Chapter 02 · Waves & Sound · Full book

Waves and Sound: Disturbances That Travel

Core intuition: Chapter Roadmap

A wave carries a disturbance — and with it energy and information — from one place to another while the medium itself stays put. This chapter builds the physics of waves from a single picture (a travelling shape) up to the subtleties of sound, the Doppler effect, and resonance.

Describing a wave — the travelling-wave function \(y(x,t) = A\sin(kx \mp \omega t)\), and the meaning of \(k\), \(\omega\), \(\lambda\), \(f\), and \(v = \omega/k = f\lambda\).
The wave equation and wave speed — why a disturbance moves, and how \(v = \sqrt{T_s/\mu}\) for strings and \(v = \sqrt{B/\rho} = \sqrt{\gamma RT/M}\) for sound emerge from the medium's stiffness and inertia.
Energy and intensity — the power a wave carries, the inverse-square law, and the logarithmic decibel scale.
Superposition — interference of coherent sources, path difference, and beats from nearly equal frequencies.
Standing waves — the normal modes of strings and air columns, harmonics, and the boundary conditions that select them.
Sound — displacement and pressure waves, and why a displacement node is a pressure antinode.
The Doppler effect and shock waves — how relative motion shifts frequency, and what happens once a source outruns its own sound.
Resonance — the resonance-tube method, end corrections, and the tuning of pipes and strings.

One idea threads through all of it: a wave is a pattern that moves, while the matter it moves through only oscillates in place.

Perplexing Questions

Perplexing questions

Before we begin, let these questions sit with you. Do not reach for formulas. You will return to each one—with full answers—before this chapter ends.

Energy Without Matter: Stand at the shore and watch an ocean wave pass. A cork floating on the surface bobs up and down but does not travel with the wave—it returns to almost exactly where it started. Yet the wave clearly carries energy: it can knock you off your feet, erode a cliff, and drive a turbine. How does energy travel from one place to another when the matter carrying it goes nowhere?
Destructive Silence: Noise-cancelling headphones play a carefully engineered sound into your ear at the same moment the ambient noise arrives. The result—if the engineering is perfect—is complete silence. Two sounds, both carrying energy, combine to produce nothing. Where does the energy go?
The Doppler Shift: A police siren emits sound at a fixed frequency—the siren itself has not changed, the speed of sound in air has not changed. Yet you hear a higher pitch as the car approaches and a lower pitch as it recedes. Nothing about the source changed during the encounter. What did?
Resonance and Destruction: A wine glass has one special pitch at which it will shatter if driven hard enough, but survives the same total energy delivered at any other frequency. Why does the frequency of the applied force matter more than the magnitude of the force itself?

Each question targets a trap that catches students every year.
By the end of this chapter, every one of them will be transparent.

What Is a Wave?

Chapter studied a single oscillator: one mass, one spring, one degree of freedom. A wave is what happens when oscillators are coupled together—when the displacement of one mass pulls on its neighbour, which pulls on its neighbour, and so on down the line.

The disturbance does not stay localised. It propagates.

From One Oscillator to a Continuum

Imagine a long row of masses connected by identical springs, all at rest. Push the first mass sideways. The spring connecting it to the second mass stretches, pulling the second mass along. By the time the second mass has begun to move, the first has already been pulled back by the third spring. The result is a pattern of displacements that moves down the row at a definite speed, long after the original push has ended.

That travelling pattern of displacements is a wave.

Core intuition: What a Wave Is and Is Not

A wave is a disturbance that propagates through a medium by coupling neighbouring elements together. The medium itself does not travel with the wave: each element oscillates about its own equilibrium. What travels is the pattern—the organised correlation between the motions of neighbouring elements. Contrast this with a thrown ball, where matter moves with the energy. In a wave, energy is transported; mass is not.

The Two Transport Mechanisms

Every travelling wave transports two things along with it: energy and momentum. Neither requires matter to flow. This is not obvious, but it is experimentally confirmed: a wave incident on an absorbing wall exerts a radiation pressure, transferring momentum to the wall despite zero net mass flow.

The depth of the idea is that nature allows two fundamentally different ways to transport energy: by moving matter (convection, projectiles) and by propagating ordered oscillations (waves). Most of classical physics, and all of electromagnetism, exploits the second mechanism.

Connection to SHM

The single-oscillator result from Chapter is the local physics underlying every wave. At any fixed point in the medium, as a wave passes, that point executes simple harmonic motion with the wave's angular frequency \(\omega\). The wave equation we will derive in Section is, at each spatial point, nothing more than the SHM equation \(\ddot{y} = -\omega^2 y\), generalised to describe how the oscillation pattern propagates in space.

