu=\omega/2\pi=60/2\pi\approx9.55\ \mathrm{Hz}$. 🔉⇢

Source: JEE-pattern

Q20 In $y=A\sin(kx-\omega t)$, the quantity $A$ represents the easy
Step solution + source
$A$ is the maximum displacement of a particle from its mean position. 🔉⇢

Source: NCERT-derived

Q21 Which equation represents a wave travelling in the $-x$ direction? medium
Step solution + source
When the $x$ and $t$ terms have the same sign, the wave moves in the $-x$ direction. 🔉⇢

Source: NCERT-derived

Q22 The maximum transverse speed of a particle in the wave $y=A\sin(kx-\omega t)$ is medium
Step solution + source
$v_y=-A\omega\cos(kx-\omega t)$, so the maximum particle speed is $A\omega$. 🔉⇢

Source: NCERT-derived

Q23 A transverse wave has amplitude $5\ \mathrm{mm}$ and frequency $100\ \mathrm{Hz}$. The maximum particle speed is about hard
Step solution + source
$v_{\max}=A\omega=A\,2\pi\nu=0.005\times2\pi\times100\approx3.14\ \mathrm{m/s}$. 🔉⇢

Source: JEE-pattern

Q24 The maximum particle acceleration for the wave $y=A\sin(kx-\omega t)$ is hard
Step solution + source
$a_y=-A\omega^{2}\sin(kx-\omega t)$, so $|a|_{\max}=A\omega^{2}$. 🔉⇢

Source: NCERT-derived

Q25 The phase difference between two points separated by a distance $\Delta x$ on a wave of wavelength $\lambda$ is medium
Step solution + source
$\Delta\phi=k\,\Delta x=(2\pi/\lambda)\,\Delta x$. 🔉⇢

Source: NCERT-derived

Q26 Two points on a wave of wavelength $2\ \mathrm{m}$ are $0.5\ \mathrm{m}$ apart. Their phase difference is hard
Step solution + source
$\Delta\phi=(2\pi/\lambda)\Delta x=(2\pi/2)(0.5)=\pi/2$. 🔉⇢

Source: JEE-pattern

Q27 The time period of the wave $y=0.01\sin(4\pi x-8\pi t)$ (SI units) is easy
Step solution + source
$\omega=8\pi$, so $T=2\pi/\omega=2\pi/8\pi=0.25\ \mathrm{s}$. 🔉⇢

Source: JEE-pattern

Q28 The SI unit of the angular wave number $k$ is easy
Step solution + source
$k=2\pi/\lambda$ has units of radians per metre. 🔉⇢

Source: NCERT-derived

Q29 For a progressive wave, the ratio of the maximum particle velocity to the wave velocity equals advanced
Step solution + source
$\dfrac{v_{\max}}{v}=\dfrac{A\omega}{\omega/k}=Ak=\dfrac{2\pi A}{\lambda}$. 🔉⇢

Source: JEE-pattern

Q30 The speed of a transverse wave on a stretched string is easy
Step solution + source
$T$ is the tension and $\mu$ the linear mass density; $v=\sqrt{T/\mu}$. 🔉⇢

Source: NCERT-derived

Q31 A string is under tension $100\ \mathrm{N}$ and has linear mass density $0.01\ \mathrm{kg/m}$. The wave speed is medium
Step solution + source
$v=\sqrt{T/\mu}=\sqrt{100/0.01}=\sqrt{10000}=100\ \mathrm{m/s}$. 🔉⇢

Source: JEE-pattern

Q32 If the tension in a string is made four times its original value, the wave speed becomes easy
Step solution + source
$v\propto\sqrt{T}$, so quadrupling $T$ multiplies $v$ by $\sqrt{4}=2$. 🔉⇢

Source: NCERT-derived

Q33 Laplace's correction assumes that sound propagation in a gas is easy
Step solution + source
Newton assumed sound propagation in air is isothermal, which gave a speed noticeably lower than experiment. Laplace corrected this by noting the compressions and rarefactions happen so rapidly that heat has no time to flow, making the process adiabatic rather than isothermal. Using the adiabatic relation raises the effective elasticity and gives a sound speed matching observation. 🔉⇢

Source: NCERT-derived

Q34 Newton's (isothermal) formula $v=\sqrt{P/\rho}$ for the speed of sound in air at STP gives about medium
Step solution + source
Newton's isothermal value is about $280\ \mathrm{m/s}$, roughly $15\%$ below the measured value. 🔉⇢

Source: NCERT-derived

Q35 Laplace's corrected formula $v=\sqrt{\gamma P/\rho}$ (with $\gamma=1.4$) gives the speed of sound in air at STP as about hard
Step solution + source
Including $\gamma$ raises Newton's value to about $332\ \mathrm{m/s}$, matching experiment. 🔉⇢

Source: NCERT-derived

Q36 The speed of sound in an ideal gas, $v=\sqrt{\gamma RT/M}$, is independent of the hard
Step solution + source
Since $P/\rho=RT/M$, pressure cancels; $v$ depends on $T$, $M$ and $\gamma$ but not on $P$ alone. 🔉⇢

Source: NCERT-derived

Q37 The speed of sound in a gas varies with the absolute temperature $T$ as medium
Step solution + source
From $v=\sqrt{\gamma RT/M}$, $v\propto\sqrt{T}$ at fixed $\gamma,M$. 🔉⇢

Source: NCERT-derived

Q38 The speed of sound in air is $332\ \mathrm{m/s}$ at $0^\circ\mathrm{C}$. At $27^\circ\mathrm{C}$ it is about hard
Step solution + source
$v\propto\sqrt{T}$: $332\sqrt{300/273}\approx348\ \mathrm{m/s}$. 🔉⇢

Source: JEE-pattern

Q39 At the same temperature and with the same $\gamma$, the speed of sound in hydrogen ($M=2$) compared with that in oxygen ($M=32$) is hard
Step solution + source
$v\propto1/\sqrt{M}$, so the ratio is $\sqrt{32/2}=\sqrt{16}=4$. 🔉⇢

Source: JEE-pattern

Q40 The speed of a wave on a string depends on medium
Step solution + source
The speed of a wave on a stretched string is fixed entirely by the medium, through the square root of tension divided by linear mass density. It does not depend on the source's frequency or amplitude; changing the frequency simply changes the wavelength so their product still equals this fixed speed. Hence only tension and mass per unit length determine it. 🔉⇢

Source: NCERT-derived

Q41 If the source frequency is increased while the wave speed on a string stays fixed, the wavelength easy
Step solution + source
Since $v=\nu\lambda$ with $v$ fixed, $\lambda\propto1/\nu$. 🔉⇢

Source: NCERT-derived

Q42 On a sonometer wire of fixed length, to double the wave speed the tension must be hard
Step solution + source
$v\propto\sqrt{T}$, so doubling $v$ requires $T\to4T$. 🔉⇢

Source: JEE-pattern

Q43 The speed of sound in air is essentially unaffected by a change in pressure at constant temperature because advanced
Step solution + source
At fixed $T$, $P/\rho=RT/M$ is constant, so $v=\sqrt{\gamma P/\rho}$ does not change with pressure. 🔉⇢

Source: NCERT-derived

Q44 A wire of density $7800\ \mathrm{kg/m^3}$ carries a longitudinal wave with $v=\sqrt{\sigma/\rho}$, where $\sigma$ is the stress. If $\sigma=3.12\times10^{8}\ \mathrm{Pa}$, the speed is advanced
Step solution + source
$v=\sqrt{\sigma/\rho}=\sqrt{3.12\times10^{8}/7800}=\sqrt{4\times10^{4}}=200\ \mathrm{m/s}$. 🔉⇢

Source: JEE-pattern

Q45 According to the principle of superposition, the resultant displacement at a point is easy
Step solution + source
For linear media, $y=y_1+y_2+\dots$ at each point and instant. 🔉⇢

Source: NCERT-derived

Q46 Two waves of equal amplitude $A$ arrive in phase at a point. The resultant amplitude is medium
Step solution + source
In-phase (constructive) superposition gives $A_{\text{res}}=A+A=2A$. 🔉⇢

Source: NCERT-derived

Q47 Two waves of amplitude $A$ arrive at a point exactly out of phase ($\Delta\phi=\pi$). The resultant amplitude is medium
Step solution + source
When two waves of equal amplitude meet exactly out of phase, with a phase difference of pi, every crest of one coincides with a trough of the other, so their displacements cancel everywhere at all times. This is complete destructive interference, and the resultant amplitude is zero. Had they been in phase instead, the amplitudes would have added to give 2A. 🔉⇢

Source: NCERT-derived

Q48 The resultant amplitude of two waves each of amplitude $A$ with a phase difference $\phi$ is hard
Step solution + source
$R=\sqrt{A^2+A^2+2A^2\cos\phi}=2A\left|\cos(\phi/2)\right|$. 🔉⇢

Source: NCERT-derived

Q49 Two coherent waves of amplitudes $3$ and $4$ units meet with a phase difference of $90^\circ$. The resultant amplitude is hard
Step solution + source
$R=\sqrt{3^2+4^2+2(3)(4)\cos90^\circ}=\sqrt{25}=5$. 🔉⇢

Source: JEE-pattern

Q50 For constructive interference, the path difference between two waves must be easy
Step solution + source
Constructive interference requires $\Delta=n\lambda$ (phase difference $2n\pi$). 🔉⇢

Source: NCERT-derived

Q51 For destructive interference, the path difference between two waves must be easy
Step solution + source
Destructive interference needs $\Delta=(n+\tfrac12)\lambda$ (phase difference an odd multiple of $\pi$). 🔉⇢

Source: NCERT-derived

Q52 Two waves of amplitude $A$ superpose with a phase difference of $120^\circ$. The resultant amplitude is hard
Step solution + source
$R=2A\cos(\phi/2)=2A\cos60^\circ=A$. 🔉⇢

Source: JEE-pattern

Q53 Two waves each of intensity $I$ interfere constructively. The maximum intensity is hard
Step solution + source
Intensity $\propto$ amplitude$^2$; amplitude doubles, so $I_{\max}=(2\sqrt{I})^2=4I$. 🔉⇢

Source: JEE-pattern

Q54 Two waves with amplitude ratio $2:1$ interfere. The ratio of maximum to minimum intensity is hard
Step solution + source
$\dfrac{I_{\max}}{I_{\min}}=\dfrac{(A_1+A_2)^2}{(A_1-A_2)^2}=\dfrac{(2+1)^2}{(2-1)^2}=9$. 🔉⇢

Source: JEE-pattern

Q55 Two sources that maintain a constant phase difference are said to be easy
Step solution + source
Two sources are called coherent when they maintain a steady, time-independent phase difference, which is what makes a stable interference pattern possible. If the phase relationship keeps changing randomly the sources are incoherent and no fixed pattern forms. Coherence is a property of the phase relationship, not of a wave being transverse or stationary, so coherent is correct. 🔉⇢

Source: NCERT-derived

Q56 The principle of superposition holds for these waves because the underlying wave equation is medium
Step solution + source
The principle of superposition works because the wave equation is linear: if two functions each satisfy it, their sum also satisfies it. This linearity lets the displacements of overlapping waves simply add algebraically at every point. If the equation were nonlinear or quadratic in displacement, such straightforward addition would fail, so linearity is exactly what guarantees superposition. 🔉⇢

Source: NCERT-derived

Q57 Two waves of equal amplitude $A$ superpose. For the resultant amplitude to equal $A$, the phase difference must be advanced
Step solution + source
$2A\cos(\phi/2)=A\Rightarrow\cos(\phi/2)=\tfrac12\Rightarrow\phi=120^\circ$. 🔉⇢

Source: JEE-pattern

Q58 Three waves of equal amplitude $A$ with phases $0$, $2\pi/3$ and $4\pi/3$ superpose. The resultant amplitude is advanced
Step solution + source
The three phasors are symmetrically placed $120^\circ$ apart and sum to zero. 🔉⇢

Source: JEE-pattern

Q59 When a wave is reflected from a rigid (fixed) boundary, it suffers a phase change of easy
Step solution + source
A fixed end forces zero displacement, so the reflected wave is inverted (phase change $\pi$). 🔉⇢

Source: NCERT-derived

Q60 When a wave is reflected from a free (open) end, the phase change is easy
Step solution + source
A wave reflecting from a free (open) boundary is not inverted, because the free end is able to move and no opposing force flips the pulse over. Therefore the reflected wave suffers no phase change, that is, a phase shift of zero. This contrasts with a rigid (fixed) end, where reflection inverts the wave and introduces a phase change of pi. 🔉⇢

Source: NCERT-derived

Q61 In a standing wave $y=2A\sin kx\cos\omega t$, nodes occur where medium
Step solution + source
Nodes are points of permanently zero amplitude, i.e. where $\sin kx=0$. 🔉⇢

Source: NCERT-derived

Q62 The distance between two consecutive nodes in a standing wave is easy
Step solution + source
In a standing wave the nodes are points that never move. Successive nodes are spaced one half wavelength apart, because the pattern repeats every half wavelength; the same spacing holds for adjacent antinodes. A full wavelength separates alternate nodes, not adjacent ones. Hence the distance between two consecutive nodes equals half the wavelength. 🔉⇢

Source: NCERT-derived

Q63 The distance between a node and the adjacent antinode is easy
Step solution + source
In a standing wave, nodes of zero amplitude and antinodes of maximum amplitude alternate, and each antinode sits midway between two nodes. The distance from a node to the neighbouring antinode is therefore one quarter of a wavelength, which is half the node-to-node spacing of half a wavelength. So the correct separation is a quarter wavelength. 🔉⇢

