Chapter 14 · Waves

14.1 Transverse and Longitudinal Waves and the Displacement Relation

A wave carries energy through a medium while each particle only oscillates about its own place. Particles move across the direction of travel in a transverse wave and along it in a longitudinal wave, and a progressive sinusoidal wave is described by $y(x,t)=a\sin(kx-\omega t+\phi)$.

When an earthquake strikes, seismometers around the world record two separate jolts. The first, the P-wave ("primary"), pushes and pulls the ground along the direction from the earthquake and races through rock at about $6\ \text{km/s}$. The second, the S-wave ("secondary"), shakes the ground from side to side, across that direction, and travels at only about $3.5\ \text{km/s}$. The longer the gap between the two, the farther away the earthquake — and since S-waves never get through the Earth's core, the pattern of where they do and do not arrive told geologists, a century ago, that the outer core is liquid.

Both jolts are waves: disturbances that travel through a medium, carrying energy, while the rock itself only shakes about its own position. This section sorts waves into two kinds — transverse (shaking across the direction of travel) and longitudinal (shaking along it) — and then writes down the equation of a sinusoidal wave travelling along a line, $y(x,t)=a\sin(kx-\omega t+\phi)$. Every particle of the medium turns out to execute the simple harmonic motion of Section 13.1, each a little behind its neighbour; the lag between particles is what makes the pattern move.

What a wave carries

Hold one end of a long rope, stretched along the floor, and flick your hand up and down once. A hump runs along the rope, reaches the far end and tugs on whoever holds it there. Yet no piece of rope travels from you to them: each piece is pulled up by the piece before it, then pulled down again, and comes back to rest where it started. What travelled was the disturbance — and with it the energy you gave the rope. Two properties of the medium make this possible: elasticity (a displaced piece is pulled back, and pulls on its neighbours) and inertia (a piece that has been set moving overshoots, so the disturbance is passed on rather than simply dying where it started).

📘 Definition 14.1 — Mechanical wave

A mechanical wave is a disturbance that travels through a material medium (a solid, liquid or gas) because of the elastic forces between neighbouring parts of the medium. Each element of the medium oscillates about its own mean position; there is no net transport of the medium, but energy and momentum are carried from one place to another. A single short disturbance is a pulse; a disturbance repeated at regular intervals is a periodic wave.

Mechanical waves need a medium — sound cannot cross the vacuum between the Moon and the Earth. Light, radio waves and X-rays are electromagnetic waves, which need no medium (Class 12 Physics); this chapter is about mechanical waves only. The medium can respond to a disturbance in two different ways.

📘 Definition 14.2 — Transverse and longitudinal waves

In a transverse wave the elements of the medium oscillate perpendicular to the direction in which the wave travels. Example: a wave on a stretched string; an S-wave in rock.

In a longitudinal wave the elements oscillate along (parallel to) the direction of travel. The medium is squeezed into compressions (regions of higher density and pressure) and stretched into rarefactions (lower density and pressure), and these regions travel along. Example: sound in air; a P-wave in rock; a pulse pushed along a slinky spring.

A transverse wave needs a medium that resists a change of shape — one with a shear modulus (Section 8.2), or a string or surface held taut by tension. A longitudinal wave needs only resistance to a change of volume, a bulk modulus. Solids have both, so they carry both kinds of wave. Liquids and gases cannot sustain a shear stress, so through their interior only longitudinal waves pass: sound in air and in water is longitudinal. Ripples on the surface of water are neither purely one nor the other — gravity and surface tension act as the restoring forces, and each bit of water moves round a small circle, partly across and partly along the direction of travel.

✏️ Example 1 — Which kind of wave?

Classify each wave as transverse, longitudinal or neither purely, and give the reason. (a) A hump sent along a washing line. (b) The sound of a bell reaching your ear. (c) A slinky spring lying on a table, with one end pushed sharply forward along its length. (d) Ripples spreading from a stone dropped in a pond. (e) An S-wave travelling through granite. Finally, (f) explain why S-waves from an earthquake are not detected on the far side of the Earth.

💡 Show solution▶

(a) Transverse: each piece of the line moves up and down, while the hump travels along the line.

(b) Longitudinal: air layers are pushed to and fro along the line from the bell to your ear, forming compressions and rarefactions.

(c) Longitudinal: the coils bunch up and spread out along the length of the spring, the direction in which the pulse travels.

(d) Neither purely: surface water moves in small circles (up–forward–down–back), so its motion has components both across and along the direction of travel.

(e) Transverse: the rock is sheared from side to side across the direction of travel, which granite (a solid, with a shear modulus) can resist.

(f) The Earth's outer core is liquid. A liquid has no shear modulus, so it cannot pass on a sideways shear: S-waves stop at the core boundary and leave an "S-wave shadow" on the far side. P-waves, being longitudinal, do pass through (bent at the boundary).

✏️ Example 2 — How far away was the earthquake?

At a seismic station the S-wave from an earthquake arrives $60\ \text{s}$ after the P-wave. Taking the speeds as $6.0\ \text{km/s}$ (P) and $3.5\ \text{km/s}$ (S), find the distance $d$ of the station from the earthquake, assuming both waves travel the same straight path at constant speed.