The parameters you know from SHM—amplitude \(A\), angular frequency \(\omega\), phase constant \(\phi\)—all reappear in wave physics. One new quantity, the wavenumber \(k\), encodes how the oscillation varies in space rather than in time. The relationship between \(k\) and \(\omega\) will give us the wave speed.

Classification of Waves

Waves are classified along two independent axes:

By the direction of oscillation relative to propagation:

Transverse waves: the medium oscillates perpendicular to the direction of propagation. A wave on a string, electromagnetic waves.
Longitudinal waves: the medium oscillates parallel to the direction of propagation. Sound in air, compression waves in a slinky.

By whether they require a material medium:

Mechanical waves: require a medium (string waves, sound, water waves). The restoring force is an elastic or pressure force within the medium.
Electromagnetic waves: require no medium. They propagate through vacuum at speed \(c = 3\times 10^8\,\mathrm{m\,s^{-1}}\). This chapter treats only mechanical waves.

The distinction between transverse and longitudinal is physical, not just notational: a solid can support both types, while a gas or liquid (which resists compression but not shear) can only sustain longitudinal waves.

What This Chapter Covers

We build the mathematical description of a sinusoidal wave in a later section, derive the wave speed for strings and sound in a later section, then treat energy and intensity, superposition, standing waves, and the Doppler effect in sequence. Sound physics—the Doppler effect, resonance in pipes and strings, beats—forms the second half of the chapter.

Misconception alert

“The medium moves with the wave.” Wrong. Each part of the medium oscillates about a fixed equilibrium. The ocean surface does not flow toward the shore with the waves: a floating cork traces a roughly circular or elliptical path and returns nearly to its starting point after each wave passes. What travels is the phase pattern, not the medium.
“A louder sound travels faster.” Wrong. The speed of a mechanical wave depends on the medium's elastic and inertial properties, not on the amplitude of the disturbance. In a given medium at a given temperature, all sounds—soft or loud, high-pitched or low—travel at the same speed. Amplitude affects intensity, not speed.

✓ Checkpoint · Before We Begin

A long horizontal slinky is stretched on a frictionless table. One end is given a single sharp push parallel to the slinky's axis. (a) Is the resulting wave transverse or longitudinal? (b) After the pulse has passed a point in the middle, does that point end up displaced from its original position?
(a) Longitudinal: the displacement (compression/rarefaction) is parallel to the direction of propagation. (b) No. Each coil of the slinky oscillates about its equilibrium and returns to it after the pulse passes; the medium does not translate with the wave.
A seismic event produces both P-waves (longitudinal, faster) and S-waves (transverse, slower) that travel through rock. S-waves cannot propagate through liquid. Using the classification in this section, explain why.
Transverse waves require the medium to exert a restoring force when sheared (displaced perpendicular to propagation). Liquids and gases resist compression but not shear—they have zero shear modulus—so they cannot provide the restoring force that sustains a transverse wave. P-waves are longitudinal (compression/expansion) and can propagate in any medium that resists compression.
In Chapter you learned that \(\omega = 2\pi f\). A wave on a string has frequency \(f = 50\,\mathrm{Hz}\) and wavelength \(\lambda = 0.60\,\mathrm{m}\). Without any new formula, argue from physical reasoning what the wave speed must be.
In one period \(T = 1/f\), the wave pattern advances by exactly one wavelength \(\lambda\), since that is the definition of wavelength. Speed \(= \lambda/T = \lambda f = 0.60 \times 50 = 30\,\mathrm{m\,s^{-1}}\). The formula \(v = f\lambda\) follows directly from the definitions of \(f\) and \(\lambda\), not from any dynamical assumption.

Transverse and Longitudinal Waves

◈ Transverse vs longitudinal

live bench

Flip a lattice between a transverse shake and a longitudinal push — same wave equation, different particle motion.

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↗ open in the Simulations page

Fig 10.1 Transverse vs. longitudinal waves: the medium oscillates perpendicular to the propagation direction (transverse) or parallel to it, producing compressions and rarefactions (longitudinal).

The classification introduced in Section deserves a more precise treatment. The direction of the medium's oscillation relative to the direction of wave propagation is not a minor notational distinction—it determines what kinds of media can support the wave, what physical quantity carries the restoring force, and how the wave interacts with boundaries.

Transverse Waves: The String as Prototype

Fix one end of a long taut string. Give the other end a single transverse flick—a displacement perpendicular to the string's length. A pulse travels along the string at speed \(v\) (derived below). The string elements between source and pulse are undisturbed; the elements that the pulse has already passed have returned to rest.

For a periodic transverse wave, drive the end sinusoidally. Each element of the string traces vertical SHM. At any instant, the string's shape is a sinusoidal snapshot in space, with spatial period \(\lambda\) (the wavelength). At any fixed position, the displacement varies sinusoidally in time with period \(T = 1/f\).