Source: NCERT-derived

Q64 For a string of length $L$ fixed at both ends carrying waves of speed $v$, the fundamental frequency is medium
Step solution + source
The fundamental has $\lambda=2L$, so $\nu_1=v/2L$. 🔉⇢

Source: NCERT-derived

Q65 A string fixed at both ends has length $1\ \mathrm{m}$ and wave speed $200\ \mathrm{m/s}$. Its fundamental frequency is medium
Step solution + source
$\nu_1=v/2L=200/(2\times1)=100\ \mathrm{Hz}$. 🔉⇢

Source: JEE-pattern

Q66 A pipe closed at one end supports medium
Step solution + source
A closed pipe has $\nu_n=(2n-1)v/4L$, i.e. odd multiples of the fundamental only. 🔉⇢

Source: NCERT-derived

Q67 The fundamental frequency of a pipe of length $L$ closed at one end is medium
Step solution + source
A closed pipe has a node at the closed end and antinode at the open end, giving $\lambda=4L$. 🔉⇢

Source: NCERT-derived

Q68 An open pipe and a closed pipe have the same length. The ratio of their fundamental frequencies (open : closed) is hard
Step solution + source
$\dfrac{v/2L}{v/4L}=2$, so open : closed $=2:1$. 🔉⇢

Source: JEE-pattern

Q69 A string of length $0.5\ \mathrm{m}$ has linear mass density $0.01\ \mathrm{kg/m}$ and tension $100\ \mathrm{N}$. Its fundamental frequency is hard
Step solution + source
$v=\sqrt{T/\mu}=100\ \mathrm{m/s}$; $\nu_1=v/2L=100/1=100\ \mathrm{Hz}$. 🔉⇢

Source: JEE-pattern

Q70 A pipe closed at one end has length $0.25\ \mathrm{m}$; the speed of sound is $330\ \mathrm{m/s}$. Its fundamental frequency is hard
Step solution + source
$\nu_1=v/4L=330/(4\times0.25)=330\ \mathrm{Hz}$. 🔉⇢

Source: JEE-pattern

Q71 A string fixed at both ends vibrates in its third harmonic. The number of antinodes is medium
Step solution + source
For a string fixed at both ends, the nth harmonic contains n loops, and each loop carries exactly one antinode at its centre. The third harmonic therefore has three loops and hence three antinodes, along with four nodes counting the two fixed ends. So the number of antinodes present in the third harmonic is three. 🔉⇢

Source: NCERT-derived

Q72 A pipe closed at one end has a fundamental frequency of $100\ \mathrm{Hz}$. Which of the following is NOT one of its resonant frequencies? advanced
Step solution + source
A closed pipe resonates only at odd multiples: $100,300,500,700,\dots$; $200$ is even and absent. 🔉⇢

Source: JEE-pattern

Q73 A sonometer wire (fixed tension) has fundamental frequency $200\ \mathrm{Hz}$. To make it produce $200\ \mathrm{Hz}$ as its second harmonic instead, its length must be advanced
Step solution + source
$\nu_1=v/2L=200\Rightarrow v=400L$. Second harmonic $=v/L'=200\Rightarrow L'=2L$. 🔉⇢

Source: JEE-pattern

Q74 Beats are produced by the superposition of two waves having easy
Step solution + source
Beats are the slow, periodic rise and fall of loudness heard when two notes of slightly different frequencies sound together. Their superposition produces an amplitude that swells and fades at a rate equal to the small frequency difference. Identical frequencies give no beats, and widely different frequencies produce beats too rapid to perceive, so slightly different frequencies are required. 🔉⇢

Source: NCERT-derived

Q75 The beat frequency of two superposed notes equals easy
Step solution + source
The beat frequency, meaning the number of loudness maxima heard per second, equals the magnitude of the difference between the two source frequencies. The sum or the average of the frequencies determines the pitch of the combined tone, not the beat rate, and the product of the frequencies has no physical meaning here. Hence the correct expression is the absolute frequency difference. 🔉⇢

Source: NCERT-derived

Q76 Two tuning forks of frequencies $256\ \mathrm{Hz}$ and $260\ \mathrm{Hz}$ are sounded together. The beat frequency is medium
Step solution + source
$|260-256|=4$ beats per second. 🔉⇢

Source: JEE-pattern

Q77 Two sources produce $5$ beats per second. If one source has frequency $200\ \mathrm{Hz}$, the other could be medium
Step solution + source
$|\nu-200|=5\Rightarrow\nu=205$ (or $195$); $205\ \mathrm{Hz}$ is listed. 🔉⇢

Source: JEE-pattern

Q78 Fork A ($512\ \mathrm{Hz}$) gives $6$ beats/s with fork B. On loading A with wax, the beats rise to $8$/s. The frequency of B is medium
Step solution + source
Wax lowers A's frequency; beats increased, so B lies below A: $B=512-6=506\ \mathrm{Hz}$. 🔉⇢

Source: JEE-pattern

Q79 Fork A ($340\ \mathrm{Hz}$) gives $5$ beats/s with fork B. Filing B (raising its frequency) reduces the beats. The frequency of B is hard
Step solution + source
Beats fall when B rises, so B was below A: $B=340-5=335\ \mathrm{Hz}$. 🔉⇢

Source: JEE-pattern

Q80 During one beat, the loudness of the resultant sound passes through easy
Step solution + source
One complete beat is a single full cycle of the loudness variation, comprising one maximum (loud) and one minimum (soft) as the two waves drift in and out of phase. The intensity swells to a peak and fades to a minimum exactly once per beat period. Therefore a single beat contains one maximum and one minimum. 🔉⇢

Source: NCERT-derived

Q81 Two waves $y_1=A\sin(2\pi\cdot256\,t)$ and $y_2=A\sin(2\pi\cdot254\,t)$ are superposed. The number of beats per second is medium
Step solution + source
Beat frequency $=|256-254|=2\ \mathrm{Hz}$. 🔉⇢

Source: JEE-pattern

Q82 Two tuning forks of $300\ \mathrm{Hz}$ and $304\ \mathrm{Hz}$ are sounded together. The beat period is hard
Step solution + source
Beat frequency $=4\ \mathrm{Hz}$, so beat period $=1/4=0.25\ \mathrm{s}$. 🔉⇢

Source: JEE-pattern

Q83 Beats are commonly used to medium
Step solution + source
Beats provide a sensitive way to tune a musical instrument to a standard frequency such as that of a tuning fork. As the instrument's note approaches the standard, the beat frequency slows; when the two frequencies match exactly the beats vanish entirely, signalling perfect tuning. This zero-beat method is far more precise than judging pitch by ear alone. 🔉⇢

Source: NCERT-derived

Q84 A note of $480\ \mathrm{Hz}$ gives $8$ beats/s with a source of frequency $\nu$. When that source is slowed, the beats rise to $12$/s. The value of $\nu$ was hard
Step solution + source
Slowing increases the beats, so $\nu$ lies below $480$: $\nu=480-8=472\ \mathrm{Hz}$. 🔉⇢

Source: JEE-pattern

Q85 A fork of unknown frequency gives $4$ beats/s with a $384\ \mathrm{Hz}$ fork. On loading the unknown fork with wax the beats drop to $2$/s. The unknown frequency is advanced
Step solution + source
Loading lowers the unknown's frequency; beats fell, so it was above $384$: $384+4=388\ \mathrm{Hz}$. 🔉⇢

Source: JEE-pattern

Q86 Two identical wires have frequencies $\nu\propto\sqrt{T}$. If tension $100\ \mathrm{N}$ gives $250\ \mathrm{Hz}$ and a second wire gives $4$ beats/s (higher frequency), its tension is about advanced
Step solution + source
$\nu_2=254\ \mathrm{Hz}$; $T_2=100(254/250)^2\approx103\ \mathrm{N}$. 🔉⇢

Source: JEE-pattern

Q87 The Doppler effect is the apparent change, due to relative motion, in the observed easy
Step solution + source
The Doppler effect is the apparent change in the frequency, and hence the pitch or colour, of a wave caused by relative motion between the source and the observer. When they approach the perceived frequency rises, and when they recede it falls. The wave's speed in the medium and its amplitude are not what the effect describes, so frequency is correct. 🔉⇢

Source: NCERT-derived

Q88 When a sound source approaches a stationary observer, the observed frequency easy
Step solution + source
As a source moves toward a stationary observer, each successive wavefront is emitted from a slightly closer position, so the crests bunch together and the wavelength shortens. Since the wave speed in the medium is unchanged, a shorter wavelength means a higher observed frequency, so the pitch increases. When the source recedes the opposite happens and the pitch drops. 🔉⇢

Source: NCERT-derived

Q89 A $500\ \mathrm{Hz}$ source moves toward a stationary observer at $34\ \mathrm{m/s}$ ($v=340\ \mathrm{m/s}$). The observed frequency is about medium
Step solution + source
$\nu'=\nu\dfrac{v}{v-v_s}=500\dfrac{340}{306}\approx556\ \mathrm{Hz}$. 🔉⇢

Source: JEE-pattern

Q90 A $500\ \mathrm{Hz}$ source recedes from a stationary observer at $34\ \mathrm{m/s}$ ($v=340\ \mathrm{m/s}$). The observed frequency is about medium
Step solution + source
$\nu'=\nu\dfrac{v}{v+v_s}=500\dfrac{340}{374}\approx455\ \mathrm{Hz}$. 🔉⇢

Source: JEE-pattern

Q91 A stationary source emits $340\ \mathrm{Hz}$; an observer moves toward it at $34\ \mathrm{m/s}$ ($v=340\ \mathrm{m/s}$). The observed frequency is medium
Step solution + source
$\nu'=\nu\dfrac{v+v_o}{v}=340\dfrac{374}{340}=374\ \mathrm{Hz}$. 🔉⇢

Source: JEE-pattern

Q92 An observer moves away from a stationary $340\ \mathrm{Hz}$ source at $34\ \mathrm{m/s}$ ($v=340\ \mathrm{m/s}$). The observed frequency is medium
Step solution + source
$\nu'=\nu\dfrac{v-v_o}{v}=340\dfrac{306}{340}=306\ \mathrm{Hz}$. 🔉⇢

Source: JEE-pattern

Q93 A $500\ \mathrm{Hz}$ source and an observer move directly toward each other, each at $34\ \mathrm{m/s}$ ($v=340\ \mathrm{m/s}$). The observed frequency is about hard
Step solution + source
$\nu'=\nu\dfrac{v+v_o}{v-v_s}=500\dfrac{374}{306}\approx611\ \mathrm{Hz}$. 🔉⇢

Source: JEE-pattern

Q94 A source and an observer move in the same direction with equal speeds along a line, the source behind the observer. The observed frequency is hard
Step solution + source
$\nu'=\nu\dfrac{v-v_o}{v-v_s}=\nu$ when $v_o=v_s$. 🔉⇢

Source: JEE-pattern

Q95 The Doppler formula for sound differs depending on whether the source or the observer moves because easy
Step solution + source
For sound the medium, air, defines a preferred reference frame, and the wave always travels at a fixed speed relative to that medium. Because of this, moving the source and moving the observer are physically different situations and give different frequency shifts. Light has no such medium, so its Doppler shift depends only on relative velocity, unlike sound. 🔉⇢

Source: NCERT-derived

Q96 A train sounds a $400\ \mathrm{Hz}$ whistle while approaching a platform at $20\ \mathrm{m/s}$ ($v=340\ \mathrm{m/s}$). The frequency heard on the platform is hard
Step solution + source
$\nu'=\nu\dfrac{v}{v-v_s}=400\dfrac{340}{320}=425\ \mathrm{Hz}$. 🔉⇢

Source: JEE-pattern

Q97 When only the source moves, the quantity that physically changes in the medium ahead of it is the medium
Step solution + source
A source moving through the medium emits each successive crest from a new position, so in the forward direction the crests are packed closer and the wavelength shrinks, while behind it they lengthen. The wave speed is set by the medium and stays fixed, so it is the wavelength that changes, which in turn alters the observed frequency. Amplitude and the source's own period are unaffected. 🔉⇢

Source: NCERT-derived

Q98 A car sounds a $660\ \mathrm{Hz}$ horn while moving at $30\ \mathrm{m/s}$ toward a wall ($v=330\ \mathrm{m/s}$). The frequency of the echo heard by the driver is advanced
Step solution + source
Wall receives $660\dfrac{330}{300}=726\ \mathrm{Hz}$; the approaching driver then hears $726\dfrac{360}{330}=792\ \mathrm{Hz}$. 🔉⇢

Source: JEE-pattern

Q99 A source approaches a stationary observer so that the observed frequency is $1.25$ times the true frequency ($v=340\ \mathrm{m/s}$). The source speed is advanced
Step solution + source
$\dfrac{v}{v-v_s}=1.25\Rightarrow v-v_s=272\Rightarrow v_s=68\ \mathrm{m/s}$. 🔉⇢

Source: JEE-pattern

Q100 For sound, the moving-observer and moving-source formulas are not identical because advanced
Step solution + source
The two cases differ because the underlying physical mechanisms differ. A moving observer runs into or away from the wavefronts, altering the relative speed at which crests are encountered, while the wavelength stays fixed. A moving source instead changes the emitted wavelength while the wave speed in the medium is unchanged. These distinct mechanisms give different classical formulas. 🔉⇢

Source: NCERT-derived

⏱️ Mock Test 30 Q · 60 min · +4 / −1 per question

Rules: Single-correct MCQ. 30 questions in 60 minutes. +4 for a correct answer, -1 for a wrong one, 0 if left blank. Attempt only when confident. Covers all seven subtopics of Waves (NCERT Class XI, Ch 14).

🎬 Video Lectures clip-indexed · NPTEL / IIT / MIT

MIT OCW / Walter Lewin, 3Blue1Brown, Khan, Veritasium. Each video is broken into clips — click a clip to play just that segment.