💡 Show solution▶

Set-up Travel times $t_S=\dfrac{d}{3.5}$ and $t_P=\dfrac{d}{6.0}$ (km, s), and $t_S-t_P=60\ \text{s}$.

Solve $d\left(\dfrac1{3.5}-\dfrac1{6.0}\right)=60$: $d\times0.1190=60$, so $d=5.0\times10^2\ \text{km}$ ($504$).

Check $t_P=\dfrac{504}{6.0}=84\ \text{s}$ and $t_S=\dfrac{504}{3.5}=144\ \text{s}$: the gap is $60\ \text{s}$ ✓. Rule of thumb for these speeds: $d\approx8.4\ \text{km}$ for every second of S–P delay. Three stations, each giving a distance, locate the earthquake where their three circles cross.

The displacement relation for a progressive wave

Stretch a long string along the $x$-axis and move its end at $x=0$ up and down in SHM. A progressive (travelling) wave moves along the string at a steady speed $v$. Let $y(x,t)$ be the transverse displacement, at time $t$, of the bit of string whose mean position is $x$. Suppose the end moves as $y(0,t)=a\sin(\phi-\omega t)$ — an SHM of amplitude $a$ and angular frequency $\omega$ (Section 13.1; a sine with a phase constant is just as good as a cosine). The disturbance takes a time $\dfrac xv$ to reach the point $x$, so that point does now exactly what the end did a time $\dfrac xv$ earlier:

$$y(x,t)=y\!\left(0,\,t-\frac xv\right)=a\sin\!\left(\phi-\omega t+\frac{\omega}{v}x\right).$$

Writing $k=\dfrac\omega v$ gives the standard form below. The same equation describes a longitudinal wave if $y$ is read as the displacement along $x$ of the layer of medium whose mean position is $x$.

📐 Result 14.1 — Displacement relation for a progressive sinusoidal wave

A sinusoidal wave of amplitude $a$ travelling in the $+x$ direction is described by

$$y(x,t)=a\sin(kx-\omega t+\phi),$$

and one travelling in the $-x$ direction by

$$y(x,t)=a\sin(kx+\omega t+\phi).$$

The argument $kx\mp\omega t+\phi$ is the phase of the wave at $x$ and $t$, and $\phi$ (the phase at $x=0$, $t=0$) is the initial phase or phase constant. Every element of the medium executes SHM with the same amplitude $a$ and angular frequency $\omega$; elements at different $x$ differ only in phase.

Two snapshots of the same equation are useful. Freeze time ($t$ fixed): $y$ against $x$ is a sine curve in space — the shape of the string at that instant. Fix a place ($x$ fixed): $y$ against $t$ is a sine curve in time — the history of one particle. The two curves have their own repeat distances, which name the constants $k$ and $\omega$.

📘 Definition 14.3 — Wavelength and wave number

The wavelength $\lambda$ of a periodic wave is the smallest distance, along the direction of travel, between two points of the medium that are in the same phase at every instant — for example from one crest to the next, or from one compression to the next. The angular wave number (or propagation constant) is

$$k=\frac{2\pi}{\lambda}\qquad(\text{unit: rad/m}),$$

since the phase $kx$ must grow by $2\pi$ over one wavelength.

📘 Definition 14.4 — Period, frequency and angular frequency of a wave

The period $T$ of a wave is the time for any one element of the medium to complete one oscillation; its frequency is $\nu=\dfrac1T$ (in hertz) and its angular frequency is

$$\omega=\frac{2\pi}{T}=2\pi\nu\qquad(\text{unit: rad/s}).$$

These are the period, frequency and angular frequency of the SHM of each element, and equal those of the source.

How fast does the pattern move? Follow one crest: it is a place where the phase has a fixed value, $kx-\omega t+\phi=\text{constant}$. As $t$ increases by $\Delta t$, $x$ must increase by $\Delta x=\dfrac{\omega}{k}\Delta t$ to keep the phase the same. So every point of the waveform moves in the $+x$ direction at $\dfrac\omega k$. (For $kx+\omega t+\phi$, $x$ must decrease: the wave moves towards $-x$.) Equivalently, in one period each particle completes one oscillation while the pattern advances by exactly one wavelength.

📐 Result 14.2 — Speed of a progressive wave

The waveform $y=a\sin(kx\mp\omega t+\phi)$ moves without change of shape at the wave speed

$$v=\frac{\omega}{k}=\frac{\lambda}{T}=\nu\lambda .$$

The frequency is set by the source; the speed is set by the medium (Section 14.2); the wavelength then follows as $\lambda=\dfrac v\nu$.

✏️ Example 3 — Reading a wave equation

A wave on a string is $y(x,t)=0.030\sin\left(4.0\pi x-200\pi t+\dfrac\pi6\right)$, with $x$, $y$ in metres and $t$ in seconds. Find (a) the amplitude; (b) the wavelength and the wave number; (c) the frequency and the period; (d) the speed and direction of the wave; (e) the displacement of the element at $x=0$ at $t=0$.

💡 Show solution▶

(a) Compare with $a\sin(kx-\omega t+\phi)$: $a=0.030\ \text{m}=3.0\ \text{cm}$.

