NCERT Solutions Class 11 Physics Chapter 14 — Waves

NCERT Class 11 Physics — Waves, Chapter 14 Exercises. All 19 questions solved.

The exercises at the end of Chapter 14 fall into three groups. Questions 14.1 to 14.7 are about the speed of a wave — on a stretched string, and of sound in air, water and tissue. Questions 14.8 to 14.13 are about reading a wave from its equation: amplitude, frequency, wavelength, speed, direction and phase, and telling a travelling wave from a stationary one. Questions 14.14 to 14.19 are about standing waves on strings, rods and air columns, and about beats. The arithmetic runs on three sets of results:

• Wave speed — on a string of tension $T$ and mass per unit length $\mu$, $v = \sqrt{T/\mu}$; for sound in a gas, $v = \sqrt{\gamma P/\rho}$; and for every wave, $v = \nu\lambda$.
• A travelling harmonic wave — $y = a\sin(kx – \omega t + \phi)$, with $k = 2\pi/\lambda$, $\omega = 2\pi\nu$ and speed $v = \omega/k$. With $kx + \omega t$ instead, it travels towards $-x$.
• Normal modes — a string fixed at both ends, or an open pipe, $\nu_n = \dfrac{nv}{2L}$ with $n = 1, 2, 3, \ldots$; a pipe closed at one end, $\nu_n = \dfrac{(2n-1)v}{4L}$. Two notes of nearly equal frequency give $|\nu_1 – \nu_2|$ beats per second.

Key insight. Almost every question here is answered by looking at where $x$ and $t$ sit in the equation, or at where the nodes must go. If $x$ and $t$ appear together, as $kx \mp \omega t$, the pattern travels, and the ratio $\omega/k$ is its speed; if they sit in separate factors, as $\sin kx\cos\omega t$, it stands still (14.8, 14.11, 14.13). For resonance, a fixed end or closed end must be a displacement node and a free or open end an antinode, and the allowed wavelengths are just the ways of fitting quarter-wavelengths between them (14.14 to 14.17).

Question 14.1

A string of mass $2.50\ \text{kg}$ is under a tension of $200\ \text{N}$. The length of the stretched string is $20.0\ \text{m}$. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

Solution. The speed of a transverse wave on a string depends only on the tension and the mass per unit length. Here $\mu = 2.50/20.0 = 0.125\ \text{kg m}^{-1}$, so

$$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{200}{0.125}} = \sqrt{1600} = 40\ \text{m s}^{-1}$$

The pulse covers the $20.0\ \text{m}$ at this steady speed, taking $t = 20.0/40 = 0.5\ \text{s}$.

Question 14.2

A stone dropped from the top of a tower of height $300\ \text{m}$ splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is $340\ \text{m s}^{-1}$? ($g = 9.8\ \text{m s}^{-2}$)

Solution. Two things take time: the stone falling down, and the sound of the splash travelling back up. The stone starts from rest, so $h = \tfrac12gt_1^2$ gives

$$t_1 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\times300}{9.8}} = 7.82\ \text{s}$$

The sound travels the $300\ \text{m}$ at a constant speed, taking $t_2 = 300/340 = 0.88\ \text{s}$. The splash is heard $t_1 + t_2 = 8.71\ \text{s}$ after the stone is let go.

Question 14.3

A steel wire has a length of $12.0\ \text{m}$ and a mass of $2.10\ \text{kg}$. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at $20\,^\circ\text{C} = 343\ \text{m s}^{-1}$.

Solution. Turning $v = \sqrt{T/\mu}$ round gives $T = \mu v^2$. The mass per unit length is $\mu = 2.10/12.0 = 0.175\ \text{kg m}^{-1}$, so

$$T = (0.175)(343)^2 = 2.06\times10^4\ \text{N}$$

Question 14.4

Use the formula $v = \sqrt{\dfrac{\gamma P}{\rho}}$ to explain why the speed of sound in air (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity.

Solution. Treat air as an ideal gas of molar mass $M$. Then $PV = nRT$ can be written in terms of the density as $P = \dfrac{\rho RT}{M}$, so $\dfrac{P}{\rho} = \dfrac{RT}{M}$ and

$$v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma RT}{M}}$$

(a) At a fixed temperature, raising the pressure raises the density in the same proportion, so $P/\rho$ — and with it $v$ — does not change. The last form shows this directly: $P$ does not appear in it.