What provides the restoring force? Tension \(T_s\) in the string. When a small segment is displaced vertically, the tension in the neighbouring strings has a net downward component that pulls it back toward the axis. The stiffer the string (higher tension), the stronger the restoring force and the faster the wave propagates. The heavier the string (higher linear mass density \(\mu\)), the more inertia each segment has and the slower it responds. We will show in Section that \(v = \sqrt{T_s/\mu}\).

Which media support transverse mechanical waves? Only solids. A transverse wave requires that displaced elements be pulled back by a restoring force perpendicular to the propagation direction. This requires the medium to resist shear—relative lateral sliding between layers. Solids have a non-zero shear modulus; liquids and gases do not. Seismic S-waves (shear waves) travel through Earth's mantle (solid) but cannot penetrate the outer core (liquid)—this is how we know the outer core is liquid.

Longitudinal Waves: The Compressions That Travel

Now give the end of the slinky (or a column of air in a tube) a push parallel to its length. The first coil compresses the second, which compresses the third, producing a region of compression (higher than equilibrium density) that travels along the medium. Between successive compressions lies a region of rarefaction (lower than equilibrium density). The alternating compressions and rarefactions form the longitudinal wave pattern.

Each element of the medium oscillates back and forth along the direction of propagation—it is displaced from its equilibrium position, then restored by the pressure difference between the compressed and rarefied regions on either side.

What provides the restoring force? The bulk modulus \(B\) of the medium: its resistance to compression. A stiffer medium (larger \(B\)) responds more forcefully to compression, propagating the wave faster. We will show \(v = \sqrt{B/\rho}\), where \(\rho\) is the mass density.

Sound in air, sound in water, and P-waves (pressure waves) in rock are all longitudinal waves. Because every material resists compression to some degree, every medium can sustain longitudinal waves.

Displacement and Pressure in a Longitudinal Wave

Fig 10.2 Displacement and pressure in a longitudinal wave are 90° out of phase: a displacement antinode is a pressure node, and a displacement node is a pressure antinode.

A subtlety of longitudinal waves, exploited heavily in JEE problems, is that there are two natural ways to describe the wave: by the displacement of medium elements, or by the pressure variation in the medium.

Let the displacement wave be: \[ s(x,t) = s_0 \cos(kx - \omega t) \] where \(s\) measures the displacement of the element at equilibrium position \(x\) along the propagation axis, and \(s_0\) is the displacement amplitude.

The pressure excess (deviation from equilibrium pressure \(p_0\)) in the medium is related to the spatial gradient of displacement: \[ \Delta p(x,t) = -B\,\frac{\partial s}{\partial x} = B k s_0 \sin(kx - \omega t) = \Delta p_{\max}\sin(kx - \omega t) \] where the pressure amplitude is: \[ \Delta p_{\max} = B k s_0 \]

Core intuition: Displacement Nodes Are Pressure Antinodes

Comparing Eqs. and : the pressure wave is \(90°\) out of phase with the displacement wave. Where displacement is maximum (antinode), the spatial gradient of displacement is zero, so pressure variation is zero—a pressure node. Where displacement is zero (node), neighbouring elements are maximally compressed or rarefied—a pressure antinode. This \(90°\) phase shift between displacement and pressure is the source of a perennial JEE confusion: the open end of a pipe is a displacement antinode but a pressure node.

Comparison Table

Property	Transverse	Longitudinal
Oscillation direction	Perpendicular to propagation	Parallel to propagation
Restoring mechanism	Shear modulus / tension	Bulk modulus
Media supported	Solids only (mechanical)	Solids, liquids, gases
Examples	String, S-wave, light (EM)	Sound, P-wave, ultrasound
Can be polarised?	Yes	No

Example 1

A sound wave in air has displacement amplitude \(s_0 = 2.0\times 10^{-8}\,\mathrm{m}\) and frequency \(f = 1000\,\mathrm{Hz}\). The speed of sound in air is \(v = 340\,\mathrm{m\,s^{-1}}\) and the bulk modulus of air is \(B = 1.42\times 10^5\,\mathrm{Pa}\). (a) Find the wavenumber \(k\) and the pressure amplitude \(\Delta p_{\max}\). (b) At a point where the displacement is at its maximum, what is the pressure variation?

Solution. (a) Wavenumber: \[ k = \frac{\omega}{v} = \frac{2\pi f}{v} = \frac{2\pi \times 1000}{340} \approx 18.5\,\mathrm{rad\,m^{-1}} \] Pressure amplitude, using : \[ \Delta p_{\max} = B k s_0 = 1.42\times 10^5 \times 18.5 \times 2.0\times 10^{-8} \approx 0.0525\,\mathrm{Pa} \] (b) Where displacement is maximum, \(s = s_0\) and \(\cos(kx - \omega t) = 1\). From it, \(\Delta p \propto \sin(kx - \omega t) = 0\). The pressure variation is zero at a displacement maximum. This is the \(90°\) phase shift: displacement antinodes are pressure nodes.

