Waves

This chapter turns the single oscillating mass of Chapter 13 into a whole medium of coupled oscillators and asks how a disturbance moves through it. It opens by placing mechanical waves (which need a medium) alongside electromagnetic and quantum matter waves, before settling on mechanical waves as its actual subject. It classifies waves as transverse or longitudinal, then builds the mathematical description of a travelling sinusoidal wave, $y(x,t) = a\sin(kx-\omega t+\varphi)$, reading off amplitude, wavelength, wave number, period, frequency and phase. From there it derives how fast a wave moves on a stretched string and how fast sound moves through gases, liquids and solids — including Newton's original, slightly wrong, formula for sound in air, and the Laplace correction that fixes it. The second half is about waves meeting: superposition, constructive and destructive interference, reflection at rigid and open boundaries, the standing waves and normal modes this produces in strings and air columns, and finally beats.

What a wave actually transports

Mechanical waves need a material medium — elastic forces bind the medium's constituents together, so displacing one part disturbs its neighbours, which disturb theirs in turn, and the disturbance propagates outward while the medium as a whole stays put. A row of spring-coupled train bogies illustrates this exactly: pull the first bogie and release it, and a jerk travels down the train without the whole train being bodily displaced. Sound in air works the same way: a compressed region pushes on its neighbour, which compresses and pushes the next, so a pulse of compression (and, right behind it, rarefaction) travels down a column of air even though no parcel of air makes the whole trip. Electromagnetic waves are the exception, needing no medium at all and travelling through vacuum at $c = 299{,}792{,}458\ \text{m/s}$ — but this chapter is concerned entirely with mechanical waves. Wind (bulk motion of air) should never be confused with a sound wave (propagation of a pressure disturbance through air that is, on average, staying put). What a wave transports is energy and a pattern of information, coupled from one oscillating element of the medium to the next through elastic restoring forces — never the material of the medium itself.

Transverse and longitudinal waves

If a medium's constituents oscillate perpendicular to the wave's direction of travel, the wave is transverse; if they oscillate parallel to it, the wave is longitudinal. A harmonic wave on a stretched string is the standard transverse example: each element of string moves straight up and down while the pattern moves along the string. Sound generated by a piston in a long air-filled pipe is the standard longitudinal example: each thin slab of air moves back and forth along the pipe's axis, alternately bunching into a compression and spreading into a rarefaction. The distinction decides which media can carry which wave. A transverse wave needs the medium to resist shearing — one layer sliding sideways relative to its neighbour — which only solids sustain. A longitudinal wave only needs resistance to a change in volume, which every elastic medium sustains: solids, liquids and gases alike. Steel can therefore carry both transverse and longitudinal waves, while air carries only longitudinal (sound) waves. Water-surface waves mix both kinds of motion: capillary waves (up to a few centimetres) are restored by surface tension, gravity waves (metres to hundreds of metres) by gravity.

The displacement relation of a travelling wave

To describe a travelling wave mathematically we need a single function of position $x$ and time $t$: at any fixed instant it gives the shape of the wave in space, and at any fixed location it gives the motion of that piece of the medium in time. For a sinusoidal transverse wave moving toward $+x$, that function is

where: $a$ is the amplitude; $k$ is the angular wave number; $\omega$ is the angular frequency; and $\varphi$ is the initial phase, the phase at $x=0$, $t=0$. Freeze time at $t=t_0$ and the argument becomes $kx+\text{constant}$, so $y$ against $x$ is a sine curve — the shape of the wave. Freeze position at $x=x_0$ instead and the argument becomes $\text{constant}-\omega t$, so $y$ varies sinusoidally in time — that piece of the medium executes simple harmonic motion, the motion of the previous chapter. As $t$ increases, $x$ must increase to keep $kx-\omega t+\varphi$ fixed, so a minus sign describes a wave travelling toward $+x$; a plus sign, $y=a\sin(kx+\omega t+\varphi)$, describes an identical wave travelling toward $-x$. For a longitudinal wave the same function applies with $y$ replaced by $s(x,t)$, displacement along the direction of propagation.