तनी हुई डोरी में अनुप्रस्थ तरंग की चाल। Bsc 2nd year 🔉⇢
Bsc study with amit

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Bsc study with amit); found via yt-dlp search 'डोरी पर तरंग की चाल speed of wave on string Hindi physics class 11', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:31Speed of a wave — segment 1जय हिंद जय भारत जय जहार आपका बीएससी स्टडी पर स्वागत है। आज का टॉपिक है बीएससी सेकंड ईयर चैप्टर तरंग का। एक समान किसी डोरी में अनुप्रस्थ तरंग की चाल। सबसे पहले आपको लिखना है किसी डोरी में अनुप्रस्थ तरंग के संचरण के लिए…speed of a wave
  • 2:31–5:01Speed of a wave — segment 2से। अब हम मान लेते हैं कि जो मूल बिंदु है वो उससे उससे x दूरी पर x दूरी पर एक अल्पांश है। एक अल्पांश पी व q है। पी व q है जिसकी मोटाई क्या है? डx या डेल्टाx इसको कई जगह dx भी ले लिया गया है। लेकिन आप दोनों में से ले…speed of a wave
  • 5:01–7:33Speed of a wave — segment 3पर एक अल्पांश पी पीq है जिसकी लंबाई डेल्टा एक्स है तथा अल्पांश pq की विस्थापन स्थिति p' q' है। माना p' तथा q पर खींची गई स्पर्श रेखाएं x अक्ष के साथ झुकाव क्रमशः तथा - डेल्टा है। जहां फाई क्या है? अति अल्प कोण है। उसके…speed of a wave
  • 7:33–10:04Speed of a wave — segment 4दिशा में है। और समान दिशा में क्या है? y अक्ष के समान दिशा में t sin - डेल्टा तो इसी ऋण विपरीत दिशा है तो माइनस अब लिखना हैकि अति अल्प है तो जो sin - डेल्टा फाई है इसको हम लगभग बराबर मान सकते हैं - डेल्टा फाई और उसको…speed of a wave
  • 10:04–12:34Speed of a wave — segment 5यहां से sin = मतलब दोनों ये भी इसके बराबर होगा। भी इसके बराबर होगा। tan इस पे ही बराबर होगा। तो का मान भी लिख सकता है। उतना ही। उसके बाद हम लोग इसके ऊपर नीचे डेल डेल्टाx और डेल्टा x का गुणा कर देते हैं। एक ही मतलब है।…speed of a wave
  • 12:34–13:14Speed of a wave — segment 6डेल्टाx² इसको आपको याद करना पड़ेगा। अब जो है समीकरण पांच है और समीकरण छ है इसकी तुलना करेंगे। दोनों की तुलना करेंगे तो जो v है और t/m है उसका मान मतलब इक्वल इक्वल हो जाएगा। अब v का मान निकालेंगे तो अंडर रूट e अपॉन एम…speed of a wave
Transverse & Longitudinal Waves | Waves | Physics | FuseSchool 🔉⇢
FuseSchool - Global Education

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📑 Clips (2)
  • 0:00–2:31Transverse and longitudinal waves — segment 1[Music] waves transfer energy from one place to another you should already know how to describe them in terms of frequency wavelength and amplitude which we looked at in another video in this video we're going to look…transverse and longitudinal waves
  • 2:31–2:58Transverse and longitudinal waves — segment 2visible light are also transverse waves so now you should be able to describe the differences between transverse waves and longitudinal waves remember it is the energy and not the slinky itself that travelstransverse and longitudinal waves
Transverse and Longitudinal Waves 🔉⇢
The Organic Chemistry Tutor

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📑 Clips (3)
  • 0:01–2:32Transverse and longitudinal waves — segment 1In this video, we're going to talk about waves. Specifically, transverse waves and longitudinal waves. But, let's focus on waves first. Waves can transfer energy and information from one place to another. It's basically…transverse and longitudinal waves
  • 2:32–5:02Transverse and longitudinal waves — segment 2water waves. Let's say if you're on the beach and you see those waves on the ocean, those are transverse waves. EM waves are also transverse. So, these are electromagnetic waves such as light waves, radio waves,…transverse and longitudinal waves
  • 5:02–5:08Transverse and longitudinal waves — segment 3have oscillations that are perpendicular to the direction of the wavetransverse and longitudinal waves
Difference between Transverse and Longitudinal Waves 🔉⇢
Najam Academy

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📑 Clips (2)
  • 0:00–2:32Transverse and longitudinal waves — segment 1difference between transfers and longitudinal waves well when the direction of oscillation of particles is at y-axis and the direction of where motion is it x-axis such wave is called transverse wave are we say that…transverse and longitudinal waves
  • 2:32–3:06Transverse and longitudinal waves — segment 2membrane of a drum it starts vibrating and sound waves are produced secondly we need to generate valves in order to transfer energy from one region to another for instance we generate wells to transfer our sound from…transverse and longitudinal waves
Breaking Down Equation Of A Wave | Asin(kx-wt) - Why this wave travels towards the +ve x-axis ? | 🔉⇢
Ami

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📑 Clips (3)
  • 0:00–2:31Progressive wave equation — segment 1okay hello guys in this video i'll you know explain you the wave equation without uh involving any of the fancy mathematics in your classes you know they might usually tell why this is true uh the reason they give the…progressive wave equation
  • 2:31–5:02Progressive wave equation — segment 2understand that by having a velocity v any point on this wave did not change the space it just got translated by t distance by t distance towards zero i mean vt distance towards the right so let's say i want to write…progressive wave equation
  • 5:02–5:11Progressive wave equation — segment 3sine kx okay so that's it that's why you come at thisprogressive wave equation
The Wave Function: Deriving y = A sin(kx − ωt) 🔉⇢
The Science Cube

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📑 Clips (5)
  • 0:00–2:30Progressive wave equation — segment 1Imagine a string in motion. At first glance, it [music] looks like a single fluid object dancing through space. But look closer. A wave is actually a coordinated effort of thousands of individual particles, each…progressive wave equation
  • 2:30–5:00Progressive wave equation — segment 2oscillation. Now let us define the specific terms in our equation. First we have the amplitude denoted as a and we know that amplitude of a wave is the magnitude of the maximum displacement of the particle from its…progressive wave equation
  • 5:00–7:31Progressive wave equation — segment 3time t=0, this equation simplifies to give us the shape of the wave as y of x and 0 equals a sin of kx. And since we just said that at distance lambda the wave would start repeating itself that the displacement y must…progressive wave equation
  • 7:31–10:06Progressive wave equation — segment 4made by a string particle as the wave moves through it. Well, we also know that t is related to the angular frequency omega as omega equals 2&lt;unk&gt;i /t. or combining these fals omega over 2 pi. Now earlier we spoke…progressive wave equation
  • 10:06–10:27Progressive wave equation — segment 5p&lt;unk&gt; over 5. Here the subtraction delays the function. This delay shifts the entire wave pattern to the right or The rule is consistent. A positive fi shifts the wave towards negativex and the negative fi shifts…progressive wave equation
Displacement Relation on Progressive Wave & Its Velocity | Class 11 Physics Ch 14 | CBSE 2025-26 🔉⇢
Magnet Brains

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📑 Clips (6)
  • 0:00–2:30Progressive wave equation — segment 1Hello everyone welcome to magnet brains my self Pratik Karna welcomes you all your education till youtube1 you will get it in detailed video format and if you are a Hindi medium student then you know what to do you also…progressive wave equation
  • 2:30–5:00Progressive wave equation — segment 2amplitude amplitude this is the word if you break it down you will understand it is made by combining two things amplitude means maximum and magnitude is value there for maximum value will be called as amplitude okay…progressive wave equation
  • 5:00–7:30Progressive wave equation — segment 3features of any wave motion so wave motion is a sort of disturbance which travels through a medium that's right sir a wave is basically a disturbance there is a pattern of disturbance okay after that a material medium…progressive wave equation
  • 7:30–10:00Progressive wave equation — segment 4the particles of the medium only vibrate simply harmonically about their mean position and they do not leave their position and they do not move with the disturbance transport does not happen they do not move along with…progressive wave equation
  • 10:00–12:31Progressive wave equation — segment 5general formula this is the general formula for velocity in os this is the general formula for velocity and osin okay now what happens here as you reach the mean position then what will be the value of displacement if…progressive wave equation
  • 12:32–15:03Progressive wave equation — segment 6what will happen here sir, the differentiable of time here, after that what happens here, x means the differentiable of linear displacement will become omega / k, is this concept completely clear in your mind, velocity…progressive wave equation
Derivation of speed of wave on string 🔉⇢
UNSW Physics

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📑 Clips (2)
  • 0:00–2:31Speed of a wave — segment 1we're now going to derive a formula for the speed of wave on the string here's the pulse that we're going to be considering is traveling from the left to the right along the screen here now imagine yourself in the…speed of a wave
  • 2:31–4:51Speed of a wave — segment 2approximation if Theta is small then sin Theta is approximately Theta so we can replace this sin Theta here with Theta that's what we've done here now what we need to do is work out this Mass here so the mass of a piece…speed of a wave
Class 11 Physics | Waves Motion | #14 Wave Speed of Transverse Wave on a Stretched String|JEE & NEET 🔉⇢
Physics Galaxy

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📑 Clips (0)

Full lecture — no clip index.

Speed of Sound in Air Physics Class 11 - Newton's Formula & Laplace Correction, Term 2 Exams 🔉⇢
Mandeep Education Academy

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📑 Clips (6)
  • 0:00–2:32Speed of a wave — segment 1Hello hello friends welcome back to Mandir Prashikshan's tube channel children today we are doing a very important derivation from class 11th physics time two chapter wells and that derivation is or you can find the…speed of a wave
  • 2:32–5:04Speed of a wave — segment 2talk about very small change, then for very small change, write dp instead of half litre, also write TV instead of delta, divided by Id, so this is rather models, secondly density, that is the density of medium, so the…speed of a wave
  • 5:04–7:35Speed of a wave — segment 3heat is produced and when the gas expands, the gas cools down, so Newton said that when this side profits like this, then the heat produced by compression will definitely be produced, so as soon as the particles move…speed of a wave
  • 7:35–10:09Speed of a wave — segment 4so he said put this middle space here you should get the speed of sound in air and you will get the value if Newton was correct it is not that Newton was finding the value of Sawan he is finding the value of Sawan speed…speed of a wave
  • 10:09–12:39Speed of a wave — segment 5reactions are so fast that there will be no time for this heat to extend. In the time you are saying that heat will come out from compression and go into transition, in that time many more compressions will take place,…speed of a wave
  • 12:39–14:59Speed of a wave — segment 6power is set, then this will become and quest to power gamma into deep equal to zero, write well, Veer Vama, West Power Mama was persuaded, it went there and became zero, what should be saved here, here Bachchan village…speed of a wave
THE PRINCIPLE OF SUPERPOSITION OF WAVES_PART 01 🔉⇢
7activestudio

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📑 Clips (2)
  • 0:07–2:42Superposition and interference — segment 1the principle of superposition of waves according to the principle overlapping waves at algebraically to produce a resultant wave or net wave in other words when any number of waves met simultaneously at a point in a…superposition and interference
  • 2:42–4:08Superposition and interference — segment 2YN equal to fub1 into x - VT + FS2 into x - VT + so on plus FN into x - VT = to Sigma I = 1 to n of fi into xus VT to illustrate the principle let us consider two ways of same angular frequency W = to of 2 piun by T…superposition and interference
Wave Superposition Introduction 🔉⇢
Flipping Physics

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📑 Clips (2)
  • 0:00–2:31Superposition and interference — segment 1Good morning. All right, everybody. Let's watch what happens when these two carts head straight toward one another. Flipping physics What just happened? Yeah, that's not possible. I mean I mean the two carts cannot just…superposition and interference
  • 2:31–5:05Superposition and interference — segment 2other instead. Okay, Bobby. How about this as an example then? Does this help show that the two wave pulses do not bounce off of one another, but instead pass through one another? Yeah, I can definitely see the wave…superposition and interference
Standing Waves on a String, Fundamental Frequency, Harmonics, Overtones, Nodes, Antinodes, Physics 🔉⇢
The Organic Chemistry Tutor

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📑 Clips (6)
  • 0:00–2:30Standing waves and normal modes — segment 1today we're going to talk about standing waves so what exactly are standard waves and how can they be created let's say if we have a string that is attached to two fixed ends and if we apply a tension force and if we…standing waves and normal modes
  • 2:30–5:01Standing waves and normal modes — segment 2so for the second one l is equal to the wavelength the subscript corresponds to the n value it tells you that there's two standard in that shape so for the next one where we have three standard waves l is going to equal…standing waves and normal modes
  • 5:01–7:36Standing waves and normal modes — segment 3so the frequency is proportional to the n value the frequency fn is basically equal to n times f1 we'll talk about how to derive this equation shortly now let's talk about the wavelength as you can see the wavelength is…standing waves and normal modes
  • 7:36–10:08Standing waves and normal modes — segment 4so now we have the equation for the wavelength we said the wavelength is equal to 2 times l so then 1 over lambda n is the reciprocal of this fraction so that's a n over 2l so let's replace this expression with n over…standing waves and normal modes
  • 10:08–12:38Standing waves and normal modes — segment 5so if you want to find the first harmonic which is the fundamental frequency n is one if you're looking for the second harmonic which is the first overtone ns2 the third harmonic or the second overtone and a string now…standing waves and normal modes
  • 12:38–15:10Standing waves and normal modes — segment 6an n value of two where we have two standard waves notice that there's three nodes one two three and two antinodes at the anti-node constructive interference occurs since we have the amplitude is at a maximum value at…standing waves and normal modes
Standing Wave Demo: Organ Pipes 🔉⇢
Physics Demos