✓ Checkpoint · Transverse and Longitudinal Waves

Can a transverse wave travel through (a) a steel rod, (b) water, (c) air? Justify each answer in one sentence using the concept of shear modulus.
(a) Yes: steel is a solid with non-zero shear modulus, capable of providing the lateral restoring force. (b) No: water is a liquid with zero shear modulus; no restoring force exists for transverse displacements. (c) No: same reason as water—gases cannot resist shear.
A sound wave in a pipe has the displacement description \(s = s_0\cos(kx - \omega t)\). Write down the pressure variation \(\Delta p\) at the same location. At which phase of the displacement cycle is the pressure variation largest in magnitude?
\(\Delta p = \Delta p_{\max}\sin(kx - \omega t)\). The pressure is largest in magnitude when \(\sin(kx-\omega t) = \pm 1\), i.e. when the displacement argument satisfies \(kx - \omega t = \pm\pi/2\)—exactly when \(\cos(kx-\omega t) = 0\), i.e. when the displacement is zero. The pressure is maximum when the displacement is zero, and zero when the displacement is at its amplitude.
The displacement amplitude of a \(500\,\mathrm{Hz}\) sound wave in air is doubled. (a) How does the pressure amplitude change? (b) How does the wave speed change?
(a) From \(\Delta p_{\max} = Bks_0\), doubling \(s_0\) doubles the pressure amplitude. (b) The wave speed \(v = \sqrt{B/\rho}\) does not depend on amplitude—it is unchanged. Speed is a property of the medium, not of the source.

Mathematical Description: The Wave Equation

◈ The progressive wave

live bench

Read y(x,t)=A sin(kx−ωt) as a snapshot and a movie at once; particle-velocity arrows track ∂y/∂t.

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↗ open in the Simulations page

We now write down the function that describes a sinusoidal travelling wave and derive the partial differential equation it satisfies. This equation—the wave equation—is one of the most important in all of physics.

The Sinusoidal Travelling Wave

Fig 10.3 Anatomy of a sinusoidal travelling wave: amplitude A, wavelength λ, and propagation at wave speed v. The faint curve is the same pattern an instant later.

Consider a transverse wave travelling in the \(+x\) direction along a string. At \(t = 0\), the string's shape is a sinusoidal snapshot: \[ y(x, 0) = A\sin(kx) \] where \(A\) is the amplitude and \(k\) is the wavenumber—the spatial analogue of angular frequency. Just as \(\omega = 2\pi/T\) counts radians per unit time, the wavenumber counts radians per unit length: \[ k = \frac{2\pi}{\lambda} \]

After time \(t\), the entire pattern has shifted by \(vt\) in the \(+x\) direction, without changing shape. The displacement of the element that was at \(x = 0\) at \(t = 0\) is now at \(x = vt\), so the value at position \(x\) at time \(t\) is what was at position \(x - vt\) at \(t = 0\): \[ \boxed{y(x,t) = A\sin(kx - \omega t)} \] where we have used \(vt = (\omega/k)t\) to write \(k(x - vt) = kx - \omega t\).

For a wave travelling in the \(-x\) direction, the sign inside flips: \[ y(x,t) = A\sin(kx + \omega t) \]

The most general form, including an initial phase \(\phi\), is: \[ y(x,t) = A\sin(kx - \omega t + \phi) \] or equivalently \(A\cos(kx - \omega t + \phi)\). The choice of sine or cosine is a matter of initial conditions.

Parameters and Their Relationships

Five parameters appear: \(A\), \(k\), \(\omega\), \(v\), and \(\phi\). They are not independent.

Name	Symbol	SI unit	Relation
Amplitude	\(A\)	m	set by source
Wavenumber	\(k\)	rad m\(^{-1}\)	\(k = 2\pi/\lambda\)
Ang. frequency	\(\omega\)	rad s\(^{-1}\)	\(\omega = 2\pi f\)
Wave speed	\(v\)	m s\(^{-1}\)	\(v = \omega/k = f\lambda\)
Phase constant	\(\phi\)	rad	set by initial conditions

The fundamental relation \(v = \omega/k = f\lambda\) follows purely from the definitions of \(k\) and \(\omega\)—it holds for any sinusoidal wave, regardless of the medium or restoring mechanism.

The Transverse Velocity of a Medium Element

Do not confuse the wave speed \(v\) (how fast the pattern moves) with the transverse velocity of a medium element (how fast the string moves up and down).