Amplitude, phase, wavelength and angular wave number

The quantity $(kx-\omega t+\varphi)$ is the phase; given amplitude $a$, the phase alone fixes the displacement at any position and instant, and $\varphi$ is simply the phase at $x=0$, $t=0$. Suitable choice of origin and starting time always lets us set $\varphi=0$ without loss of generality. The wavelength $\lambda$ is the minimum distance between two points of the same phase — equivalently, the distance between two consecutive crests. Since sine repeats every $2\pi$ radians,

where: $k$, the angular wave number, has SI unit radian per metre and represents $2\pi$ times the number of complete waves per unit length. A larger $k$ means a shorter wavelength.

Period, angular frequency and frequency

Fixing $x=0$ and taking $\varphi=0$ gives $y(0,t) = -a\sin(\omega t)$, a pure simple harmonic oscillation in time. The period $T$, the time for one complete oscillation, relates to $\omega$ by

where: $\omega$ has SI unit radian per second, and $\nu$, the frequency, is measured in hertz. The source of the disturbance fixes $\nu$ directly; once the medium (and hence the speed) is fixed, the speed relation below then determines the wavelength — not the reverse.

The speed of a travelling wave

To find how fast a wave moves, track a fixed point of phase — a crest — and see how its position changes with time. Working through the condition that $kx-\omega t+\varphi$ stays constant as a crest advances (done in full in the first derivation below) gives the universal speed relation for any progressive wave,

where: $v$ is the rate at which the pattern — not any particle of the medium — advances along $x$. In exactly one period $T$, the pattern moves forward by exactly one wavelength $\lambda$. Crucially, the speed of a mechanical wave is set entirely by the medium's inertial property (mass density) and elastic property (Young's, shear or bulk modulus) — never by the wave's own wavelength or frequency. Given a speed fixed by the medium and a frequency chosen by the source, $v=\lambda\nu$ then fixes the wavelength.

Speed of a transverse wave on a stretched string

For a wave on a string, the restoring force comes from the tension $T$, and the relevant inertia is the linear mass density $\mu = m/L$. An exact formula follows from Newton's laws applied to a string element, a calculation kept outside this chapter's scope; NCERT instead reaches the same result by dimensional analysis: the only combination of $T$ (dimension $[\text{MLT}^{-2}]$) and $\mu$ (dimension $[\text{ML}^{-1}]$) with the dimension of speed is $\sqrt{T/\mu}$, times an undetermined constant that the exact calculation shows equals $1$:

where: $v$ depends only on tension and mass per unit length, both properties of the string itself. A tighter string (larger $T$) speeds the wave up; a heavier string (larger $\mu$) slows it down.

Speed of sound: Newton's formula and the Laplace correction

Sound is longitudinal, so the relevant restoring property is resistance to compression, captured by the bulk modulus $B = -\Delta P/(\Delta V/V)$. The same dimensional argument gives the speed of a longitudinal wave in a fluid,

and, since a solid bar under purely longitudinal strain obeys the same argument with Young's modulus $Y$ in place of $B$, the speed in a solid bar is $v=\sqrt{Y/\rho}$. where: $B$ or $Y$ measures resistance to compression or stretching, and $\rho$ is mass density. Liquids and solids carry sound faster than gases not because they are lighter — they are far denser — but because their moduli are enormously larger still, so $B/\rho$ ends up bigger despite the bigger $\rho$. For an ideal gas, Newton assumed compressions happen slowly enough for heat to equalise temperature — an isothermal process, for which $PV=\text{constant}$ gives $B=P$ directly, so $v=\sqrt{P/\rho}$. This under-predicts the measured speed of sound in air by about 15 percent. Laplace pointed out that sound oscillates far too fast for meaningful heat exchange, so the process is adiabatic, governed by $PV^\gamma=\text{constant}$, giving a larger modulus $B_{\text{ad}}=\gamma P$ and the corrected formula

where: $\gamma = C_p/C_v$ ($=7/5$ for air, predominantly diatomic). This correction, worked out in full in a worked example below, raises the predicted speed of sound in air at STP from Newton's $280\ \text{m/s}$ to $331\ \text{m/s}$, matching the measured value almost exactly.