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📑 Clips (2)
  • 0:00–2:31Standing waves and normal modes — segment 1okay this is a demonstration of Organ Pipe modes of vibration uh you may already be aware that a longer Organ Pipe this is open on one end and also open on the other end and this is where the air is introduced then…standing waves and normal modes
  • 2:31–3:48Standing waves and normal modes — segment 2sound the uh as I mentioned before all these all these pipes that I've shown you so far are open on both ends if you ask what happens when you close one end and I'm just going to put my hand across it to close this end…standing waves and normal modes
Beat Frequency Physics Problems 🔉⇢
The Organic Chemistry Tutor

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  • 0:01–2:35Beats — segment 1in this video we're going to talk about how to solve physics problems associated with the beak frequency whenever two sound waves with a frequency that's very close to each other whenever they're close together they can…beats
  • 2:35–3:40Beats — segment 2so let's focus on the 415 frequency if a beat frequency of 5 hertz is produced with this sound and a tuning fork that means the frequency of the 24 could be two numbers it could be 415 minus 5 which is 410 or it could…beats
TO FIND FREQUENCY OF TUNING FORK AFTER SOUNDED WITH ANOTHER TUNING FORK OF KNOWN FREQUENCY ?? 🔉⇢
PRAMOR MORE'S PHYSICS TUTORIALS

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📑 Clips (4)
  • 0:00–2:31Beats — segment 1First for students in English it is difficult which song is 100g concept profile and loading happens so I do n't understand when weight increases or decreases final answer what to write sir that job is a very simple…beats
  • 2:31–5:02Beats — segment 2but in the question he said that it should be increased, som, see this Edison should have been big that power total disbursement no matter how much he would have left 238 anything Dwivedi increased employment brother, I…beats
  • 5:02–7:33Beats — segment 3Decrease volume February loading beating a small business is equal to right and loading decrease take a back 512 with that six subscribe to school we did now half subscribe so subscribe so understand 500 600 800…beats
  • 7:33–10:04Beats — segment 4and what is the loading and what has happened between the loading and the grinding and the healthy will increase in the first 5 seconds why did you increase the fielding also and it is small first then you gave it 985…beats
What is Doppler Effect | Sound Waves | Extraclass.com 🔉⇢
Extraclass Official

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  • 0:06–2:37Doppler effect — segment 1what does the dopplers effect have you ever noticed that the sound of an ambulance siren changes as it passes by your scooty on the road the siren sounds louder and more shrill as the ambulance approaches your scooty…doppler effect
  • 2:37–5:07Doppler effect — segment 2on stationary observer when a source moving with velocity we s produces a sound wave of frequency F and the waves travel towards an observer with velocity V then it has the wavelength lambda not equal to V divided by F…doppler effect
  • 5:07–7:35Doppler effect — segment 3solve an example to understand this concept better question a train is moving on a straight track with speeds 20 meters per second it is blowing a whistle and the frequency of 1,000 Hertz the percentage change in…doppler effect
Doppler Effect Formula Made Easy 🔉⇢
Shack Solem

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  • 0:00–2:30Doppler effect — segment 1Hello everyone. Uh in this video I'm going to talk a little bit about the Doppler effect. Okay, I've written down the formula already. It's the frequency of the observer is equal to velocity of the sound plus or minus…doppler effect
  • 2:30–2:56Doppler effect — segment 2velocity of the of the source will be zero. But even if it isn't, you know when to add when to subtract. you know, if they're uh coming closer to each other, add these guys and subtract these guys. When they're going…doppler effect
The Doppler Effect: what does motion do to waves? 🔉⇢
Alt Shift X

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  • 0:00–2:31Doppler effect — segment 1the Doppler effect in Sheldon's words it's the apparent change in the frequency of a wave caused by relative motion between the source of the wave and the Observer the Doppler effect is perhaps best explained visually…doppler effect
  • 2:31–3:03Doppler effect — segment 2stars or anything that you can see change color depending on their relative motion to you of course you can't see this minute difference with your eyes but astronomers with the right equipment can use this effect to…doppler effect
अनुप्रस्थ तरंग किसे कहते है | अनुदैर्ध्य तरंग kya hai | Transverse wave and Longitudinal waves Hindi 🔉⇢
RKR STUDY

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📑 Clips (4)
  • 0:00–2:30Transverse and longitudinal waves — segment 1इससे पिछले वाले वीडियो में हम सभी ने ध्वनि के बारे में अध्ययन किया था ध्वनि के टॉपिक को समझने के दौरान हम सबों को एक टर्म निकल के आया था वह था आपका तरंग तो तरंग क्या होता है इसके बारे में आज की इस वीडियो के माध्यम से हम…transverse and longitudinal waves
  • 2:30–5:01Transverse and longitudinal waves — segment 2जाकर के ये आपको इस तरह का चीजें देखने को मिलेगा और इसी को हम लोग क्या कहते हैं इसी को हम लोग तरंग कहते हैं ठीक है अब ये तरंग जो होता है वो कितने प्रकार के होती है इसको थोड़ा सा जान लीजिए तो तरंग आपके दो प्रकार के होते…transverse and longitudinal waves
  • 5:01–7:31Transverse and longitudinal waves — segment 3आगे की ओर बढ़ना शुरू कर देता है अब गौर करिएगा एक ऊपर नीचे की ओर इसका कण यहां पर क्या करेगा कंपन शुरू करेगा इसका डायरेक्शन ऊपर नीचे रहेगा और दूसरा क्या होता है दूसरे प्रकार का कि क्षैतिज दिशा में एकदम सीधा आगे की ओर…transverse and longitudinal waves
  • 7:31–9:23Transverse and longitudinal waves — segment 4कहते हैं और इसका जो उदाहरण है वही आपके स्प्रिंग वाले तरंग हो जाएंगे प्रकाश के तरंग हो जाएंगे वायु में उत्पन्न ध्वनि की तरंग हो जाएंगे ठीक है अब यहां पर दो-तीन महत्त्वपूर्ण बातें हैं अधारित रंग और अनुप्रस्थ रंग से…transverse and longitudinal waves
अनुप्रस्थ और अनुदैर्ध्य तरंग | Anuprasth Aur Anudharya Tarang | Longitudinal and Transverse Waves 🔉⇢
Alpha Affairs

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📚 Teaches: Tier-1 educational channel (Alpha Affairs); found via yt-dlp search 'अनुप्रस्थ और अनुदैर्ध्य तरंगें transverse longitudinal waves Hindi physics class 11', oEmbed-verified live.

📑 Clips (3)
  • 0:00–2:32Transverse and longitudinal waves — segment 1आज की इस वीडियो में हम पढ़ेंगे एक ऐसा टॉपिक जो क्लास 6थ से लगाकर 12थ तक पूछा जाता है और कॉम्पिटिटिव एग्ज़ाम में भी चाहे वह एसएससी हो, रेलवे हो या स्टेट पुलिस का एग्जाम हो सभी एग्जाम्स में इससे क्वेश्चन देखने को मिल जाते…transverse and longitudinal waves
  • 2:32–5:02Transverse and longitudinal waves — segment 2तरंगे क्या होती है ये अनुप्रस्थ तरंगे चलिए आगे चलते हैं इनमें हम अंतर देखते हैं दोनों तरंग में अनुप्रस्थ तरंग क्या होती है अनुदैर्ध्य तरंग क्या होती है उनमें क्या अंतर होता है क्योंकि यहां से सबसे ज्यादा क्वेश्चन पूछे…transverse and longitudinal waves
  • 5:02–5:23Transverse and longitudinal waves — segment 3था हमारा टॉपिक अनुप्रस्थ और अनुदैर्ध्य तरंगे क्या होती है? क्या उनमें अंतर होता है? बड़े ही आसान भाषा में हमने इसके बारे में जाना। उम्मीद करते हैं सभी चीजें आपको क्लियर हो गई होगी। तो अगर आपको यह वीडियो अच्छी लगी हो तो…transverse and longitudinal waves
L-2, प्रगामी तरंगें (Progressive Waves) | अध्याय-15, तरंगें (Waves) Class 11th Physics 🔉⇢
Learn and Share

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Learn and Share); found via yt-dlp search 'प्रगामी तरंग समीकरण progressive wave equation Hindi physics class 11', oEmbed-verified live.

📑 Clips (6)
  • 0:03–2:34Progressive wave equation — segment 1कर दो मेरे प्यारे बच्चो नमस्कार आप चित्र या फिर से आप सबका स्वागत करते हैं आपके अपने चैनल लगा लिंक शेयर पर और हम बात करेंगे क्लास इलेवेंथ फिजिक्स के बारे में चैप्टर नंबर चल रहा है चैप्टर नंबर 15 में का अंतिम चैप्टर…progressive wave equation
  • 2:34–5:05Progressive wave equation — segment 2है कुछ तुम पहली पढ़कर आए हैं आयाम क्या होती है आवृत्ति क्या होती है तरंगदैधर्य क्या होती है वह मेघा मतलब हमारा जीवन परिचय कोणीय आवृत्ति उसके बारे में भी शिफ्ट करें लेकिन कुछ कलांतर पड़ा था पन्ने और प्रारंभ करना पड़ा था…progressive wave equation
  • 5:05–7:36Progressive wave equation — segment 3इसका मात्रक क्या ऐसे ही मात्रक इसका मात्रक बच्चों मीटर मीटर में आपने मीटर सेंटीमीटर लेकिन क्योंकि हमारे बात करें इसके मीटर फैशन ऐसा ही मात्रक है बट इसके बाद बात करते हैं अपन त रंग बदलने के बारे में तो एक्सप्रेशन टो…progressive wave equation
  • 7:36–10:06Progressive wave equation — segment 4तक तो चलो चलते हैं कि Idea Music एप्लीकेशन के बारे में बात करेंगे बच्चों वह वापस आवृत्ति टिकरा बेसिक एजुकेशन क्लास नाइंथ वाली है हैं और अपने से पहले चैप्टर में पड़ा है इसमें इंग्लिश बोलते हैं फ्रीक्वेंसी हुआ था ठीक है…progressive wave equation
  • 10:06–12:37Progressive wave equation — segment 5इंग्लिश में देव नंबर है कि वे नंबर क्या होता है बच्चों व नंबर होता है मेरे पास एक काम की लंबाई में तरंग घरों की संख्या मालू यह मैंने लिया एक मीटर की लंबाई एक मीटर की लंबाई में कितने तरंगदैधर्य आगे इसको बोलते हैं अपन…progressive wave equation
  • 12:37–15:11Progressive wave equation — segment 6पन्ने पूर्णिया वृद्धि ओमेगा 3 बच्चों के पास इसको बोलते पर रेगुलर फ्रिकवेंसी इन इंग्लिश में को बच्चों रेगुलर है कि फ्रीक्वेंसी है मैं तो कुछ नहीं जो अपने पहले थोड़ा सा ओमेगा किसके बराबर होता बच्चों ठीक है बेटा ठीक है…progressive wave equation
Progressive wave equation #waves #progressivewaves 🔉⇢
Physics Study With Pradeep Sir

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Physics Study With Pradeep Sir); found via yt-dlp search 'प्रगामी तरंग समीकरण progressive wave equation Hindi physics class 11', oEmbed-verified live.

📑 Clips (3)
  • 0:00–2:31Progressive wave equation — segment 1हेलो हेलो फ्रेंड स्वागत है आपका आश्रम देखने वाले प्रगामी तरंग समीकरण किस प्रकार से निर्मित की जा सकती है तो हम जानते हैं कि जब भी कोई करंट कंपन करता है तो वहां से तरंग उत्पन्न होना शुरू हो जाती है अब यहां पर हम देखते…progressive wave equation
  • 2:31–5:03Progressive wave equation — segment 2चक्कर में जो कुल तरंगों के बीच में जो अंतर होता है वह दर्शक अब हम यह चाहिए कि बिंदु से दूसरे बिंदु है उसमें कितना लांघ कर दो यह बराबर होगा टू पर ही पोंदलू डू इनटू इट्स का मल्टिप्लिकेशन कर दें क्योंकि हम सिर्फ एक दूरी पर…progressive wave equation
  • 5:03–5:37Progressive wave equation — segment 3यहां पर समझने वाली बात यह होती है कि जो लोग ने अपने क्वेश्चन आता है वह प्रगामी तरंग कि तरह हमारे पास आ जाती है तो यह वीडियो बहुत ही एक इंपोर्टेंट वीडियो हो सकता आपके लिए क्योंकि प्रगामी तरंग समीकरण अक्सर जाम में पूछने…progressive wave equation
Chapter 15 (Class 11) Velocity of Transverse Waves in a Stretched String 🔉⇢
KC SIR (RES)