(b) $k=4.0\pi\ \text{rad/m}=12.6\ \text{rad/m}$, so $\lambda=\dfrac{2\pi}{k}=\dfrac{2\pi}{4.0\pi}=0.50\ \text{m}$.

(d) $v=\dfrac\omega k=\dfrac{200\pi}{4.0\pi}=50\ \text{m/s}$; the signs of $kx$ and $\omega t$ are opposite, so it travels towards $+x$.

(e) $y(0,0)=0.030\sin\dfrac\pi6=0.015\ \text{m}=1.5\ \text{cm}$.

Check $\nu\lambda=100\times0.50=50\ \text{m/s}$, the same as $\dfrac\omega k$ ✓.

✏️ Example 4 — Writing the equation of a sound wave

A sound wave of frequency $500\ \text{Hz}$ and displacement amplitude $2.0\ \mu\text{m}$ travels through air at $340\ \text{m/s}$ in the $-x$ direction. At $t=0$ the layer of air at $x=0$ is at its mean position and moving in the $+x$ direction. Write the displacement relation $y(x,t)$.

💡 Show solution▶

Constants $\omega=2\pi\times500=1000\pi=3.14\times10^3\ \text{rad/s}$; $\lambda=\dfrac{340}{500}=0.68\ \text{m}$; $k=\dfrac{2\pi}{0.68}=9.24\ \text{rad/m}$.

Form Travelling towards $-x$: $y=a\sin(kx+\omega t+\phi)$. (Here $y$ is a displacement along $x$: sound is longitudinal.)

Phase At $x=0$: $y=a\sin(\omega t+\phi)$. $y(0,0)=0$ needs $\sin\phi=0$; the layer moving towards $+x$ means $y$ is increasing, which picks $\phi=0$ (with $\phi=\pi$ it would be decreasing).

Answer $y(x,t)=2.0\times10^{-6}\sin\left(9.24x+1000\pi t\right)$, in metres and seconds.

Check $\dfrac\omega k=\dfrac{3141.6}{9.240}=340\ \text{m/s}$ ✓.

Both rows obey the same equation $y=a\sin(kx\mp\omega t)$ (shown in slow motion). Top: beads on a string, displaced across the direction of travel. Bottom: layers of a gas (or coils of a slinky), displaced along it — watch the compressions (C) and rarefactions (R) travel. The red element is at $x=1.0\ \text{m}$: follow it, and notice that it never leaves its own neighbourhood. Change $\lambda$, $\nu$ and the direction.

amplitude $a$:0.12 m$\lambda$:1.5 m$\nu$:0.50 Hz

wave speed $v=\nu\lambda$—

red element: displacement—

max particle speed $\omega a$—

$k$, $\omega$—

Phase difference and the motion of the particles

At a given instant, two elements a distance $\Delta x$ apart have phases $kx_1-\omega t+\phi$ and $kx_2-\omega t+\phi$. Their phase difference (Section 13.1) is $k\Delta x$, the same at every instant. The element farther along the direction of travel repeats the motion of the nearer one a time $\dfrac{\Delta x}{v}$ later — it lags by that time. At one place, the phase changes by $\omega\Delta t$ in a time $\Delta t$.

📐 Result 14.3 — Phase difference in a progressive wave

Two points a distance $\Delta x$ apart along a progressive wave differ in phase by

$$\Delta\phi=k\,\Delta x=\frac{2\pi}{\lambda}\,\Delta x ,$$

and the phase at one point changes by $\Delta\phi=\omega\,\Delta t=\dfrac{2\pi}{T}\Delta t$ in a time $\Delta t$. Points one wavelength apart (or any whole number of wavelengths) are in phase; points $\dfrac\lambda2$ apart (or any odd number of half-wavelengths) are in antiphase.

Each element of the medium is itself an oscillator, so it has a velocity and an acceleration of its own — quite different from the speed $v$ of the wave. At a fixed $x$, $y=a\sin(kx-\omega t+\phi)$ is an SHM in time, and the velocity and acceleration of SHM (Section 13.2) apply. The rate of change of $\sin(\theta_0-\omega t)$ with $t$ is $-\omega\cos(\theta_0-\omega t)$ (you will prove this in mathematics). To avoid a clash with the wave speed, write $v_y$ for the velocity of an element.

📐 Result 14.4 — Velocity and acceleration of an element of the medium

For the wave $y=a\sin(kx-\omega t+\phi)$, the element at $x$ has velocity and acceleration

$$v_y=-\omega a\cos(kx-\omega t+\phi),$$ $$a_y=-\omega^2a\sin(kx-\omega t+\phi)=-\omega^2y .$$

The maximum particle speed is $\omega a$ and the maximum particle acceleration is $\omega^2a$. The wave speed $v=\dfrac\omega k$ is a different quantity: the ratio $\dfrac{\omega a}{v}=ka=\dfrac{2\pi a}{\lambda}$, which is small for the gentle waves of everyday experience.

Which way is an element moving? On a snapshot of a wave travelling towards $+x$, imagine the whole curve slid a little to the right. An element on a stretch where the string rises ahead of it (positive slope) finds a lower part of the curve arriving, so it is moving down; on a falling stretch it is moving up; at a crest or a trough it is momentarily at rest. For a wave travelling towards $-x$ every one of these directions reverses.