Superposition and interference

When two wave pulses travelling in opposite directions cross, each retains its own identity afterward, but while overlapping, the net displacement at every point is the algebraic sum of the two — positive and negative displacements can partly or fully cancel. This is the principle of superposition: for two disturbances present simultaneously,

where: each wave behaves as if the other were not present. For two harmonic waves of equal amplitude $a$, the same $k$ and $\omega$, differing by a constant phase $\varphi$, superposition produces a single resultant wave (derived below) of amplitude $A(\varphi) = 2a\cos(\varphi/2)$: largest, $2a$, when the waves are in step ($\varphi=0$, constructive interference), and zero everywhere and always when exactly out of step ($\varphi=\pi$, destructive interference). Superposition is the mathematical engine behind every interference phenomenon in this chapter, from beats to standing waves.

Reflection, standing waves and normal modes

When a travelling wave meets a rigid boundary, the displacement there must remain zero at all times, so the incident and reflected waves can only sum to zero continuously if the reflected wave is inverted — a phase change of $\pi$. The same follows dynamically: the pulse pushes on the wall, and by Newton's third law the wall pushes back, launching a reversed pulse. At a free (open) boundary, nothing constrains the displacement to zero, so the reflected wave matches the incident wave in phase, with no reversal. When a wave train reflects back and forth between two boundaries, the incident and reflected waves overlap continuously and build a steady standing wave. Adding a wave travelling in $+x$ to an identical one travelling in $-x$ gives

where: unlike a travelling wave, $kx$ and $\omega t$ appear separately, not combined — the amplitude $2a\sin(kx)$ varies with position but is fixed in time at each point, while every element oscillates in phase with the same $\omega$. Points where $\sin(kx)=0$ are nodes (always zero displacement); points where $|\sin(kx)|=1$ are antinodes (largest displacement). Both are spaced $\lambda/2$ apart. For a string of length $L$ fixed at both ends, $x=0$ and $x=L$ must both be nodes, restricting wavelengths to $L=n\lambda/2$ and giving the normal-mode frequencies

where: $n=1$ is the fundamental; $n=2$ is the second harmonic, and so on — a freely vibrating string is generally a superposition of several harmonics, with the plucking point deciding which are strongest. For an air column closed at one end and open at the other, the closed end must be a node while the open end is an antinode, restricting the length to $L=(n+\tfrac12)\lambda/2$ and giving

where: the fundamental ($n=0$) is $v/4L$, and only odd multiples of it — $\nu, 3\nu, 5\nu,\dots$ — are possible; even harmonics simply do not fit. A pipe open at both ends has an antinode at each end instead, which allows every harmonic, $\nu_n=nv/2L$, just like the string, since matching antinode-to-antinode is mathematically identical to matching node-to-node.

Beats

Two harmonic sound waves of slightly different frequency, heard together, do not simply blend into one steady tone. A listener hears a note near their average frequency, but its loudness audibly waxes and wanes — beats, which musicians use to tune two strings by listening for the beat to slow down and vanish. Superposing two waves of equal amplitude and nearby angular frequencies $\omega_1$ and $\omega_2$ at a fixed location (worked out in full below) gives a resultant whose amplitude itself oscillates slowly, producing an audible loudness that rises and falls at the beat frequency

where: $\nu_1$ and $\nu_2$ are the two original frequencies. Two tones of $11\ \text{Hz}$ and $9\ \text{Hz}$ beat at exactly $2\ \text{Hz}$ — a case NCERT illustrates with its own figure of the superposed waveforms. The phenomenon is not confined to sound: temple pillars carved from a single block of stone, such as the musical pillars at the Nellaiappar temple in Tamil Nadu, produce the rhythmic "taal" of Indian classical music by exactly this same beating of two close natural frequencies when struck together.

Board level - decoding a travelling wave equation (NCERT Example 14.2)

A wave travelling along a string is described by $y(x,t) = 0.005\sin(80.0x - 3.0t)$, SI units. Find (a) the amplitude, (b) the wavelength, (c) the period and frequency, and (d) the displacement $y$ at $x=30.0\ \text{cm}$, $t=20\ \text{s}$.