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (KC SIR (RES)); found via yt-dlp search 'डोरी पर तरंग की चाल speed of wave on string Hindi physics class 11', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:50Speed of a wave — segment 1गुड मॉर्निंग मित्रों आज सभी मित्रों का गवर्नमेंट व्लॉग्स यूट्यूब चैनल पे हार्दिक अभिनंदन हम क्लास 11th का चैप्टर नंबर 15 का अध्ययन कर रहे हैं और उसमें भी आज हम अध्ययन करेंगे तानी हुई डोरी में अनुप्रस्थ रंग का वेद…speed of a wave
  • 2:50–5:27Speed of a wave — segment 2इसको ऊपर किया है तो इसमें तनाव जरूर लगेगा तो मानते हैं की इसमें तनाव जो है वो 37 से θ को बना रहा है इट मिंस यह हो गया इसको हमें वियोजन करें तो जिधर को बनता है उधर हम तो लेते हैं टी कोस थीटा बनाती है तो इधर जाएगा साइन…speed of a wave
  • 5:27–7:58Speed of a wave — segment 3कितनी होगी कैपिटल एम ये हो जाएगा डेल हेल्पर केंद्र की ओर लगने वाला कुल तनाव टूटी साइन थीटा इस अल्फाज पर केंद्र की ओर लगने वाला कुल तनाव टूटी साइन थीटा आवश्यक अभिकेंद्रीय बाल प्रधान करता है आवश्यक अभिकेंद्रीय बाल प्रदान…speed of a wave
  • 8:00–10:57Speed of a wave — segment 4थीटा अति अल्प है अति अल्प है तो साइन थीटा को हम लगभग यह थीटा है और यह 90 है अब यह दोनों समांतर को है सिमिलरली यहां पर कितना देगा तू थीटा तो हम क्या जानते हैं कौन बराबर चाप बाते त्रिज्या कितना है आर क्या करेंगे समीकरण एक…speed of a wave
  • 10:57–13:30Speed of a wave — segment 5रंग का वेज है डियर स्टूडेंट्स इस चित्रा का वेज आया है इससे साफ स्पष्ट है की जो अनुप्रस्थ तरंग का वेज होता है वह आयाम और तरंग धैर्य पर निर्भर नहीं करता है क्योंकि जो वेज है उसके बराबर में एन तो आयाम है और ना ही लिंडा है…speed of a wave
  • 13:30–16:16Speed of a wave — segment 6एम पर तो स्टूडेंट्स हमने देखा की तानी भी डोरी में अनुप्रस्थ तरंग का वेज का फॉर्मूला और वह फॉर्मूला किन-किन भौतिक राशियों पर निर्भर करता है इसका भी हमने अध्ययन किया अब नेक्स्ट हम अध्ययन करेंगे anudair तरंग का चाल भी दिशा…speed of a wave
Principle of Superposition of Waves, Wave Optics, Sound 🔉⇢
gajendra singh rathore

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (gajendra singh rathore); found via yt-dlp search 'तरंगों का अध्यारोपण superposition of waves Hindi physics class 11', oEmbed-verified live.

📑 Clips (2)
  • 0:00–2:31Superposition and interference — segment 1झाला कि तरंग प्रकाशिकी क्लास ट्वेल्थ फिजिक्स और इलेवेंथ फिजिक्स में ध्वनि दोनों के अंदर यह तरंगों के अध्यारोपण का सिद्धांत आता है तो इस पर एक शर्ट क्वेश्चन जो है वह बन सकता है छोटा प्रश्न 2 अंक गंवा की तरंगों के…superposition and interference
  • 2:31–3:28Superposition and interference — segment 2बहुत अच्छे से आप इसको समझ लीजिए दूसरी महत्वपूर्ण बात यह है कि वह जो परिणाम में विस्थापन आएगा वह कई बार बढ़कर दिखेगा कईं बार वह अधिक आयु बढ़कर कब का यह अपना दो तरंगों की बात करें अगर वह दोनों तरंगे सम्मान कला में है तो…superposition and interference
Principle of SUPERPOSITION of Waves class 11th chapter 15th Physics in Hindi 🔉⇢
CONCEPTUAL PHYSICS Vishal Gangwani

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (CONCEPTUAL PHYSICS Vishal Gangwani); found via yt-dlp search 'तरंगों का अध्यारोपण superposition of waves Hindi physics class 11', oEmbed-verified live.

📑 Clips (4)
  • 0:00–2:31Superposition and interference — segment 1हेलो फ्रेंड्स आज मैं बात करने वाले हैं सुपरपोजिशन प्रिंसिपल ऑफ वेल्स के बारे में डिफरेंट सुपरपोजिशन का मतलब क्या हुआ है इसी पोजीशन जहां पर दो या दो से ज्यादा कई सैनिकों मल्टीपल व्यवसाय करके मिले उस पोजीशन को सुपरपोजिशन…superposition and interference
  • 2:31–5:04Superposition and interference — segment 2इसका मैथमेटिकल पाठ के अंदर जाना पड़ेगा हमें कैंसिल करनी पड़ेगी दो वैक्स जिसमें पहली बेबी हम लोग नजर करिए वन इज इक्वल टू में यह साइन के एक मैंने समझा कि हमें पता है कि एक ट्रैवलिंग वेव इक्वेशन है अच्छी तरीके से ठीक है…superposition and interference
  • 5:04–7:35Superposition and interference — segment 3पर के एक्ट - ओमेगा टीईएस फाइबर टू ठीक है यह चीज मगर भटूरे भटूरे पड़ा हुआ है ठीक है अब देखिए अब यहां पर थोड़ा ध्यान दीजिए कि अगर इस ही की वैल्यू में जीरो रख फूफा यानि कि हमें पता फेस चेंज एनी सैक्रिफिस चेंज की वैल्यू…superposition and interference
  • 7:35–8:41Superposition and interference — segment 4जब साफ करें मिलती हैं तो वह अलग-अलग तरीके के रिजल्ट्स क्रिएट करती है जैसे कि मैं सबसे इंट्रस्टिंग रिजल्ट्स जो है वह इंटरफ्रेंस है इसको आप लोग कहते हैं हिंदी में व्यतिकरण ऐसे जो रिजल्ट है वह मेनली आते टॉरेंट वेब सही है…superposition and interference
11th physics ch-30( अप्रगामी तरंगे: वायु - स्तंभों के कम्पन) 🔉⇢
physics lecture

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (physics lecture); found via yt-dlp search 'अप्रगामी तरंगें स्वरमापी organ pipe standing waves Hindi physics class 11', oEmbed-verified live.

📑 Clips (6)
  • 0:01–2:40Standing waves and normal modes — segment 1हेलो स्टूडेंट्स आपका न्यू चैप्टर है आप आगामी तरंगे इसमें आपकी फर्स्ट डेफिनेशन है pragami तरंगे इसमें dekhiae जब दो एक जैसी अनुप्रस्थ अथवा anudaidhya pragami तरंगे किसी बढ़ मध्य में एक ही चाल से परंतु विपरीत दिशाओं में…standing waves and normal modes
  • 2:40–5:11Standing waves and normal modes — segment 2समीकरण एक्स एक्स की दिशा में चल रही है धनात्मक दिशा में चल रही है इधर ठीक है यह जब dhirsi आएगा तबियत टाइप से फिक्स होगी इस वक्त क्या है यह फ्री है यह नहीं है ठीक है और paraavarti तारा XX के समय अब क्या है वह उसके पहले…standing waves and normal modes
  • 5:11–7:41Standing waves and normal modes — segment 30.02 एक्स ठीक है सिर्फ व्यक्त तरंग की विपरीत दिशा में चलने वाली परंतु अन्य सभी बातों में समांतर इनकी समीकरण लिखिए संयोजन से जो अपरा मित्र बन रही है उसकी समीकरण में लिखनी है और हमसे पूछा है दो निकटतम nispando के बीच की…standing waves and normal modes
  • 7:41–10:30Standing waves and normal modes — segment 4आई तू होती है dekhiae यहां पर एंड यहां पर ठीक तो इनके बीच की दूरी क्या है यहां से यहां से यहां तक के बीच की दूरी क्या होती है नहीं आएगा तेरे बस का नहीं है सबसे पहले प्रदर्शित है डोरी में nishkand कहां पर स्थित है ए इस…standing waves and normal modes
  • 10:58–13:45Standing waves and normal modes — segment 5आगे dekhiae डोरी में तरंग का paraavartan ढीला से होता है आपकी तरह की क्वेश्चन हमारी यह हो गई माइंस होता है एक्स अपॉन 50 क्योंकि यहां पर लेम्डा की वैल्यू क्या है ओनली 53 नंदा की वैल्यू 50 है ठीकstanding waves and normal modes
  • 14:22–16:59Standing waves and normal modes — segment 6लिसन करेंगे इसके बराबर होगा तो यहां से हमारी क्या है लेम्डा की वैल्यू ए जाएगी जब यहां से कैलकुलेशन करेंगे 2 * 3.14 करेंगे 6.28 निकल जाएगा बीच की दूरी कितनी होती है लैंड हमारा बंद है अपॉन में हमने क्या किया तू कर दिया…standing waves and normal modes
L-6, अप्रगामी तरंगें (Standing waves) | अध्याय-15, तरंगें (Waves) Class 11th Physics 🔉⇢
Learn and Share

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Learn and Share); found via yt-dlp search 'अप्रगामी तरंगें स्वरमापी organ pipe standing waves Hindi physics class 11', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:32Standing waves and normal modes — segment 1कर दो कि वह बड़ी बहन फिल्म अजय को हुआ था मेरे प्यारे मित्रों नमस्कार आप की जगह फिर से आप सबका स्वागत करते हैं आपके अपने चैनल लर्निंग शेयर पर और हम बात करेंगे क्लास इलेवेंथ इसके बारे में चैप्टर चल रहा है चैप्टर नंबर…standing waves and normal modes
  • 2:32–5:02Standing waves and normal modes — segment 2बच्चे बोले हमने रिबन का कोर्स लूट लिया तो हमारी 11वीं की फिजिक्स बड़ी है तो घबराने की जरूरत नहीं है क्लास ट्वेल्थ फिजिक्स जो है वह कम ही को रिलेटेड है क्लास इलेवेंथ पर क्योंकि इंडिया का जो सिलेबस डेवलप किया गया है फॉर द…standing waves and normal modes
  • 5:02–7:33Standing waves and normal modes — segment 3बोलते हैं स्टैंडिंग वेनसडे जनवरी 20 अब इसका नाम कैसे पड़ा कहां से आया क्यों आया वह सबके बारे में विपन डिटेल में बात करेंगे तो आज का हमारा घोल कह सिर्फ और सिर्फ और प्रगामी तरंग ऊपर फोकस करना ठीक है चल है है तो यहां पर…standing waves and normal modes
  • 7:33–10:04Standing waves and normal modes — segment 4हम लोग तरंग दैर्घ्य ठीक है तुम्हें पता है कि रंगों में रंग यहां पर यह क्रॉस में क्वांटिटी करता जरूर बजरंग वे सजना वे सजना वे सजना वे सजना रोवता कंगना के जरिए आते आते आते यह तरह-तरह यहां से यहां से व्यवसाई के यहां से…standing waves and normal modes
  • 10:04–12:36Standing waves and normal modes — segment 5नो डिले इन नोड है और यहां पर सेंट इन उन्हें वापस फ्रेंड बन जाता है ठीक है तो जब हम किसी भी तरंगों को भेजते हैं किसी रस्सी से तो तरंग जब नोट करके आती है तो परावर्तन और अपवर्तन के किसी हिंदू कि होती है जिस बिंदु पर मिलती…standing waves and normal modes
  • 12:36–15:07Standing waves and normal modes — segment 6कि अध्यारोपित है नहीं होती है तो कि उस स्थान पर कि उस अ स्थान पर गण कि यहां तो गण को अधिकतम संयम से गति करता है में पाया हमसे गति करता है कि यह देखकर वहां पर मैक्सिमम ड्यूटी साथ एडिसन करने लग जाता है तो उसको पर प्रतिबंध…standing waves and normal modes
Chapter 15 (Class 11) Beats 🔉⇢
KC SIR (RES)

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (KC SIR (RES)); found via yt-dlp search 'विस्पंद beats frequency Hindi physics class 11', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:33Beats — segment 1गुड इवनिंग मित्रों एक साथ एक ही दिशा में संचित होकर अध्याय रूप अध्यारोपण करती है तो अध्यारोपण क्षेत्र में एक स्थान पर समय के साथ परिवर्तित होती हैं इस घटना को विश्व बंद कहते हैं है इसमें जब तीव्रता अधिकतम होती है या…beats
  • 2:33–5:04Beats — segment 2बदलती है की एक सेकंड में एक सेकंड में जितनी बार अधिकतम या न्यूनतम एक सेकंड में तीव्रता जितनी बार अधिकतम या न्यूनतम होती है उसे विस्पंद आवृति कहते [संगीत] है जब लगभग समान आवर्ती की दो ध्वनि तरंगे एक स्थान पर एक ही दिशा…beats
  • 5:04–7:45Beats — segment 3मैंने क्या किया एक दूसरी समान आयाम की तरंग और लेता हूं जो मेरी ग्रीन से है यह थोड़ा जमेगा नहीं रखूंगा तो इसको यह आपका समय टी है अब दूसरी में तरंग ली देखो दोनों तरंगों का कला को समान था और देखो इस स्थिति में दोनों का…beats
  • 7:45–10:18Beats — segment 4प्रणाम तीव्रता लिखो एक ही बात है हमें जानते हैं की आयाम का वर्ग तीव्रता के समरूप होती है या तीव्रता आयाम के वृत्त के समानुपाती होती है यह हम पहले से ही जानते हैं तेरी प्रणामी तीव्रता निकलती है तब भी ठीक है और परिणामी…beats
  • 10:18–13:03Beats — segment 5समान आवृत्ति की दो ध्वनि तरंगी जिनका आयाम समान था और जिनका माना माना दोनों तरंगों का तरंगों की आवृत्ति क्रम शहर n1 कॉम आया और दोनों कला को पर अध्यारोपित हैं तो तो विस्थापन के समीकरण विस्थापन के समीकरण निम्नलिखित होंगे 2…beats
  • 13:03–15:41Beats — segment 6जब दो या दो से अधिक tarangiya इसका मतलब है की उनके visthapanon का योग प्रणामी विस्थापन के बराबर होता है अब मैन लेते हैं की इन दोनों का प्रणाम विस्थापन वही है तो ए बराबर हो जाएगा y1 प्लस Y2 पाई एन तू टी बेस्ट फ्रेंड्स एक…beats
L-8, विस्पंद (Beats) | अध्याय-15, तरंगें (Waves) Class 11th Physics 🔉⇢
Learn and Share