(b) $v\propto\sqrt T$, so the speed rises with temperature.

(c) Water vapour has a molar mass of $18$, less than that of nitrogen ($28$) or oxygen ($32$). Humid air has some of its nitrogen and oxygen replaced by water vapour, so at the same pressure and temperature its average molar mass, and its density, are lower, and $v$ is higher.

Question 14.5

You have learnt that a travelling wave in one dimension is represented by a function $y = f(x, t)$ where $x$ and $t$ must appear in the combination $x – vt$ or $x + vt$, i.e. $y = f(x \pm vt)$. Is the converse true? Examine if the following functions for $y$ can possibly represent a travelling wave: (a) $(x – vt)^2$, (b) $\log\left[(x + vt)/x_0\right]$, (c) $1/(x + vt)$.

Solution. The converse is not true. Any function of $x – vt$ keeps its shape and slides along at speed $v$, but that alone does not make it a wave that a medium can carry. $y$ is the displacement of a real particle of the medium, so it must be finite everywhere and at all times.

(a) $(x – vt)^2$ grows without limit as $|x – vt|$ grows: far from the point $x = vt$ the displacement would be enormous.

(b) $\log[(x + vt)/x_0]$ has no value at all where $(x + vt)/x_0$ is zero or negative, and it runs off to $-\infty$ as $x + vt \to 0$.

(c) $1/(x + vt)$ dies away far from $x = -vt$, but at $x = -vt$ it is infinite. That point exists at every instant — it simply moves along at speed $v$ — so the displacement is infinite somewhere at every moment.

Note on the book’s answer. The key agrees that the converse is false and states the right test — an acceptable wave must be finite everywhere and at all times — but then says that only function (c) passes it. It does not: $1/(x + vt)$ is infinite at $x = -vt$. By the key’s own test all three fail. Function (c) is the least bad, finite and dying away everywhere except at one moving point, which is presumably why it was singled out.

Question 14.6

A bat emits ultrasonic sound of frequency $1000\ \text{kHz}$ in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is $340\ \text{m s}^{-1}$ and in water $1486\ \text{m s}^{-1}$.

Solution. The frequency is set by the source: every layer of the medium oscillates in step with the one that drives it, so the frequency does not change on reflection or on crossing into the water. The speed depends on the medium, so the wavelength $\lambda = v/\nu$ changes with it. With $\nu = 10^6\ \text{Hz}$:

(a) The reflected sound travels back through air: $\lambda = 340/10^6 = 3.4\times10^{-4}\ \text{m}$.

(b) The transmitted sound travels in water: $\lambda = 1486/10^6 = 1.49\times10^{-3}\ \text{m}$.

Question 14.7

A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is $1.7\ \text{km s}^{-1}$? The operating frequency of the scanner is $4.2\ \text{MHz}$.

Solution. $\lambda = \dfrac{v}{\nu} = \dfrac{1.7\times10^3}{4.2\times10^6} = 4.05\times10^{-4}\ \text{m}$. A wavelength of about $0.4\ \text{mm}$ is what lets the scanner pick out small structures: a wave cannot resolve details much smaller than its own wavelength.

Question 14.8

A transverse harmonic wave on a string is described by

$$y(x, t) = 3.0\sin(36\,t + 0.018\,x + \pi/4)$$

where $x$ and $y$ are in cm and $t$ in s. The positive direction of $x$ is from left to right. (a) Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation? (b) What are its amplitude and frequency? (c) What is the initial phase at the origin? (d) What is the least distance between two successive crests in the wave?

Comparing with $y = a\sin(\omega t + kx + \phi)$: $a = 3.0\ \text{cm}$, $\omega = 36\ \text{rad s}^{-1}$, $k = 0.018\ \text{rad cm}^{-1}$ and $\phi = \pi/4$.

Question 14.8 (a)

Solution. $x$ and $t$ appear together, in the single combination $36t + 0.018x = 0.018\,(x + 2000t)$, so the whole shape slides along as time passes: it is a travelling wave. A function of $x + vt$ moves towards $-x$, since a given phase is found at a smaller $x$ as $t$ increases. The speed is

$$v = \frac{\omega}{k} = \frac{36}{0.018} = 2000\ \text{cm s}^{-1} = 20\ \text{m s}^{-1}$$

Question 14.8 (b)

Solution. The amplitude is the coefficient of the sine, $3.0\ \text{cm}$. The frequency is $\nu = \dfrac{\omega}{2\pi} = \dfrac{36}{2\pi} = 5.7\ \text{Hz}$.