For the wave \(y(x,t) = A\sin(kx - \omega t)\), the transverse velocity of the element at position \(x\) is: \[ v_y(x,t) = \frac{\partial y}{\partial t} = -A\omega\cos(kx - \omega t) \]

This is SHM in time, with amplitude \(A\omega\). The maximum transverse speed of any element is \(v_{y,\max} = A\omega\)— exactly the \(v_{\max} = A\omega\) formula from an earlier chapter.

The transverse acceleration is: \[ a_y(x,t) = \frac{\partial^2 y}{\partial t^2} = -A\omega^2\sin(kx - \omega t) = -\omega^2 y(x,t) \]

Again: each element satisfies \(a_y = -\omega^2 y\), the SHM condition. A travelling wave is a collection of coupled SHM oscillators, each executing the same motion but shifted in phase.

Deriving the Wave Equation

Fig 10.4 A string element of length Δx and mass μΔx. The tension Tₛ at each end makes small angles θ₁, θ₂ with the horizontal; their net vertical component drives the element — giving the wave equation.

We now derive the partial differential equation that \(y(x,t)\) satisfies—the wave equation.

Birth of the formula: The Wave Equation from Newton's Second Law

Consider a small segment of a taut string of length \(\Delta x\), with linear mass density \(\mu\). The segment has mass \(\mu\,\Delta x\). The string makes small angles with the horizontal, so we use the small-angle approximation: \(\sin\theta \approx \tan\theta = \partial y/\partial x\). The net upward force on the segment comes from the tension \(T_s\) at the two ends. The vertical components are: \[ F_y = T_s\left.\frac{\partial y}{\partial x}\right|_{x+\Delta x} - T_s\left.\frac{\partial y}{\partial x}\right|_{x} \approx T_s\,\frac{\partial^2 y}{\partial x^2}\,\Delta x \] Newton's second law (\(F = ma\)) for the segment: \[ T_s\,\frac{\partial^2 y}{\partial x^2}\,\Delta x = \mu\,\Delta x\,\frac{\partial^2 y}{\partial t^2} \] Dividing by \(\mu\,\Delta x\): \[ \boxed{\frac{\partial^2 y}{\partial t^2} = \frac{T_s}{\mu}\,\frac{\partial^2 y}{\partial x^2}} \] Comparing with the general wave equation form \(\partial^2 y/\partial t^2 = v^2\,\partial^2 y/\partial x^2\), we read off: \[ v = \sqrt{\frac{T_s}{\mu}} \]

Equation is Newton's second law applied to an infinitesimal string element. The left side is the element's acceleration; the right side is (tension) \(\times\) (curvature of the string), which is the restoring force per unit mass. The wave equation is not a new law—it is \(F = ma\) in disguise.

Verification. Substitute \(y = A\sin(kx - \omega t)\) into : \[ \frac{\partial^2 y}{\partial t^2} = -A\omega^2\sin(kx-\omega t) \qquad \frac{\partial^2 y}{\partial x^2} = -Ak^2\sin(kx-\omega t) \] Equation requires \(\omega^2 = (T_s/\mu)\,k^2\), giving \(\omega/k = \sqrt{T_s/\mu} = v\). ✓

The Phase of the Wave

The argument \((kx - \omega t + \phi)\) is called the phase of the wave. A surface of constant phase—a wavefront—moves at the wave speed.

Setting \(kx - \omega t = \text{const}\) and differentiating: \[ k\,dx - \omega\,dt = 0 \implies \frac{dx}{dt} = \frac{\omega}{k} = v \]

Wavefronts travel at speed \(v\).

Two points separated by distance \(\Delta x\) in the propagation direction have a phase difference: \[ \Delta\phi = k\,\Delta x = \frac{2\pi\,\Delta x}{\lambda} \]

Points separated by one wavelength are in phase (\(\Delta\phi = 2\pi\)); points separated by half a wavelength are in antiphase (\(\Delta\phi = \pi\)). This is the spatial analogue of the temporal relationship \(\Delta\phi = \omega\,\Delta t\).

Example 2

A transverse sinusoidal wave on a string is described by: \[ y(x,t) = 0.040\sin(5.0x - 200t)\,\mathrm{m} \] where \(x\) is in metres and \(t\) in seconds. (a) Find the amplitude, wavenumber, angular frequency, wave speed, frequency, and wavelength. (b) Find the maximum transverse speed and maximum transverse acceleration of any element of the string. (c) Find the phase difference between two points separated by \(0.30\,\mathrm{m}\) along the string. (d) Write down the equation for a wave of the same amplitude and frequency travelling in the \(-x\) direction.