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Learn and Share); found via yt-dlp search 'विस्पंद beats frequency Hindi physics class 11', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:31Beats — segment 1ब्लुटूथ शब्द साफ ही गया प्यारे बच्चों नमस्कार आप पिंपल या फिर से आप सबका स्वागत करते हैं आपकी अपनी इस चैनल तो सबसे पहले आप सबको महाशिवरात्रि की हार्दिक शुभकामनाए आप सब बहुत बढ़िया है मजे में तो हम बात करेंगे क्लास…beats
  • 2:31–5:03Beats — segment 2क्लास में अपने जैसा कि मैंने आपको बताया कि बंद पैक और खुले वाइट बारे में बात की थी खुले पाइप में आपको यह बता दिया कि सभी प्रकार की आवृत्त्या उच्च उत्पन्न करना संभव है लेकिन अगर आप एक बंध आपके बारे में बात करें तो एक तरफ…beats
  • 5:03–7:40Beats — segment 3लुट जिनकी आवृत्ति में थे जिनकी आवृत्ति में कि थोड़ा सांतरो थोड़ा सा अंतर हो को ज्यादा नहीं होना चाहिए थोड़ा से इधर होना चाहिए थोड़ा सा किधर हो कि सलमान मध्य में के समान शिक्षा में के समान दिशा में कि चलते हुए में चलते…beats
  • 7:40–10:12Beats — segment 4कि ऐसे हम क्या बोलते हैं जो तरंगे उत्पन्न उसको बीच बोलते हैं अब देखो बीच बनाएंगे कैसे बीच की परमिशन आपको थोड़ा को समझ क्यों बोलना क्या चाहते हैं मामलों को पहली तरह है मेरे पास आप कुछ इस तरीके की है हमने पहली तरंग भी…beats
  • 10:12–12:42Beats — segment 5तक में इसके बावजूद को किसका आया देखो कितनी बार चेंज हुआ सबसे व्यायाम बहुत ज्यादा हृदय का हुआ फिर और कॉपर और गमों पर जीरो पर पहुंचा फिर धीरे-धीरे वापस इसका सेवन ढेर लग गया बढ़ने लग गया तो ऐसा क्यों आपको जैसे यहां पर अगर…beats
  • 12:42–15:15Beats — segment 6मैं इसको बोलते हिंदी में स्वरित्र दुसरे स्वरित्र ठीक है कि वे उक्त व्यक्ति ने बच्चों के पास दूसरा सब्सक्राइब कर लें अच्छा ठीक है यह पहला ट्यून फॉर के बच्चों और दूसरा ट्यून फॉर पीएम यहां पर इंडिविजुअल आफ बजे का इस्तेमाल…beats
| डॉप्लर प्रभाव क्या है?| ऐनीमेशन की मदद से समझे बेहद आसान तरीके में | हिन्दी में 🔉⇢
Easy Naukri

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Easy Naukri); found via yt-dlp search 'डॉप्लर प्रभाव doppler effect sound Hindi physics class 11', oEmbed-verified live.

📑 Clips (0)

Full lecture — no clip index.

Doppler Effect (डॉप्लर प्रभाव) by Neeraj Sir #Physics #ncert #sciencemagnet 🔉⇢
Science Magnet

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Science Magnet); found via yt-dlp search 'डॉप्लर प्रभाव doppler effect sound Hindi physics class 11', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:32Doppler effect — segment 1झाल हेलो हेलो फ्रेंड्स वेलकम टू साइंस मैग्नेट दोस्तों कैसे हैं आप लोग सभी आई हॉप सब बहुत अच्छे हैं शानदार है जानदार है तो दोस्तों इस सेशन में हम लोग डिस्कस करने वाले हैं डॉपलर इफेक्ट जिसे आप लोग हिंदी में डॉपलर प्रभाव…doppler effect
  • 2:32–5:03Doppler effect — segment 2इसको आप लोग बोलते हैं क्रस्ट और हिंदी में आप लोग उसको क्या बोलते हैं इसके अंदर बोलते हैं यह होता है नीचे वाला पार्ट होता है आप लोगों का टिफिन जिसे आप लोग हिंदी में बोलेंगे अगर तो आप लोग ध्यान से रखेंगे आपको एक तरफ एक…doppler effect
  • 5:03–7:33Doppler effect — segment 3को सुनाई देगी कहानी क्लियर है ठीक है कि पुरुषों जैसे मान लीजिए एंबुलेंस है तो यहां पर आप लोग देखिए इससे साउंड और है उसके कोई ना कोई तो फ्रीक्वेंसी होगी मैं मान लेता हूं फॉर एग्जामपल आप लोग यह मानकर चलिए 500 वर्ड्स 568…doppler effect
  • 7:33–10:04Doppler effect — segment 4मंजू एक चलो फ्रीक्वेंसी उससे कम या अधिक सुनाई देगा ही आप लोग डॉक्टर पैक बोलते हैं और यह नाम डॉपलर इफेक्ट को दिया गया क्योंकि इसके बारे में हम लोगों को किसने बताया था क्रिस्चियन डॉप्लर ने बताया था इस साईट के बारे में…doppler effect
  • 10:04–12:36Doppler effect — segment 5पड़ेगा तभी तो यह वैल्यू कांस्टेंट आएगी तो यहां पर जब आप लोग के पास मारा तो फ्रीक्वेंसी क्या होगी ज्यादा होगी फ्रीक्वेंसी ज्यादा का मतलब पेट से ज्यादा पीछे ज्यादा का मतलब आवाज कैसे पतली तो जब आपके तरफ कोई चीज आ रही है तो…doppler effect
  • 12:36–15:10Doppler effect — segment 6फ्रीक्वेंसी जिसका साउंड पे गुस्सा आता है ना एक्चुअल यानि वास्तविक आप रोटी लेकिन हम लोगों को तो कुछ और आवृत्ति की जीवनी सुनाई दिया जाएगा ना तो हम लोग को तो यह निकालना है कि हम क्वेकर सुनाई देगा हमको एक्चुअल से मतलब नहीं…doppler effect
Constructive and Destructive interference | Physics | Khan Academy 🔉⇢
Khan Academy Physics

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Khan Academy Physics); found via yt-dlp search 'interference of waves constructive destructive Khan Academy', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:30Superposition and interference — segment 1so imagine you've got a wave source this could be a little oscillator that's creating a wave on a string or a little paddle that goes up and down that creates waves on water or a speaker that creates sound waves it…superposition and interference
  • 2:30–5:00Superposition and interference — segment 2see what happens i'm gonna overlap these two waves and we'll perform the same analysis i don't really even need the backdrop now because look at i've got one and negative one one and negative one zero zero and zero zero…superposition and interference
  • 5:00–7:31Superposition and interference — segment 3constructive interference is to take two wave sources that start in phase and just put them right next to each other and a way to get destructive is to take two wave sources that are pi shifted out of phase and put them…superposition and interference
  • 7:31–10:02Superposition and interference — segment 4that was when they were right next to each other you got constructive when this difference is equal to one wavelength we also got constructive when it was two wavelengths we got constructive it turns out any integer…superposition and interference
  • 10:02–12:34Superposition and interference — segment 5distance to get to the detector so x1 and x2 are going to be equal you subtract them you'd get zero this time the zero is giving us destructive instead of constructive so let's see what happens if we move this forward…superposition and interference
  • 12:34–13:45Superposition and interference — segment 6second condition over here to figure out whether you get constructive or destructive if neither of them get a phase shift or interestingly if both of them get a phase shift you could use this one because you can imagine…superposition and interference
Standing waves on strings | Physics | Khan Academy 🔉⇢
Khan Academy Physics

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Khan Academy Physics); found via yt-dlp search 'standing waves on a string nodes and antinodes physics', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:31Standing waves and normal modes — segment 1if you've got a medium and you disturb it you can create a wave and if you create a wave in a medium that has no boundaries in other words a medium that's so big this wave basically never meets the boundary then there's…standing waves and normal modes
  • 2:31–5:03Standing waves and normal modes — segment 2reflected upside down that doesn't matter too much for our purposes but every time it's going to reflect it flips its direction and it keeps bouncing now let's say instead of sending in a single pulse we send in a whole…standing waves and normal modes
  • 5:03–7:33Standing waves and normal modes — segment 3right or left this peak is just going to move up and down so a lot of times when we draw these standing waves we draw a dash line underneath here that mirrors the Bold line because all this peak's going to do is go from…standing waves and normal modes
  • 7:33–10:04Standing waves and normal modes — segment 4cuz there's nothing happening there there's no motion and these maximum displacement points are the constructive points we should give those a name what do you think we call those if you guessed anti Noe then you're…standing waves and normal modes
  • 10:04–12:34Standing waves and normal modes — segment 5it as 20 M over 3 and we'll keep going here I'll draw the rest this is the fourth harmonic how big is this wavelength well this wavelength covers half of the string so this wavelength is going to be half the length of…standing waves and normal modes
  • 12:34–13:26Standing waves and normal modes — segment 6on all instruments with a string both ends are fixed so recapping when you confine a wave into a given region the wave will reflect off the boundaries and overlap with itself causing constructive and destructive…standing waves and normal modes
Beat frequency | Physics | Khan Academy 🔉⇢
Khan Academy

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Khan Academy); found via yt-dlp search 'beats phenomenon two frequencies physics explained', oEmbed-verified live.

📑 Clips (5)
  • 0:00–2:31Beats — segment 1what's up everybody I want to talk to you about beat frequency and to do so let me talk to you about this air displacement versus time graph so this is going to give you the displacement of the air molecules for any…beats
  • 2:31–5:01Beats — segment 2between these two peaks it's hard to see it's almost the same but this Red Wave has a slightly longer period if you can see the time between Peaks is a little longer than the time between Peaks for the Blue Wave and you…beats
  • 5:01–7:31Beats — segment 3frequency let me play just a slightly different frequency I'll play 443 Hertz and you're probably like that just sounds like the exact same thing I can't tell the difference between the two but if I play them both…beats
  • 7:31–10:02Beats — segment 4now I should say to be clear we're playing two different sound waves our ears really just sort of going to hear one total wave so these waves overlap you can do this whole analysis using Wave interference you write down…beats
  • 10:02–11:48Beats — segment 5445 htz playing a little sharp or it might be 435 HZ might be playing a little flat so it would have to tune to figure out how it can get to the point where there'd be zero beat frequency cuz when there's zero beat…beats

💬 Doubt Solving Ask-any-doubt RAG + FAQ

🤖 Ask any doubt RAG · NCERT-grounded, cited

Answers are grounded in tier-1 material (NCERT / JEE PYQ) and cited — the AI will say so if a doubt isn't in the indexed syllabus.