Step 1. Compare with $y(x,t)=a\sin(kx-\omega t)$: $a=0.005\ \text{m}=5\ \text{mm}$, $k=80.0\ \text{rad/m}$, $\omega=3.0\ \text{rad/s}$.

Step 2. Wavelength: $\lambda=2\pi/k = 2\pi/80.0 = 0.0785\ \text{m} = 7.85\ \text{cm}$.

Step 3. Period and frequency: $T=2\pi/\omega=2\pi/3.0=2.09\ \text{s}$, $\nu=1/T\approx0.48\ \text{Hz}$.

Step 4. At $x=0.300\ \text{m}$, $t=20\ \text{s}$: phase $= 80.0(0.300)-3.0(20) = -36.0\ \text{rad}$. Adding $12\pi\approx37.70\ \text{rad}$ (six full turns) gives $1.70\ \text{rad}$, so $y=0.005\sin(1.70)\approx4.96\times10^{-3}\ \text{m}$. Answer: amplitude $5\ \text{mm}$, wavelength $7.85\ \text{cm}$, period $2.09\ \text{s}$, frequency $0.48\ \text{Hz}$, $y\approx5\ \text{mm}$.

Board level - speed of a transverse wave on a steel wire (NCERT Example 14.3)

A steel wire $0.72\ \text{m}$ long has mass $5.0\times10^{-3}\ \text{kg}$, under tension $60\ \text{N}$. Find the wave speed.

Step 1. $\mu = m/L = (5.0\times10^{-3})/(0.72) = 6.9\times10^{-3}\ \text{kg/m}$.

Step 2. $v=\sqrt{T/\mu}=\sqrt{60/(6.9\times10^{-3})}=\sqrt{8696}$.

Step 3. $v \approx 93\ \text{m/s}$. Answer: the transverse wave speed is $93\ \text{m/s}$.

Competitive level - speed of sound in air: Newton versus Laplace (JEE/NEET pattern, NCERT Example 14.4 extended)

One mole of air (mass $29.0\times10^{-3}\ \text{kg}$) occupies $22.4$ litres at STP, $P\approx1.01\times10^{5}\ \text{Pa}$. Estimate the speed of sound using Newton's and the Laplace-corrected formula ($\gamma_{\text{air}}=7/5$), and state which matches the measured $331\ \text{m/s}$.

Step 1. Density: $\rho_0 = (29.0\times10^{-3})/(22.4\times10^{-3}) = 1.29\ \text{kg/m}^3$.

Step 2. Newton (isothermal, $B=P$): $v=\sqrt{P/\rho_0} = \sqrt{78{,}295} \approx 280\ \text{m/s}$.

Step 3. This is about 15 percent low, signalling the isothermal assumption is wrong: sound oscillates too fast for heat exchange, so the process is adiabatic, $B_{\text{ad}}=\gamma P$.

Step 4. Laplace: $v=\sqrt{\gamma P/\rho_0} = \sqrt{109{,}613} \approx 331\ \text{m/s}$, matching experiment. Answer: Newton gives $280\ \text{m/s}$ (15 percent low); Laplace gives $331\ \text{m/s}$, matching measurement, because sound compressions are adiabatic, not isothermal. Timing note: about ninety seconds if $\rho_0=1.29\ \text{kg/m}^3$ is already a standing number. The idea being tested is why $\gamma$ appears — real compressions happen too fast for heat exchange, so always use $\gamma P$, never $P$, when asked for the accurate speed of sound in a gas.

Competitive level - beats reveal which sitar string is out of tune (JEE/NEET pattern, NCERT Example 14.6)

Strings A and B produce $5$ beats per second. Tension in B is slightly increased and the beat frequency drops to $3\ \text{Hz}$. If $\nu_A=427\ \text{Hz}$, find the original $\nu_B$.

Step 1. Raising tension always raises frequency ($\nu=v/2L$, $v=\sqrt{T/\mu}$ grows with $T$), so raising B's tension can only raise $\nu_B$.