Solution. (a) Reading from \(y = A\sin(kx - \omega t)\): \[ A = 0.040\,\mathrm{m},\quad k = 5.0\,\mathrm{rad\,m^{-1}},\quad \omega = 200\,\mathrm{rad\,s^{-1}} \] Wave speed: \(v = \omega/k = 200/5.0 = 40\,\mathrm{m\,s^{-1}}\). Frequency: \(f = \omega/2\pi = 200/(2\pi) \approx 31.8\,\mathrm{Hz}\). Wavelength: \(\lambda = 2\pi/k = 2\pi/5.0 \approx 1.26\,\mathrm{m}\). (Check: \(v = f\lambda = 31.8 \times 1.26 \approx 40\,\mathrm{m\,s^{-1}}\). ✓) (b) Maximum transverse speed: \[ v_{y,\max} = A\omega = 0.040 \times 200 = 8.0\,\mathrm{m\,s^{-1}} \] Maximum transverse acceleration: \[ a_{y,\max} = A\omega^2 = 0.040 \times (200)^2 = 1600\,\mathrm{m\,s^{-2}} \] (c) Phase difference: \[ \Delta\phi = k\,\Delta x = 5.0 \times 0.30 = 1.5\,\mathrm{rad} \] (d) Replace \(kx - \omega t\) with \(kx + \omega t\): \[ y(x,t) = 0.040\sin(5.0x + 200t)\,\mathrm{m} \]

✓ Checkpoint · The Wave Equation

A wave on a string is described by \(y = 0.02\cos(3x + 120t)\,\mathrm{m}\) (\(x\) in m, \(t\) in s). (a) State the direction of propagation. (b) Find the wave speed. (c) Find the transverse velocity of the element at \(x = 0\) at \(t = 0\).
(a) Negative \(x\)-direction: the sign inside is \(+\), so \(kx + \omega t\), meaning the pattern moves in \(-x\). (b) \(v = \omega/k = 120/3 = 40\,\mathrm{m\,s^{-1}}\). (c) \(v_y = \partial y/\partial t = -0.02 \times 120\sin(0) = 0\). At \(t = 0\), \(x = 0\): the element is at its maximum displacement (\(\cos 0 = 1\)) and is instantaneously at rest, as in SHM.
Show that the function \(y = A\sin(kx)\cos(\omega t)\) also satisfies the wave equation \(\partial^2 y/\partial t^2 = v^2\,\partial^2 y/\partial x^2\), provided \(v = \omega/k\). What kind of wave does this function represent?
\(\partial^2 y/\partial t^2 = -A\omega^2\sin(kx)\cos(\omega t) = -\omega^2 y\). \(v^2\,\partial^2 y/\partial x^2 = v^2\times(-Ak^2\sin(kx)\cos(\omega t)) = -v^2k^2 y\). These are equal when \(\omega^2 = v^2k^2\), i.e. \(v = \omega/k\). ✓. This function represents a standing wave: it does not translate in either direction but oscillates in place. This is the superposition of two travelling waves of equal amplitude moving in opposite directions, as we will show in a later section.
The tension in a guitar string is quadrupled while its length and mass are kept fixed (so \(\mu\) is unchanged). How does the wave speed change, and what happens to the frequency of the fundamental standing wave?
From \(v = \sqrt{T_s/\mu}\): quadrupling \(T_s\) doubles \(v\). The fundamental frequency of a string fixed at both ends is \(f_1 = v/(2L)\). Doubling \(v\) doubles \(f_1\). The string sounds one octave higher.

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Wavelength of a fork in two media

A tuning fork of frequency \(f = 512\,\mathrm{Hz}\) is sounded first in air (\(v_{\text{air}} = 340\,\mathrm{m\,s^{-1}}\)) and then in water (\(v_{\text{water}} = 1480\,\mathrm{m\,s^{-1}}\)). Find the wavelength in each medium.

Solution. Given: \(f = 512\,\mathrm{Hz}\), \(v_{\text{air}} = 340\,\mathrm{m\,s^{-1}}\), \(v_{\text{water}} = 1480\,\mathrm{m\,s^{-1}}\).
Find: \(\lambda_{\text{air}}\), \(\lambda_{\text{water}}\).
Setup: Frequency is fixed by the source, so it is identical in both media; only the speed—and hence the wavelength \(\lambda = v/f\)—changes.
Solve: \[ \lambda_{\text{air}} = \frac{340}{512} \approx 0.664\,\mathrm{m}, \qquad \lambda_{\text{water}} = \frac{1480}{512} \approx 2.89\,\mathrm{m}. \] Answer: \(\boxed{\lambda_{\text{air}} \approx 0.664\,\mathrm{m},\quad \lambda_{\text{water}} \approx 2.89\,\mathrm{m}}\)
Check: The wavelengths scale as the speeds: \(\lambda_{\text{water}}/\lambda_{\text{air}} = 4.35\), matching \(v_{\text{water}}/v_{\text{air}} = 1480/340 = 4.35\). ✓

Example 2 · Wavenumber and angular frequency

A sound wave has wavelength \(\lambda = 0.50\,\mathrm{m}\) and frequency \(f = 680\,\mathrm{Hz}\). Find the wavenumber \(k\), the angular frequency \(\omega\), and the wave speed.