Frequently-asked doubts

Does the medium travel along with the wave?
No. In a mechanical wave the particles of the medium only oscillate about their fixed equilibrium positions; it is the pattern of disturbance — and with it energy and momentum — that moves forward. A cork on rippling water bobs up and down but is not carried across the pond, and the air in front of a loudspeaker vibrates in place rather than blowing toward you. The wave equation $\partial^2 y/\partial t^2=v^2\,\partial^2 y/\partial x^2$ describes the propagation of the shape, not of matter.
What is the difference between particle velocity and wave velocity?
They are entirely different quantities. The wave velocity $v=\omega/k=\nu\lambda$ is the constant speed at which the waveform advances, fixed by the medium. The particle velocity $v_{particle}=\partial y/\partial t=-A\omega\cos(kx-\omega t)$ is the transverse (or longitudinal) speed of an individual element, which oscillates and reaches a maximum of $A\omega$ at the equilibrium position and zero at the extremes. A given particle speeds up and slows down each cycle while the wave itself glides on at constant $v$.
Why does the wave speed depend on the medium but not on the source frequency or amplitude?
For a mechanical wave the speed is set by a competition between a restoring (elastic) property and an inertial property of the medium — $v=\sqrt{T/\mu}$ on a string, $v=\sqrt{\gamma P/\rho}$ in a gas. Neither expression contains frequency or amplitude, so a loud note and a soft note, a high pitch and a low pitch, all travel at the same speed. Changing the source changes the wavelength that is laid down (since $\lambda=v/\nu$), but not the propagation speed.
Why is there a phase change of $\pi$ on reflection at a rigid end but none at a free end?
At a rigid (fixed) end the medium cannot move, so the wall must exert a force that exactly cancels the incoming displacement; by Newton's third law this returns an inverted pulse — a crest comes back as a trough, a phase change of $\pi$. At a free end the element is free to overshoot, and the pulse reflects the same way up, with no phase change. These boundary rules are what force a displacement node at a closed end and an antinode at an open end.
Why does a pipe closed at one end produce only odd harmonics?
The closed end must be a displacement node and the open end a displacement antinode. The shortest standing wave that fits is a quarter wavelength ($L=\lambda/4$), and every higher mode that also has a node at one end and an antinode at the other adds half a wavelength, giving allowed frequencies $\nu_n=(2n-1)v/4L=\nu_1,3\nu_1,5\nu_1,\dots$. The even harmonics would require an antinode at the closed end, which the boundary forbids, so they simply cannot form.
What do the $\pm$ signs mean in the Doppler formula $\nu'=\nu\dfrac{v\pm v_o}{v\mp v_s}$?
Choose the signs so that motion which reduces the separation raises the pitch. In the numerator use $+v_o$ when the observer moves toward the source and $-v_o$ when moving away. In the denominator use $-v_s$ when the source moves toward the observer (this makes the denominator smaller and $\nu'$ larger) and $+v_s$ when it moves away. A reliable check: any approach must give $\nu'>\nu$ and any recession $\nu'<\nu$.
What is the difference between beats and interference?
Both come from superposition, but beats are interference in time while (spatial) interference is interference in space. Beats arise from two waves of slightly different frequencies at the same place, producing a loudness that rises and falls at $\nu_{beat}=|\nu_1-\nu_2|$. Spatial interference arises from two coherent waves of the same frequency arriving at different places with different path differences, giving a fixed pattern of loud and quiet points. Time-varying loudness at one point means beats; a fixed pattern across space means interference.
Is the Doppler effect for a moving source the same as for a moving observer?
No, not for sound. A moving observer changes the rate at which unchanged wavefronts are encountered, giving $\nu'=\nu(v\pm v_o)/v$, whereas a moving source alters the wavelength of the fronts it emits, giving $\nu'=\nu v/(v\mp v_s)$. For the same speed the two produce different frequency shifts, and the difference grows at higher speeds. The asymmetry exists because sound needs a medium that defines a rest frame; for light there is no medium and only the relative velocity matters.
Why is Newton's calculated speed of sound about $15\%$ too low, and how did Laplace fix it?
Newton assumed the compressions and rarefactions are isothermal, so the elastic modulus is just $P$ and $v=\sqrt{P/\rho}\approx280\ \mathrm{m\,s^{-1}}$ for air. In reality the oscillations are so rapid that no heat flows, so the process is adiabatic and the correct modulus is $\gamma P$. Laplace's formula $v=\sqrt{\gamma P/\rho}$ multiplies Newton's value by $\sqrt{\gamma}=\sqrt{1.4}=1.18$, giving $331\ \mathrm{m\,s^{-1}}$, in agreement with experiment.
Can transverse mechanical waves travel through the body of a gas or liquid?
No. A transverse wave needs a restoring force against shearing, and fluids in bulk have no shear elasticity, so they cannot sustain transverse mechanical waves internally — only longitudinal (pressure) waves. This is why sound in air and water is longitudinal, and why an earthquake's transverse S-waves are stopped by the Earth's liquid outer core while its longitudinal P-waves pass through. (Surface waves on a liquid are a separate, gravity-driven case.)
Does frequency or wavelength change when a wave crosses into a new medium?
The frequency is fixed by the source and is conserved across a boundary — the far medium is driven at exactly the rate the near medium delivers oscillations. What changes is the speed (a property of the new medium) and therefore the wavelength, through $\lambda=v/\nu$. Sound passing from air into water keeps its frequency but its wavelength lengthens because the speed jumps from about $340$ to $1480\ \mathrm{m\,s^{-1}}$.
How can a wave carry energy if the particles just stay put on average?
Each oscillating element does work on its neighbour ahead of it, handing energy along the medium in a relay while itself only moving back and forth. Over one cycle a particle returns to where it started, but during that cycle it has passed on kinetic and potential energy. For a harmonic wave the time-averaged power transmitted is $\langle P\rangle=\tfrac12\mu\omega^2A^2v$, showing energy flows continuously even though no matter is transported.
Are overtones and harmonics the same thing?
Not in general. A harmonic is a mode whose frequency is an integer multiple of the fundamental; an overtone is simply any mode above the fundamental. For a string or an air column the overtones happen to coincide with harmonics, so people use the words loosely. But for a drumhead or a bell the higher modes are not integer multiples of the fundamental, so those overtones are inharmonic and the object has no single clear pitch.
What decides the pitch and the loudness of a sound?
Pitch is determined by the frequency $\nu$ — a higher frequency is a higher pitch — while loudness is determined by the intensity, which scales as the square of the amplitude, $I\propto A^2$ (and also as $\omega^2$). The two are independent: you can play a note louder without altering its pitch by increasing the amplitude at fixed frequency. Confusing 'higher' (pitch) with 'louder' (amplitude) is a common slip.

Trap-answer taxonomy

Trap: Confusing particle velocity with wave velocity

Students set the speed at which a string element moves equal to the wave speed, or plug $A\omega$ into $v=\nu\lambda$.

Fix: Keep them separate: wave speed $v=\omega/k$ is constant and set by the medium; particle velocity $\partial y/\partial t=-A\omega\cos(kx-\omega t)$ oscillates with peak $A\omega$. They are equal only by coincidence of numbers, never by principle.

Trap: Thinking the medium is carried forward by the wave

Because a wave clearly 'moves', students imagine air or water is transported along with it, especially in longitudinal sound.

Fix: Only the disturbance (energy) propagates; each particle oscillates about a fixed point and returns. Picture a cork bobbing on ripples or a fixed dust speck vibrating in a sound field — neither drifts downstream.

Trap: Using Newton's isothermal formula for sound in gases

Applying $v=\sqrt{P/\rho}$ gives an answer about $15\%$ too low because it wrongly assumes the compressions are isothermal.

Fix: Sound compressions are adiabatic, so always use Laplace's $v=\sqrt{\gamma P/\rho}=\sqrt{\gamma RT/M}$. The extra factor $\sqrt{\gamma}$ is exactly the missing $\approx18\%$.

Trap: Getting the Doppler signs backwards

Students memorise $\nu'=\nu(v\pm v_o)/(v\mp v_s)$ but pick the wrong signs, especially for the source in the denominator.

Fix: Reason physically instead of memorising: any approach must raise the pitch and any recession lower it. Choose signs so an approaching source shrinks the denominator (raising $\nu'$) and an approaching observer enlarges the numerator.

Trap: Expecting even harmonics in a closed pipe

Students apply the open-pipe series $\nu_n=nv/2L$ to a one-end-closed pipe and predict $2\nu_1,4\nu_1,\dots$.

Fix: A closed pipe forces a node at the closed end and an antinode at the open end, so only odd harmonics survive: $\nu_n=(2n-1)v/4L=\nu_1,3\nu_1,5\nu_1,\dots$. Its fundamental is also half that of an open pipe of the same length.

Trap: Treating beats and interference as the same effect

Students blur the two because both arise from superposition, and try to use path difference to explain beats.

Fix: Beats are interference in time from two slightly different frequencies at one place ($\nu_{beat}=|\nu_1-\nu_2|$); spatial interference is from two equal frequencies at different path differences. Time-varying loudness versus a fixed spatial pattern is the discriminator.

Trap: Forgetting frequency is conserved across a boundary

When a wave enters a new medium students change the frequency along with the speed, or hold the wavelength fixed.

Fix: The source sets the frequency, which stays constant across any boundary; the speed changes with the medium and the wavelength adjusts via $\lambda=v/\nu$. Only $v$ and $\lambda$ change on refraction, never $\nu$.

Trap: Assuming wave speed depends on amplitude or frequency

Students think a louder or higher-pitched sound travels faster, or that a bigger pluck sends a faster pulse down a string.

Fix: For linear mechanical waves the speed depends only on the medium ($v=\sqrt{T/\mu}$ or $\sqrt{\gamma P/\rho}$). Amplitude and frequency do not appear, so they change the wavelength or energy carried, never the speed.

🚪 Dive Deeper Mystery room · 41 discoveries

Discovered 0 / 41

JEE Advanced archive — DISCOVER → ATTEMPT → REVEAL

A steel wire of length $0.72$ m has a mass of $5.0\ \mathrm{g}$ and is under a tension of $60$ N. Find the speed of a transverse wave on it.

Attempt, then reveal full solution
Linear mass density $\mu=m/L=\dfrac{5.0\times10^{-3}}{0.72}=6.94\times10^{-3}\ \mathrm{kg\,m^{-1}}$. Then $v=\sqrt{T/\mu}=\sqrt{\dfrac{60}{6.94\times10^{-3}}}=\sqrt{8.64\times10^{3}}\approx93\ \mathrm{m\,s^{-1}}$. Only $T$ and $\mu$ enter — the wave's frequency and amplitude are irrelevant to its speed.

NCERT XI Example 14.3 (adapted)

Estimate the speed of sound in air at STP using Laplace's formula. Take $P=1.01\times10^5$ Pa, $\rho=1.29\ \mathrm{kg\,m^{-3}}$, $\gamma=1.4$.

Attempt, then reveal full solution
$v=\sqrt{\gamma P/\rho}=\sqrt{\dfrac{1.4\times1.01\times10^{5}}{1.29}}=\sqrt{1.096\times10^{5}}\approx331\ \mathrm{m\,s^{-1}}$. Newton's isothermal value $\sqrt{P/\rho}\approx280\ \mathrm{m\,s^{-1}}$ is $15\%$ low; the factor $\sqrt{\gamma}=\sqrt{1.4}=1.18$ supplies the correction.

NCERT XI Example 14.4

A pipe $30.0$ cm long is open at both ends. Which harmonic mode resonates a $1.1$ kHz source? Take $v=330\ \mathrm{m\,s^{-1}}$. Will it resonate the same source if one end is closed?

Attempt, then reveal full solution
Open pipe: $\nu_n=nv/2L=n\dfrac{330}{2(0.30)}=550n$ Hz. Setting $550n=1100$ gives $n=2$, the second harmonic. Closed pipe: $\nu_n=(2n-1)\dfrac{v}{4L}=(2n-1)275$ Hz, giving $275,825,1375,\dots$ Hz. Since $1100$ is not an odd multiple of $275$, a one-end-closed pipe of the same length will not resonate the $1.1$ kHz source.

NCERT XI Example 14.5

Two sitar strings A and B playing the note 'Dha' are slightly out of tune and produce $5$ beats per second. The tension in B is increased slightly and the beat frequency rises to $7$ Hz. If A is $427$ Hz, what was the original frequency of B?

Attempt, then reveal full solution
Beats give $|\nu_A-\nu_B|=5$, so $\nu_B=427\pm5=422$ or $432$ Hz. Increasing B's tension raises $\nu_B$. If $\nu_B$ were $432$, raising it moves it further from $427$ and increases beats — consistent. If $\nu_B$ were $422$, raising it would first decrease beats. Since the beat frequency increased, $\nu_B=432$ Hz.

NCERT XI Example 14.6 (adapted)

A wave travelling along a string is described by $y(x,t)=0.005\sin(80.0x-3.0t)$ in SI units. Find the amplitude, wavelength, period and speed.

Attempt, then reveal full solution
Comparing with $y=A\sin(kx-\omega t)$: amplitude $A=0.005\ \mathrm{m}=5\ \mathrm{mm}$; $k=80.0\ \mathrm{rad\,m^{-1}}$ so $\lambda=2\pi/k=7.85\times10^{-2}\ \mathrm{m}=7.85\ \mathrm{cm}$; $\omega=3.0\ \mathrm{rad\,s^{-1}}$ so $T=2\pi/\omega=2.09\ \mathrm{s}$; speed $v=\omega/k=3.0/80.0=3.75\times10^{-2}\ \mathrm{m\,s^{-1}}$.

NCERT XI Example 14.2

A source of frequency $500$ Hz moves toward a stationary observer at $30\ \mathrm{m\,s^{-1}}$. Take $v=340\ \mathrm{m\,s^{-1}}$. Find the observed frequency. What if instead the observer moves toward the stationary source at $30\ \mathrm{m\,s^{-1}}$?

Attempt, then reveal full solution
Source approaching: $\nu'=\nu\dfrac{v}{v-v_s}=500\dfrac{340}{340-30}=500\dfrac{340}{310}=548.4$ Hz. Observer approaching: $\nu'=\nu\dfrac{v+v_o}{v}=500\dfrac{340+30}{340}=500\dfrac{370}{340}=544.1$ Hz. The two are not equal — the Doppler effect for sound is asymmetric between source and observer motion.

NCERT XI §14.8 (authored)

A car sounding a horn of $400$ Hz approaches a large wall at $10\ \mathrm{m\,s^{-1}}$. Take $v=340\ \mathrm{m\,s^{-1}}$. What beat frequency does the driver hear between the direct horn and the echo?

Attempt, then reveal full solution
The wall receives $\nu_1=400\dfrac{v}{v-v_c}=400\dfrac{340}{330}$ (source approaching), then re-emits it; the driver, now a moving observer approaching that stationary re-emitter, hears $\nu_2=\nu_1\dfrac{v+v_c}{v}=400\dfrac{340}{330}\cdot\dfrac{350}{340}=400\dfrac{350}{330}=424.2$ Hz. Beat frequency $=|424.2-400|\approx24$ Hz.

JEE-style (authored); NCERT XI §14.8

A string of length $L$ fixed at both ends vibrates in its third harmonic. How many nodes and antinodes are present, and what is the wavelength?

Attempt, then reveal full solution
For the $n$th harmonic $L=n\lambda/2$, so the third harmonic has $\lambda=2L/3$. A string fixed at both ends in its $n$th harmonic has $n+1$ nodes (including the two ends) and $n$ antinodes. For $n=3$: $4$ nodes and $3$ antinodes.

NCERT XI §14.6 (authored)

Compare the speed of sound in helium and in oxygen at the same temperature. Take $\gamma_{He}=5/3$, $M_{He}=4$; $\gamma_{O_2}=7/5$, $M_{O_2}=32$ (g/mol).

Attempt, then reveal full solution
$v=\sqrt{\gamma RT/M}$, so $\dfrac{v_{He}}{v_{O_2}}=\sqrt{\dfrac{\gamma_{He}/M_{He}}{\gamma_{O_2}/M_{O_2}}}=\sqrt{\dfrac{(5/3)/4}{(7/5)/32}}=\sqrt{\dfrac{0.4167}{0.04375}}=\sqrt{9.52}\approx3.09$. Sound is about three times faster in helium — why a helium-filled voice sounds high-pitched.