✏️ Example 5 — Phase difference along a sound wave

A loudspeaker emits a $250\ \text{Hz}$ note into air in which sound travels at $300\ \text{m/s}$ (a cold day). Find (a) the wavelength; (b) the phase difference between two layers of air $0.40\ \text{m}$ apart along the direction of travel; (c) the smallest separation of two layers that are a quarter-cycle ($\dfrac\pi2$) out of phase; (d) the change of phase at one place in $1.0\ \text{ms}$.

💡 Show solution▶

(a) $\lambda=\dfrac v\nu=\dfrac{300}{250}=1.2\ \text{m}$.

(b) $\Delta\phi=\dfrac{2\pi}{1.2}\times0.40=\dfrac{2\pi}{3}\ \text{rad}=120^\circ$ (the farther layer lags).

(d) $\Delta\phi=2\pi\nu\,\Delta t=2\pi\times250\times1.0\times10^{-3}=\dfrac\pi2\ \text{rad}$.

Check In $1.0\ \text{ms}$ the wave moves $300\times0.0010=0.30\ \text{m}=\dfrac\lambda4$ — the same quarter-cycle seen in space and in time ✓. Interactive 2 starts with these values.

✏️ Example 6 — Particle speed versus wave speed

A wave on a rope is $y=0.050\sin(2.0x-40t)$ (SI units). Find (a) the wave speed; (b) the maximum speed of an element of the rope, and compare; (c) the velocity and (d) the acceleration of the element at $x=0.50\ \text{m}$ at $t=0.10\ \text{s}$.

💡 Show solution▶

(a) $v=\dfrac\omega k=\dfrac{40}{2.0}=20\ \text{m/s}$, towards $+x$.

(b) $\omega a=40\times0.050=2.0\ \text{m/s}$ — one tenth of the wave speed ($ka=0.10$).

(c) Phase $=2.0\times0.50-40\times0.10=-3.0\ \text{rad}$. $v_y=-\omega a\cos(-3.0)=-2.0\times(-0.990)=+2.0\ \text{m/s}$ ($1.98$): upward, almost the maximum, because the element is close to its mean position.

(d) $y=0.050\sin(-3.0)=-7.06\times10^{-3}\ \text{m}$, so $a_y=-\omega^2y=-1600\times(-7.06\times10^{-3})=+11\ \text{m/s}^2$ ($11.3$), upward — back towards the mean position.

Check $\left(\dfrac{v_y}{\omega a}\right)^2+\left(\dfrac{y}{a}\right)^2=0.980+0.020=1$ ✓, as for any SHM.

A sound wave travels towards $+x$ at $300\ \text{m/s}$ (played very slowly). Top: a snapshot of the displacement against $x$, with two layers $P$ and $Q$ a distance $\Delta x$ apart. Bottom: the history of each layer against time. Slide $\Delta x$ until $Q$ moves exactly in step with $P$, then exactly opposite to it. What fraction of a wavelength is each?

separation $\Delta x$:0.40 mfrequency $\nu$:250 Hz

wavelength $\lambda$—

$\Delta x/\lambda$—

phase difference $k\Delta x$—

Q lags P by—

⚠️ Common mistakes

1. "The medium travels with the wave." Only the disturbance and its energy travel; each element oscillates about its own mean position.

2. Confusing the wave speed with the particle speed. $v=\dfrac\omega k$ is the speed of the pattern and is fixed by the medium; $v_y$ changes from $0$ to $\omega a$ during each oscillation and depends on the amplitude.

3. Reading the direction of travel. Opposite signs of the $x$ and $t$ terms mean travel towards $+x$ — and that includes $a\sin(\omega t-kx)$, which is $-a\sin(kx-\omega t)$. The same signs mean travel towards $-x$.

4. $k$ is not the wavelength. $k=\dfrac{2\pi}{\lambda}$; in $\sin(4.0\pi x-\dots)$ the wavelength is $0.50\ \text{m}$, not $4.0\pi\ \text{m}$ or $\dfrac1{4.0\pi}\ \text{m}$.

5. Transverse waves in a gas. Through the bulk of a gas or liquid only longitudinal waves travel; a fluid has no shear rigidity.

6. Changing the frequency changes the speed. In a given medium the speed is fixed; doubling $\nu$ halves $\lambda$.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — any shape that travels, and the slope rule

In JEE Main, NEET and JEE Advanced (as part of progressive waves); not required by the CBSE syllabus.

The argument that gave Result 14.1 works for any shape of pulse, not only a sine curve. If the end of the string moves as $y(0,t)=g(t)$, the point $x$ repeats it a time $\dfrac xv$ later, so the displacement is a function of $x-vt$ only:

$$y(x,t)=f(x-vt)\ \ (\text{towards }+x),\qquad y(x,t)=f(x+vt)\ \ (\text{towards }-x).$$

So $y=\dfrac{0.20}{1+(x-3t)^2}$ is a pulse moving towards $+x$ at $3\ \text{m/s}$, while $y=\sin(x^2-3t)$ is not a travelling wave at all (its shape changes). Comparing the rates of change of $f(x-vt)$ with $t$ and with $x$ gives the slope rule, much used in JEE Advanced:

$$v_y=-v\times(\text{slope of the string, }\tfrac{\partial y}{\partial x}).$$

For the sine wave the slope is $ka\cos(kx-\omega t+\phi)$, and $-\dfrac\omega k\times ka\cos(\dots)$ reproduces Result 14.4. The slope rule turns the "slide the snapshot" argument above into a formula, and it shows that an element moves fastest where the string is steepest — at the mean position, not at a crest.