Solution. Given: \(\lambda = 0.50\,\mathrm{m}\), \(f = 680\,\mathrm{Hz}\).
Find: \(k\), \(\omega\), \(v\).
Setup: \(k = 2\pi/\lambda\), \(\omega = 2\pi f\), and \(v = \omega/k\).
Solve: \[ k = \frac{2\pi}{0.50} \approx 12.6\,\mathrm{rad\,m^{-1}}, \quad \omega = 2\pi(680) \approx 4.27\times 10^{3}\,\mathrm{rad\,s^{-1}}, \] \[ v = \frac{\omega}{k} = \frac{4.27\times 10^{3}}{12.6} \approx 340\,\mathrm{m\,s^{-1}}. \] Answer: \(\boxed{k \approx 12.6\,\mathrm{rad\,m^{-1}},\ \omega \approx 4.27\times 10^{3}\,\mathrm{rad\,s^{-1}},\ v \approx 340\,\mathrm{m\,s^{-1}}}\)
Check: Independently, \(v = f\lambda = 680 \times 0.50 = 340\,\mathrm{m\,s^{-1}}\), matching \(\omega/k\). ✓

Example 3 · Speed of sound from first principles

Find the speed of sound in air at \(T = 300\,\mathrm{K}\), treating air as an ideal diatomic gas: \(\gamma = 1.40\), \(R = 8.314\,\mathrm{J\,mol^{-1}K^{-1}}\), molar mass \(M = 28.9\,\mathrm{g\,mol^{-1}}\).

Solution. Given: \(\gamma = 1.40\), \(R = 8.314\,\mathrm{J\,mol^{-1}K^{-1}}\), \(T = 300\,\mathrm{K}\), \(M = 0.0289\,\mathrm{kg\,mol^{-1}}\).
Find: \(v\).
Setup: For an ideal gas \(v = \sqrt{\gamma RT/M}\), with \(M\) in \(\mathrm{kg\,mol^{-1}}\).
Solve: \[ v = \sqrt{\frac{1.40 \times 8.314 \times 300}{0.0289}} = \sqrt{1.21\times 10^{5}} \approx 348\,\mathrm{m\,s^{-1}}. \] Answer: \(\boxed{v \approx 348\,\mathrm{m\,s^{-1}}}\)
Check: Since \(v \propto \sqrt{T}\), scaling to \(0^\circ\mathrm{C}\): \(v(273\,\mathrm{K}) = 348\sqrt{273/300} \approx 331\,\mathrm{m\,s^{-1}}\), the standard textbook value of the speed of sound at \(0^\circ\mathrm{C}\). ✓

Example 4 · Intensity and the decibel scale

A sound has intensity \(I = 3.2\times 10^{-6}\,\mathrm{W\,m^{-2}}\) (reference \(I_0 = 10^{-12}\,\mathrm{W\,m^{-2}}\)). (a) Find its sound level \(\beta\). (b) By how much does \(\beta\) rise if the intensity is multiplied by \(100\)?

Solution. Given: \(I = 3.2\times 10^{-6}\,\mathrm{W\,m^{-2}}\), \(I_0 = 10^{-12}\,\mathrm{W\,m^{-2}}\).
Find: \(\beta\); the change for \(I \to 100\,I\).
Setup: \(\beta = 10\log_{10}(I/I_0)\) decibels.
Solve: \[ \beta = 10\log_{10}\!\left(\frac{3.2\times 10^{-6}}{10^{-12}}\right) = 10\log_{10}(3.2\times 10^{6}) \approx 65\,\mathrm{dB}. \] A factor of \(100\) adds \(10\log_{10}(100) = 20\,\mathrm{dB}\), giving \(85\,\mathrm{dB}\).
Answer: \(\boxed{\beta \approx 65\,\mathrm{dB};\quad \text{then } 85\,\mathrm{dB}}\)
Check: Each factor of \(10\) in intensity adds exactly \(10\,\mathrm{dB}\); \(\times 100\) is two decades, so \(+20\,\mathrm{dB}\)—consistent. ✓

Example 5 · Beats and the wax test

Fork A (\(f_A = 256\,\mathrm{Hz}\)) and fork B together produce \(4\) beats per second. When a small blob of wax is added to B, the beat rate falls to \(2\) per second. Find \(f_B\).