JEE-style (authored); NCERT XI §14.4

The equation of a standing wave on a string is $y=0.04\sin(5\pi x)\cos(200\pi t)$ (SI). Find the positions of the nodes and the frequency of vibration.

Attempt, then reveal full solution
Nodes occur where $\sin(5\pi x)=0$, i.e. $5\pi x=m\pi\Rightarrow x=m/5=0,0.2,0.4,\dots\ \mathrm{m}$, spaced $0.2$ m apart. The time factor $\cos(200\pi t)$ gives $\omega=200\pi$, so $\nu=\omega/2\pi=100$ Hz. (The component travelling-wave speed is $v=\omega/k=200\pi/5\pi=40\ \mathrm{m\,s^{-1}}$.)

JEE-style (authored); NCERT XI §14.6

A harmonic wave on a string has $A=2.0$ mm, $\nu=50$ Hz on a string with $\mu=0.02\ \mathrm{kg\,m^{-1}}$ under tension $80$ N. Find the average power transmitted.

Attempt, then reveal full solution
$v=\sqrt{T/\mu}=\sqrt{80/0.02}=\sqrt{4000}=63.2\ \mathrm{m\,s^{-1}}$; $\omega=2\pi\nu=314\ \mathrm{rad\,s^{-1}}$. Average power $\langle P\rangle=\tfrac{1}{2}\mu\omega^2A^2v=\tfrac{1}{2}(0.02)(314)^2(2.0\times10^{-3})^2(63.2)=\tfrac12(0.02)(9.86\times10^4)(4\times10^{-6})(63.2)\approx0.25\ \mathrm{W}$.

JEE-style (authored); Rayleigh

A tuning fork of unknown frequency gives $4$ beats/s with a standard $256$ Hz fork. On loading the unknown fork with a little wax the beat frequency drops to $2$/s. Find the unknown frequency.

Attempt, then reveal full solution
Beats give the unknown as $256\pm4=252$ or $260$ Hz. Loading with wax always lowers a fork's frequency. If the fork were $252$, lowering it moves it further from $256$ and beats would rise to $>4$; if it were $260$, lowering it moves it toward $256$ and beats fall — matching the observed drop to $2$. So the unknown frequency is $260$ Hz.

JEE-style (authored); NCERT XI §14.7

For the fundamental mode of a $1.0$ m open pipe and a $1.0$ m closed pipe (one end closed), find the ratio of their fundamental frequencies. Take $v=340\ \mathrm{m\,s^{-1}}$.

Attempt, then reveal full solution
Open pipe fundamental: $\nu_{open}=v/2L=340/2.0=170$ Hz. Closed pipe fundamental: $\nu_{closed}=v/4L=340/4.0=85$ Hz. Ratio $\nu_{open}:\nu_{closed}=2:1$ — the closed pipe sounds an octave lower for the same length, and its overtones are only $85,255,425,\dots$ Hz (odd harmonics).

JEE-style (authored); NCERT XI §14.6

A transverse harmonic wave $y=3.0\sin(36t+0.018x+\pi/4)$ (cm, s) travels on a string. State its direction of travel, amplitude and speed.

Attempt, then reveal full solution
The combination $(36t+0.018x)$ has the same sign on $t$ and $x$, so the wave moves in the $-x$ direction. Amplitude $A=3.0$ cm. Here $\omega=36\ \mathrm{rad\,s^{-1}}$, $k=0.018\ \mathrm{rad\,cm^{-1}}$, so $v=\omega/k=36/0.018=2000\ \mathrm{cm\,s^{-1}}=20\ \mathrm{m\,s^{-1}}$.

NCERT XI Exercise 14.8 (adapted)

A steel rod $100$ cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations is $2.53$ kHz. What is the speed of sound in steel?

Attempt, then reveal full solution
A rod clamped at the middle has a displacement node at the clamp and antinodes at both free ends, so the fundamental has $L=\lambda/2$, i.e. $\lambda=2L=2.0$ m. Then $v=\nu\lambda=2.53\times10^3\times2.0=5.06\times10^3\ \mathrm{m\,s^{-1}}$.

NCERT XI Exercise 14.18 (adapted)

Two waves $y_1=A\sin(kx-\omega t)$ and $y_2=A\sin(kx+\omega t)$ superpose. Show that the result is a standing wave and find the amplitude of oscillation at $x$.

Attempt, then reveal full solution
Adding, $y=y_1+y_2=A[\sin(kx-\omega t)+\sin(kx+\omega t)]=2A\sin kx\cos\omega t$, using $\sin C+\sin D=2\sin\tfrac{C+D}{2}\cos\tfrac{C-D}{2}$. There is no $(kx\mp\omega t)$ argument, so the pattern does not travel — it is a standing wave. The amplitude at $x$ is $2A\sin kx$: zero at nodes ($\sin kx=0$) and maximum $2A$ at antinodes.

NCERT XI §14.6 (derivation)

A whistle of frequency $1000$ Hz is whirled in a horizontal circle of radius $1.0$ m at angular speed $10\ \mathrm{rad\,s^{-1}}$. A distant listener in the plane hears frequencies between which limits? Take $v=340\ \mathrm{m\,s^{-1}}$.

Attempt, then reveal full solution
Tangential speed $v_s=\omega r=10\times1.0=10\ \mathrm{m\,s^{-1}}$. Maximum frequency when the whistle moves straight toward the listener: $\nu_{max}=1000\dfrac{340}{340-10}=1030.3$ Hz. Minimum when moving away: $\nu_{min}=1000\dfrac{340}{340+10}=971.4$ Hz. The listener hears the pitch swing between about $971$ and $1030$ Hz.

JEE-style (authored); NCERT XI §14.8

A string under tension $T$ has fundamental $\nu_1$. By what factor must the tension change to raise the fundamental by one octave (double it), keeping length and $\mu$ fixed?

Attempt, then reveal full solution
$\nu_1=\dfrac{1}{2L}\sqrt{T/\mu}\propto\sqrt{T}$. To double $\nu_1$ we need $\sqrt{T'/T}=2$, so $T'=4T$: the tension must be quadrupled. This steep dependence is why over-tightening a string to raise its pitch risks snapping it.

JEE-style (authored); Mersenne

The displacement of a medium particle in a sound wave is $s=6.0\times10^{-6}\cos(1900t-5.7x)$ (SI). Find the frequency, wavelength and speed of the wave.

Attempt, then reveal full solution
$\omega=1900\ \mathrm{rad\,s^{-1}}\Rightarrow\nu=\omega/2\pi=302$ Hz. $k=5.7\ \mathrm{rad\,m^{-1}}\Rightarrow\lambda=2\pi/k=1.10$ m. Speed $v=\omega/k=1900/5.7=333\ \mathrm{m\,s^{-1}}$, consistent with sound in air.

JEE-style (authored); NCERT XI §14.3

Show why the speed of sound in an ideal gas is independent of pressure at fixed temperature.

Attempt, then reveal full solution
Start from $v=\sqrt{\gamma P/\rho}$. For an ideal gas $PV=nRT$ gives $P/\rho=RT/M$. Substituting, $v=\sqrt{\gamma RT/M}$, which contains no $P$. Increasing pressure at constant $T$ raises $\rho$ in proportion, so the ratio $P/\rho$ — and hence $v$ — is unchanged. Only temperature and molar mass matter.

NCERT XI §14.4 (derivation)

A pipe closed at one end resonates at consecutive frequencies $350$ Hz and $450$ Hz with no resonance between them. Find the fundamental frequency and the pipe length. Take $v=350\ \mathrm{m\,s^{-1}}$.

Attempt, then reveal full solution
A closed pipe has frequencies $(2n-1)\nu_1$. Consecutive resonances differ by $2\nu_1$, so $2\nu_1=450-350=100\Rightarrow\nu_1=50$ Hz. Then $350=(2n-1)50\Rightarrow2n-1=7$ (the 7th harmonic) — consistent. Length: $\nu_1=v/4L\Rightarrow L=v/4\nu_1=350/(4\times50)=1.75$ m.

JEE-style (authored); NCERT XI §14.6

Two identical sources emit sound of frequency $680$ Hz in phase. A listener stands so that the path difference from the two sources is $0.25$ m. Take $v=340\ \mathrm{m\,s^{-1}}$. Is the interference constructive or destructive there?

Attempt, then reveal full solution
Wavelength $\lambda=v/\nu=340/680=0.50$ m. Path difference $=0.25\ \mathrm{m}=\lambda/2$, an odd multiple of half a wavelength, so the waves arrive antiphase and interfere destructively — the listener hears a minimum.

JEE-style (authored); Young

A wave pulse on a light string reflects off a junction with a much heavier string. Describe the reflected pulse. What if it reflected off a junction with a much lighter string?

Attempt, then reveal full solution
A heavier (denser) string behaves like a nearly rigid boundary, so the reflected pulse is inverted (phase change $\pi$). A much lighter string behaves like a nearly free end, so the reflected pulse is upright (no phase change). Part of the energy is also transmitted in each case, but the sign of the reflected pulse is fixed by whether the boundary is 'stiffer' or 'freer'.

NCERT XI §14.6 (concept)

The fundamental frequency of a sonometer wire is $200$ Hz. If the vibrating length is halved and the tension is made $9$ times larger, what is the new fundamental frequency?

Attempt, then reveal full solution
$\nu_1=\dfrac{1}{2L}\sqrt{T/\mu}$. Halving $L$ multiplies $\nu_1$ by $2$; making $T$ nine times larger multiplies it by $\sqrt{9}=3$. Combined factor $2\times3=6$, so the new frequency is $6\times200=1200$ Hz.

JEE-style (authored); Mersenne

Sound of frequency $340$ Hz enters from air ($v_{air}=340\ \mathrm{m\,s^{-1}}$) into water ($v_{water}=1480\ \mathrm{m\,s^{-1}}$). Find the wavelength in each medium.

Attempt, then reveal full solution
Frequency is set by the source and does not change on crossing a boundary. In air $\lambda_{air}=v_{air}/\nu=340/340=1.0$ m. In water $\lambda_{water}=v_{water}/\nu=1480/340=4.35$ m. The wavelength stretches because the speed rises while $\nu$ stays fixed.

JEE-style (authored); NCERT XI §14.4

An observer moving at $34\ \mathrm{m\,s^{-1}}$ recedes from a stationary $600$ Hz source; separately the source instead moves toward a stationary observer at $34\ \mathrm{m\,s^{-1}}$. Compare the two observed frequencies. Take $v=340\ \mathrm{m\,s^{-1}}$.

Attempt, then reveal full solution
Observer receding (source at rest): $\nu'=600\dfrac{v-v_o}{v}=600\dfrac{306}{340}=540$ Hz. Source approaching (observer at rest): $\nu''=600\dfrac{v}{v-v_s}=600\dfrac{340}{306}=666.7$ Hz. The magnitudes of the shift differ ($60$ Hz down vs $66.7$ Hz up) — again the source/observer asymmetry.

JEE-style (authored); NCERT XI §14.8

A string vibrates so that the distance between a node and the next antinode is $0.10$ m and the frequency is $500$ Hz. Find the speed of the component travelling waves.

Attempt, then reveal full solution
The node-to-adjacent-antinode distance is a quarter wavelength: $\lambda/4=0.10\Rightarrow\lambda=0.40$ m. Then $v=\nu\lambda=500\times0.40=200\ \mathrm{m\,s^{-1}}$.

JEE-style (authored); NCERT XI §14.6

Given $v=\sqrt{\gamma RT/M}$, by how much does the speed of sound in air change when the temperature rises from $27^\circ$C to $47^\circ$C? Take $v_{27}=347\ \mathrm{m\,s^{-1}}$.

Attempt, then reveal full solution
$v\propto\sqrt{T}$ with $T$ in kelvin. $T_1=300$ K, $T_2=320$ K, so $\dfrac{v_2}{v_1}=\sqrt{320/300}=\sqrt{1.0667}=1.0328$. Thus $v_2=347\times1.0328\approx358\ \mathrm{m\,s^{-1}}$, an increase of about $11\ \mathrm{m\,s^{-1}}$ — close to the $0.6\ \mathrm{m\,s^{-1}}$ per $^\circ$C rule ($0.6\times20=12$).

JEE-style (authored); NCERT XI §14.4

📊 Rank Predictor JoSAA/MCC-calibrated

Disclaimer: These bands are approximate and illustrative, built from publicly reported JoSAA 2023-24 closing-rank trends. Actual ranks depend on the number of candidates, paper difficulty and normalisation in a given year, and vary by category and shift. Use them for orientation, not as a guarantee.
What this does: Waves is a compact, high-yield chapter for JEE Main, typically contributing one to two questions each year, and together with oscillations it forms a substantial block of the mechanics-and-waves section. In JEE Advanced it appears mostly inside multi-concept problems that chain wave motion with simple harmonic motion or with the resonance of strings and air columns. The bands below map an approximate overall JEE Main percentile to a JoSAA closing-rank range, to help you gauge where a given performance sits. They are indicative only.
How to read it: enter your score on a full chapter mock below. The tool maps it — via historical JEE marks→percentile→JoSAA closing-rank data — to the percentile and All-India-Rank band a student at that level typically lands in. It is a calibration signal for THIS chapter's mastery, not a full-exam rank.
Chapter-mock scorePercentile bandProjected AIR band
99.5+ percentile99.5+$\lt 1500$
99.0-99.5 percentile99.0-99.5$1500-4000$
98.0-99.0 percentile98.0-99.0$4000-9000$
95.0-98.0 percentile95.0-98.0$9000-25000$
90.0-95.0 percentile90.0-95.0$25000-55000$
80.0-90.0 percentile80.0-90.0$55000-120000$
$\lt 80$ percentile$\lt 80$$\gt 120000$

JoSAA 2023-24 closing-rank trends (indicative)

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