Exercises

Use SI units; phases are in radians. Give answers to 2–3 significant figures.

Kinds of waves.

1 State, with a reason, whether transverse waves can travel (a) along a stretched steel wire, (b) through air, (c) through the steel of a railway line.
2 Which of these is not a mechanical wave: sound in water, ripples on a pond, light from the Sun, a seismic P-wave? Why?
3 A cork floats on a pond as ripples pass it. Does the cork travel with the ripples? Describe its motion.
4 In a longitudinal wave, what is the distance between the centre of a compression and the centre of the next rarefaction, in terms of $\lambda$?
5 Find the wavelength in air of a $680\ \text{Hz}$ note if sound travels at $340\ \text{m/s}$.
6 Crests of a water wave are $4.0\ \text{m}$ apart, and a crest passes the end of a pier every $2.0\ \text{s}$. Find the speed of the wave.
7 The S-wave from an earthquake arrives $2.0$ minutes after the P-wave. Taking $8.0\ \text{km/s}$ (P) and $4.5\ \text{km/s}$ (S), find the distance to the earthquake.
8 Name the two properties a medium must have to carry a mechanical wave, and say what each does.

💡 Show solutions▶

1 (a) Yes: the tension pulls a displaced piece of wire back sideways. (b) No: a gas has no shear rigidity, so only longitudinal waves pass through it. (c) Yes: solid steel has a shear modulus (it carries both kinds).

2 Light: it is an electromagnetic wave and needs no material medium (it crosses empty space).

3 No. It bobs up and down and slightly to and fro (round a small loop) about the same place: the energy passes, the water (and cork) stays.

4 $\dfrac\lambda2$ (compression to compression is $\lambda$).

5 $\lambda=\dfrac{340}{680}=0.50\ \text{m}$.

6 $\lambda=4.0\ \text{m}$, $T=2.0\ \text{s}$: $v=\dfrac\lambda T=2.0\ \text{m/s}$.

7 $d\left(\dfrac1{4.5}-\dfrac1{8.0}\right)=120$: $d\times0.0972=120$, $d=1.2\times10^3\ \text{km}$ ($1234$).

8 Elasticity — a restoring force that pulls a displaced element back and couples it to its neighbours; inertia — so that a moving element overshoots and passes the disturbance on.

Reading and writing the displacement relation.

1 For $y=0.020\sin(2\pi x-100\pi t)$, find the amplitude, wavelength, frequency, period, speed and direction of travel.
2 For $y=5.0\sin(0.40\pi x+8.0\pi t)$, with $x$, $y$ in cm and $t$ in s, find $a$, $\lambda$, $\nu$, $v$ and the direction of travel.
3 Write the equation of a wave of amplitude $4.0\ \text{cm}$, wavelength $2.0\ \text{m}$ and speed $10\ \text{m/s}$ travelling towards $+x$, if the element at $x=0$ is at $y=0$ and moving downwards at $t=0$.
4 For $y=0.10\sin(4.0x-20t)$, find the displacement of the element at $x=0.50\ \text{m}$ at $t=0.20\ \text{s}$.
5 A wave is written $y=3.0\sin\left(\dfrac{2\pi t}{0.040}-\dfrac{2\pi x}{0.80}\right)$ (SI). Find its speed and direction.
6 Show that $y=a\cos(kx-\omega t)$ is a wave with the same speed and direction as $a\sin(kx-\omega t)$, and find the value of $\phi$ for which it equals $a\sin(kx-\omega t+\phi)$.
7 Find $k$ and $\omega$ for a $256\ \text{Hz}$ sound wave in air, where $v=343\ \text{m/s}$.
8 Show that $kx$ and $\omega t$ are dimensionless and that $\dfrac\omega k$ has the dimensions of speed.

💡 Show solutions▶

1 $a=0.020\ \text{m}$; $k=2\pi$, $\lambda=1.0\ \text{m}$; $\omega=100\pi$, $\nu=50\ \text{Hz}$, $T=0.020\ \text{s}$; $v=\dfrac{100\pi}{2\pi}=50\ \text{m/s}$ towards $+x$.

2 $a=5.0\ \text{cm}$; $\lambda=\dfrac{2\pi}{0.40\pi}=5.0\ \text{cm}$; $\nu=4.0\ \text{Hz}$; $v=\dfrac{8.0\pi}{0.40\pi}=20\ \text{cm/s}$ towards $-x$ (same signs).

3 $k=\dfrac{2\pi}{2.0}=\pi\ \text{rad/m}$, $\omega=vk=10\pi\ \text{rad/s}$. At $x=0$, $y=a\sin(\phi-\omega t)$; $y=0$ and decreasing at $t=0$ gives $\phi=0$: $y=0.040\sin(\pi x-10\pi t)$ m.