Solution. Given: \(f_A = 256\,\mathrm{Hz}\); \(4\) beats/s; loading B with wax \(\Rightarrow\) \(2\) beats/s.
Find: \(f_B\).
Setup: The beat rate is \(|f_A - f_B|\), so \(f_B = 252\) or \(260\,\mathrm{Hz}\). Adding mass lowers a fork's frequency; the correct root is the one for which lowering \(f_B\) reduces the beat rate.
Solve: If \(f_B = 260\,\mathrm{Hz}\), wax lowers it toward \(256\), so beats decrease—consistent. If \(f_B = 252\,\mathrm{Hz}\), wax lowers it away from \(256\), so beats would increase—contradiction. Hence \(f_B = 260\,\mathrm{Hz}\).
Answer: \(\boxed{f_B = 260\,\mathrm{Hz}}\)
Check: A wax blob lowering \(260\,\mathrm{Hz}\) by about \(2\,\mathrm{Hz}\) gives \(|256 - 258| = 2\) beats, exactly the observed drop; the \(252\,\mathrm{Hz}\) root is excluded. ✓

Problem bank

Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.

✎ Free practice questions

Reading the Wave Function
A wave on a string is described by \(y = 0.030\sin(4.0x - 160t)\,\mathrm{m}\) (\(x\) in m, \(t\) in s). Find: (a) amplitude, (b) wavelength, (c) frequency, (d) wave speed, (e) maximum transverse speed of any element.
(a) \(0.030\,\mathrm{m}\). (b) \(\lambda = 2\pi/k = 2\pi/4.0 \approx 1.57\,\mathrm{m}\). (c) \(f = \omega/(2\pi) = 160/(2\pi) \approx 25.5\,\mathrm{Hz}\). (d) \(v = \omega/k = 160/4.0 = 40\,\mathrm{m\,s^{-1}}\). (e) \(v_{y,\max} = A\omega = 0.030\times160 = 4.8\,\mathrm{m\,s^{-1}}\).
String Speed
A steel wire of mass \(8.0\,\mathrm{g}\) and length \(1.6\,\mathrm{m}\) is under a tension of \(200\,\mathrm{N}\). Find the speed of transverse waves on the wire.
\(\mu = 8.0\times10^{-3}/1.6 = 5.0\times10^{-3}\,\mathrm{kg\,m^{-1}}\). \(v = \sqrt{T_s/\mu} = \sqrt{200/(5.0\times10^{-3})} = \sqrt{40000} = 200\,\mathrm{m\,s^{-1}}\).
Sound Level
The intensity of a sound wave is \(3.0\times10^{-5}\,\mathrm{W\,m^{-2}}\). Find the sound level in dB. (\(I_0 = 10^{-12}\,\mathrm{W\,m^{-2}}\).)
\(\beta = 10\log_{10}(3.0\times10^{-5}/10^{-12}) = 10\log_{10}(3.0\times10^7) = 10(7 + \log_{10}3.0) = 10(7.477) \approx 74.8\,\mathrm{dB}\).
Power of a Wave
A wave \(y = 0.02\sin(10x - 400t)\,\mathrm{m}\) travels on a string with linear density \(\mu = 4.0\times10^{-2}\,\mathrm{kg\,m^{-1}}\). Find: (a) the wave speed and verify \(v = \omega/k\), (b) the average power transmitted by the wave.
(a) \(v = \omega/k = 400/10 = 40\,\mathrm{m\,s^{-1}}\). Check: \(v = \sqrt{T_s/\mu}\) requires \(T_s = v^2\mu = 1600 \times 4.0\times10^{-2} = 64\,\mathrm{N}\). ✓ (b) \(P = \tfrac{1}{2}\mu\omega^2 A^2 v = \tfrac{1}{2}(4.0\times10^{-2})(400)^2(0.02)^2(40) = 51.2\,\mathrm{W}\).
Coupled Doppler: Moving Observer and Moving Source
An ambulance moves east at \(v_A = 30\,\mathrm{m\,s^{-1}}\) emitting a siren at \(f_s = 900\,\mathrm{Hz}\). A listener moves west (toward the ambulance) at \(v_L = 15\,\mathrm{m\,s^{-1}}\) (\(v_s = 340\,\mathrm{m\,s^{-1}}\)). (a) Find the frequency heard as they approach. (b) After they pass, both continue on; find the frequency heard. (c) Find the beat frequency if both signals could be heard at once (e.g. one directly, one reflected from a wall behind the listener).
(a) Source toward listener, listener toward source: \(f_1 = 900\,(340+15)/(340-30) = 900\times355/310 \approx 1031\,\mathrm{Hz}\). (b) Both receding: \(f_2 = 900\,(340-15)/(340+30) = 900\times325/370 \approx 790\,\mathrm{Hz}\). (c) \(f_B = |f_1 - f_2| = |1031 - 790| = 241\,\mathrm{Hz}\).