Step 2. Since $\nu_{\text{beat}}=|\nu_A-\nu_B|=5\ \text{Hz}$ only fixes the size of the gap, either $\nu_B=432\ \text{Hz}$ or $\nu_B=422\ \text{Hz}$.

Step 3. If $\nu_B$ started at $432\ \text{Hz}$, raising it would widen the gap, increasing the beat frequency — contradicting the observed decrease. If it started at $422\ \text{Hz}$, raising it narrows the gap toward $3\ \text{Hz}$ — consistent.

Step 4. So $\nu_B=422\ \text{Hz}$. Answer: the original frequency of B is $422\ \text{Hz}$. Timing note: under a minute once you know the trick — beats never reveal the sign of the difference, only its size, so combine the direction of the deliberate change with the direction of the observed effect to break the tie. Write both candidates and eliminate one; do not guess.

• Treating a wave as bulk transport of the medium. A wave carries energy and a pattern, not matter; never say the air itself travels from speaker to ear.
• Confusing wave velocity $v$ with particle velocity $\partial y/\partial t$. Wave speed is how fast the pattern advances; particle velocity is how fast one bit of the medium moves about its mean position — different in both size and formula.
• Getting the sign between $kx$ and $\omega t$ backwards. $kx-\omega t$ travels toward $+x$; $kx+\omega t$ travels toward $-x$.
• Forgetting the $\pi$ phase reversal at a rigid boundary. Fixed end: $y_r=-a\sin(kx-\omega t)$. Free end: $y_r=+a\sin(kx-\omega t)$, no reversal.
• Miscounting harmonics in a pipe closed at one end. Only odd harmonics, fundamental $v/4L$, not $v/2L$ as for a string or open-open pipe.
• Skipping the Laplace correction. Sound compressions are adiabatic, not isothermal; use $B=\gamma P$, not $B=P$, unless Newton's historical estimate is specifically requested.
• Believing beats alone tell you which frequency is higher. $\nu_{\text{beat}}=|\nu_1-\nu_2|$ gives only the size of the mismatch; use a deliberate frequency change to break the tie.
• Assuming wave speed depends on frequency. For these non-dispersive media, $v$ is fixed by the medium alone; changing frequency changes $\lambda$ (via $v=\lambda\nu$), never $v$.

• Classify a described wave as transverse or longitudinal, and state which media can carry each and why.
• Read $a$, $k$, $\omega$, $\varphi$ off a wave equation $y(x,t)=a\sin(kx\mp\omega t+\varphi)$, and state the direction of propagation from the sign between $kx$ and $\omega t$.
• Convert between $\lambda$, $k$, $T$, $\omega$, $\nu$ using $\lambda=2\pi/k$, $T=2\pi/\omega$, $\nu=1/T$, and combine into $v=\omega/k=\lambda\nu$.
• Calculate wave speed on a string ($v=\sqrt{T/\mu}$) and speed of sound in a fluid, solid bar or ideal gas ($v=\sqrt{B/\rho}$, $\sqrt{Y/\rho}$, $\sqrt{\gamma P/\rho}$).
• Explain why Laplace's adiabatic correction is needed over Newton's isothermal assumption, and reproduce the STP calculation giving $331\ \text{m/s}$.
• Superpose two waves differing only in phase and find the resultant amplitude $2a\cos(\varphi/2)$, identifying constructive and destructive interference.
• State and justify the phase change on reflection at a rigid versus open boundary, and derive $y=2a\sin(kx)\cos(\omega t)$ from two oppositely travelling waves.
• Find the normal-mode frequencies of a string fixed at both ends, a pipe closed at one end, and a pipe open at both ends, and identify nodes, antinodes and harmonic number from a given frequency.
• Calculate a beat frequency and use a deliberate frequency change to determine which of two sources was originally higher or lower.

Ask the tutor: If you want to see the full Newton's-second-law derivation of the string wave equation, or more practice picking the right harmonic series for open versus closed pipes, ask the tutor to walk through it with you.