4 Phase $=2.0-4.0=-2.0\ \text{rad}$: $y=0.10\sin(-2.0)=-0.091\ \text{m}$.

5 $\omega=\dfrac{2\pi}{0.040}=50\pi$, $k=\dfrac{2\pi}{0.80}=2.5\pi$: $v=20\ \text{m/s}$. Opposite signs: towards $+x$ (it is $-3.0\sin(kx-\omega t)$).

6 $\cos\theta=\sin\left(\theta+\dfrac\pi2\right)$, so $a\cos(kx-\omega t)=a\sin\left(kx-\omega t+\dfrac\pi2\right)$: same $k$ and $\omega$, hence same $v=\dfrac\omega k$ and direction; $\phi=\dfrac\pi2$.

7 $\omega=2\pi\times256=1.61\times10^3\ \text{rad/s}$; $\lambda=\dfrac{343}{256}=1.34\ \text{m}$, $k=\dfrac{2\pi}{1.34}=4.69\ \text{rad/m}$.

8 $[k]=[\text{L}^{-1}]$ (radians are dimensionless) and $[x]=[\text{L}]$; $[\omega]=[\text{T}^{-1}]$, $[t]=[\text{T}]$. So $kx$, $\omega t$ are dimensionless — as the argument of a sine must be — and $\left[\dfrac\omega k\right]=[\text{L}\,\text{T}^{-1}]$.

III

Phase difference and the motion of the particles.

1 Find the phase difference between two points $0.60\ \text{m}$ apart on a wave of wavelength $2.4\ \text{m}$.
2 Two points $5.0\ \text{cm}$ apart on a $400\ \text{Hz}$ wave differ in phase by $\dfrac\pi3$. Find the wavelength and the wave speed.
3 By how much does the phase at one point of a $50\ \text{Hz}$ wave change in $5.0\ \text{ms}$?
4 For $y=0.020\sin(5.0x-100t)$, find the maximum particle speed and the wave speed.
5 For the same wave, find the velocity and acceleration of the element at $x=0$ at $t=0$.
6 A wave travels along a string towards $+x$. At one instant the element $P$ is at $y=0$ and the string slopes upwards to the right of $P$. Is $P$ moving up or down? What if the wave travels towards $-x$?
7 On a $200\ \text{Hz}$ wave, the nearest two points that are always in antiphase are $0.75\ \text{m}$ apart. Find the wavelength and the speed.
8 The elements of a medium carrying a wave have a maximum speed of $0.30\ \text{m/s}$ and an amplitude of $1.5\ \text{mm}$, and the wavelength is $0.20\ \text{m}$. Find $\omega$, $\nu$ and the wave speed.

💡 Show solutions▶

1 $\Delta\phi=\dfrac{2\pi}{2.4}\times0.60=\dfrac\pi2\ \text{rad}$ ($90^\circ$).

2 $\dfrac{2\pi}{\lambda}\times0.050=\dfrac\pi3$: $\lambda=0.30\ \text{m}$; $v=400\times0.30=1.2\times10^2\ \text{m/s}$.

3 $2\pi\times50\times0.0050=\dfrac\pi2\ \text{rad}$.

4 $\omega a=100\times0.020=2.0\ \text{m/s}$; $v=\dfrac{100}{5.0}=20\ \text{m/s}$.

5 Phase $0$: $v_y=-\omega a\cos0=-2.0\ \text{m/s}$ (downwards, at full speed); $y=0$, so $a_y=-\omega^2y=0$.

6 Towards $+x$: sliding the curve right brings the lower part (from the left) to $P$, so $P$ moves down (slope rule: $v_y=-v\times\text{slope}\lt0$). Towards $-x$: $P$ moves up.

7 $\dfrac\lambda2=0.75\ \text{m}$: $\lambda=1.5\ \text{m}$; $v=200\times1.5=3.0\times10^2\ \text{m/s}$.

8 $\omega=\dfrac{0.30}{1.5\times10^{-3}}=200\ \text{rad/s}$; $\nu=\dfrac{200}{2\pi}=32\ \text{Hz}$ ($31.8$); $v=\nu\lambda=6.4\ \text{m/s}$.

Applications and harder problems.

1 (Modelling) A medical ultrasound scanner uses $2.0\ \text{MHz}$ waves in soft tissue, where sound travels at $1540\ \text{m/s}$. Find the wavelength. Why do scanners use such high frequencies?
2 (Modelling) A bat hunts with $50\ \text{kHz}$ calls in air ($340\ \text{m/s}$). Find the wavelength, and comment on the size of the insects it can detect.
3 (Modelling) Over the deep ocean a tsunami travels at about $200\ \text{m/s}$ with a period of $15$ minutes. Find its wavelength, and explain why a ship at sea hardly notices a tsunami of amplitude $1\ \text{m}$ passing beneath it.
4 A wave is $y=0.030\sin\left(\dfrac\pi2x-4\pi t+\dfrac\pi3\right)$ (SI). Find its wavelength, frequency and speed, and the displacement and velocity of the element at $x=1.0\ \text{m}$ at $t=0$.
5 A wave of wavelength $0.80\ \text{m}$ travels towards $+x$. At $t=0$ the element at $x=0$ is at $y=+a$. Find the smallest positive $x$ at which an element is at $y=0$ and moving upwards at $t=0$.
6 Show that for any sinusoidal wave $\dfrac{\text{maximum particle speed}}{\text{wave speed}}=\dfrac{2\pi a}{\lambda}$. For what amplitude would the elements move faster than the wave?

💡 Show solutions▶

1 $\lambda=\dfrac{1540}{2.0\times10^6}=7.7\times10^{-4}\ \text{m}=0.77\ \text{mm}$. A wave cannot show detail much smaller than its wavelength, so a short wavelength (high frequency) gives a sharp image.

2 $\lambda=\dfrac{340}{5.0\times10^4}=6.8\ \text{mm}$: comparable to small insects, so the echoes reveal moths and midges a few millimetres across.

3 $\lambda=vT=200\times900=1.8\times10^5\ \text{m}=180\ \text{km}$. The sea surface rises by $1\ \text{m}$ over a horizontal distance of $45\ \text{km}$ (a quarter-wavelength) and does so over $3.75$ minutes — far too gentle to feel.

4 $\lambda=\dfrac{2\pi}{\pi/2}=4.0\ \text{m}$, $\nu=2.0\ \text{Hz}$, $v=8.0\ \text{m/s}$. Phase $=\dfrac\pi2+\dfrac\pi3=\dfrac{5\pi}6$: $y=0.030\sin\dfrac{5\pi}6=0.015\ \text{m}$; $v_y=-4\pi\times0.030\cos\dfrac{5\pi}6=+0.33\ \text{m/s}$ (upwards).

5 $y=a\cos(kx-\omega t)$, so at $t=0$ $y=a\cos kx$ and $v_y=a\omega\sin kx$. $y=0$ with $v_y\gt0$ needs $kx=\dfrac\pi2$: $x=\dfrac\lambda4=0.20\ \text{m}$. (The crest at $x=0$ is moving towards it ✓.)

6 $\dfrac{\omega a}{\omega/k}=ka=\dfrac{2\pi a}{\lambda}$. It exceeds $1$ if $a\gt\dfrac{\lambda}{2\pi}\approx0.16\lambda$ — a very steep wave, far outside the small-amplitude waves this chapter assumes.

🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks Distinguish between transverse and longitudinal waves, giving one example of each. Why can only longitudinal waves travel through a gas?

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Difference Transverse: elements oscillate perpendicular to the direction of travel (crests and troughs), e.g. a wave on a stretched string. Longitudinal: elements oscillate along the direction of travel (compressions and rarefactions), e.g. sound in air. [1]

Gas A transverse wave needs the medium to resist a sideways shear; a gas has no shear modulus, only a bulk modulus, so it can pass on only compressions and rarefactions. [1]

Exam tip Name the elastic property (shear modulus / bulk modulus) — "gases are not rigid" alone may lose the mark.

E2. Board · 3 marks A wave is described by $y=0.050\sin(20\pi t-0.50\pi x)$, with $x$, $y$ in metres and $t$ in seconds. Find (a) its wavelength, frequency and speed, and the direction of travel; (b) the phase difference between two points $1.0\ \text{m}$ apart; (c) the maximum speed of an element of the medium.

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(a) $k=0.50\pi$: $\lambda=\dfrac{2\pi}{0.50\pi}=4.0\ \text{m}$; $\omega=20\pi$: $\nu=10\ \text{Hz}$; $v=\nu\lambda=40\ \text{m/s}$, towards $+x$ (opposite signs). [1]

(b) $\Delta\phi=k\Delta x=0.50\pi\times1.0=\dfrac\pi2\ \text{rad}$. [1]

Exam tip The form $\sin(\omega t-kx)$ travels towards $+x$ just like $\sin(kx-\omega t)$; only the initial phase differs (by $\pi$).

E3. NEET Two points $0.80\ \text{m}$ apart on a wave of frequency $120\ \text{Hz}$ differ in phase by $\dfrac\pi2$. The speed of the wave is (a) $96\ \text{m/s}$ (b) $192\ \text{m/s}$ (c) $384\ \text{m/s}$ (d) $768\ \text{m/s}$

💡 Show solution▶

Idea $\dfrac\pi2$ is a quarter-cycle, so $0.80\ \text{m}=\dfrac\lambda4$ and $\lambda=3.2\ \text{m}$.

Answer $v=\nu\lambda=120\times3.2=384\ \text{m/s}$: (c).

Exam tip (a) takes $\lambda=0.80\ \text{m}$; (b) treats $\dfrac\pi2$ as half a cycle; (d) doubles instead of halving. Write $\Delta\phi=\dfrac{2\pi}{\lambda}\Delta x$ and solve — don't guess the fraction.

E4. JEE Main (Numerical value) A transverse wave on a string is $y=0.020\sin(4.0x-80t)$ in SI units. The maximum speed of an element of the string is $n\times10^{-2}$ times the speed of the wave. Find $n$.

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Idea $\dfrac{\omega a}{\omega/k}=ka$.

Numbers $ka=4.0\times0.020=0.080$ (check: $\omega a=1.6\ \text{m/s}$, $v=20\ \text{m/s}$, ratio $0.080$ ✓).

Answer $n=8$.

Exam tip The ratio $ka=\dfrac{2\pi a}{\lambda}$ does not involve $\omega$ — no need to find either speed separately.

E5. JEE Advanced (One or more correct) A wave on a string is $y=a\sin(kx+\omega t+\phi)$. Which statements are correct? (A) It travels towards $-x$ with speed $\dfrac\omega k$. (B) Elements a distance $\dfrac\lambda2$ apart always have equal and opposite displacements. (C) The maximum speed of an element equals the wave speed if $a=\lambda$. (D) Of two elements $\dfrac\lambda4$ apart, whenever one is at an extreme position the other is at its mean position.

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(A) Same signs of $kx$ and $\omega t$: constant phase needs $x$ to decrease as $t$ grows, at rate $\dfrac\omega k$ ✓.

(B) Phase difference $k\dfrac\lambda2=\pi$: $\sin(\theta+\pi)=-\sin\theta$ ✓.

(D) Phase difference $\dfrac\pi2$: when $\sin\theta=\pm1$, $\sin\left(\theta\pm\dfrac\pi2\right)=\cos\theta=0$ ✓.

Answer (A), (B), (D).

Exam tip Translate each distance into a phase difference first ($\lambda\to2\pi$, $\tfrac\lambda2\to\pi$, $\tfrac\lambda4\to\tfrac\pi2$); every statement then becomes a trigonometric identity.

E6. NEET At a seismic station the S-wave from an earthquake arrives $40\ \text{s}$ after the P-wave. If the P- and S-waves travel at $7.0\ \text{km/s}$ and $4.0\ \text{km/s}$, the earthquake is about (a) $160\ \text{km}$ (b) $280\ \text{km}$ (c) $373\ \text{km}$ (d) $440\ \text{km}$ away.

💡 Show solution▶

Idea $\dfrac d{4.0}-\dfrac d{7.0}=40$: $d\times\dfrac{3}{28}=40$.

Answer $d=\dfrac{1120}{3}=373\ \text{km}$: (c).

Exam tip (a) and (b) are the distances the S- and P-waves cover in $40\ \text{s}$; (d) adds the speeds. The delay depends on the difference of the travel times.

🔬 Research–Teaching NexusResearch-tutored · optional, about 60 minutes

Research question. Can you locate a real earthquake yourself, using nothing but the delay between its P- and S-waves at three seismic stations?

Why it matters. Every earthquake alert — from the National Center for Seismology in India to the United States Geological Survey and Japan's early-warning system, which can stop bullet trains before the strong shaking arrives — starts with exactly this calculation. Longitudinal P-waves outrun transverse S-waves, and the gap between them is a ruler for distance.

What to do. Use a free, public seismogram viewer (for example the IRIS "Wilber" service or the Raspberry Shake station map) and choose a recent earthquake of magnitude 5 or more.

Pick three stations at different distances and directions, $200$–$1500\ \text{km}$ from the event. On each seismogram, mark the first arrival (P) and the later, larger sideways arrival (S); read the delay $\Delta t$.
Assume crustal speeds $v_P=6.0\ \text{km/s}$ and $v_S=3.5\ \text{km/s}$ and find each distance from $d=\dfrac{\Delta t}{1/v_S-1/v_P}$.
On a map (paper, or an online map with a circle tool), draw a circle of radius $d$ round each station. Where do the three circles meet?
Compare your epicentre with the official one. Estimate how far off you are and why.
Look at a station on the far side of the Earth (more than about $104^\circ$ of arc away). Can you find an S-wave? What does this tell you about the core?

Tools: A web browser, a map with a circle or ruler tool, a calculator or spreadsheet.

Hand in: The three seismograms with P and S marked; a table of $\Delta t$ and $d$; your map with the three circles; a paragraph comparing your result with the official epicentre.

Going further. Use the official epicentre to measure $v_P$ and $v_S$ from the arrival times at many stations. Plot arrival time against distance: are the graphs straight? (Deeper paths travel through faster rock, so they curve.)

Instructor note: with $6.0$ and $3.5\ \text{km/s}$, $d\approx8.4\ \text{km}$ per second of delay, so a $60\ \text{s}$ delay is about $500\ \text{km}$ (Example 2). Students typically land within $50$–$100\ \text{km}$ of the official epicentre; the main errors are picking the S onset (it arrives in the P-wave's tail) and the depth of the focus. The S-wave shadow beyond about $104^\circ$ is the classic evidence for the liquid outer core.

🧪 Self-check quiz

0/5 correct

Q1. In a longitudinal wave, the elements of the medium oscillate

Q2. A wave is $y=0.040\sin(2.0x-10t)$ in SI units. Its speed is

Q3. Two points on a progressive wave are half a wavelength apart. Their oscillations are

Q4. Transverse mechanical waves cannot travel through

Q5. A source in a given medium doubles its frequency. The wavelength of the waves it sends out

← Chapter 14 — Waves 14.2 The Speed of a Travelling Wave →