Question 14.8 (c)

Solution. At the origin, $x = 0$, and at the start, $t = 0$, the argument of the sine is just $\pi/4$.

Question 14.8 (d)

Solution. Successive crests are one wavelength apart:

$$\lambda = \frac{2\pi}{k} = \frac{2\pi}{0.018} = 349\ \text{cm} \approx 3.5\ \text{m}$$

Question 14.9

For the wave described in Exercise 14.8, plot the displacement ($y$) versus ($t$) graphs for $x = 0$, $2$ and $4\ \text{cm}$. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?

Solution. Fixing $x$ in the wave equation leaves the motion of that one point:

$$x = 0:\ \ y = 3.0\sin(36t + 0.785) \qquad x = 2\ \text{cm}:\ \ y = 3.0\sin(36t + 0.821) \qquad x = 4\ \text{cm}:\ \ y = 3.0\sin(36t + 0.857)$$

Each is simple harmonic motion with amplitude $3.0\ \text{cm}$ and period $T = 2\pi/36 = 0.175\ \text{s}$. Moving $2\ \text{cm}$ along the string adds $0.018\times2 = 0.036\ \text{rad}$ to the phase, so each point runs $0.036/36 = 0.001\ \text{s}$ ahead of the one on its left. The point at $x = 4\ \text{cm}$ leads, as it should: the wave travels from right to left and reaches it first.

The three graphs look identical at this scale, and that is itself the lesson: points $2\ \text{cm}$ apart on a wave of wavelength $349\ \text{cm}$ move almost in step. What changes from point to point is only the phase.

Question 14.10

For the travelling harmonic wave

$$y(x, t) = 2.0\cos2\pi(10t – 0.0080\,x + 0.35)$$

where $x$ and $y$ are in cm and $t$ in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of (a) $4\ \text{m}$, (b) $0.5\ \text{m}$, (c) $\lambda/2$, (d) $3\lambda/4$.

Solution. At a fixed instant, the phase changes with position at the rate $k = 2\pi\times0.0080\ \text{rad cm}^{-1}$, so two points $\Delta x$ apart differ in phase by $\Delta\phi = k\,\Delta x$. The wavelength is $\lambda = 2\pi/k = 1/0.0080 = 125\ \text{cm}$. The distances must be in centimetres, to match $k$:

(a) $\Delta x = 400\ \text{cm}$: $\Delta\phi = 2\pi(0.0080)(400) = 6.4\pi\ \text{rad}$.

(b) $\Delta x = 50\ \text{cm}$: $\Delta\phi = 2\pi(0.0080)(50) = 0.8\pi\ \text{rad}$.

(c) $\Delta x = \lambda/2$: $\Delta\phi = \dfrac{2\pi}{\lambda}\cdot\dfrac\lambda2 = \pi\ \text{rad}$.

(d) $\Delta x = 3\lambda/4$: $\Delta\phi = \dfrac{2\pi}{\lambda}\cdot\dfrac{3\lambda}{4} = \dfrac{3\pi}{2}\ \text{rad}$.

Note on the book’s answer. The key prints $\pi/2$ for (d). Phases that differ by $2\pi$ describe the same motion, so a lag of $3\pi/2$ is the same as a lead of $\pi/2$: the key has described the same relationship from the other point’s side. $3\pi/2$ is what $k\,\Delta x$ gives, and it matches the key’s own (a), where $6.4\pi$ is left as it is rather than being reduced to $0.4\pi$.

Question 14.11

The transverse displacement of a string (clamped at its both ends) is given by

$$y(x, t) = 0.06\sin\left(\frac{2\pi}{3}x\right)\cos(120\pi t)$$

where $x$ and $y$ are in m and $t$ in s. The length of the string is $1.5\ \text{m}$ and its mass is $3.0\times10^{-2}\ \text{kg}$. Answer the following: (a) Does the function represent a travelling wave or a stationary wave? (b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave? (c) Determine the tension in the string.

Question 14.11 (a)

Solution. Here $x$ and $t$ sit in separate factors. The shape $\sin(2\pi x/3)$ never moves; the factor $\cos(120\pi t)$ only scales it up and down. The points where $\sin(2\pi x/3) = 0$ — at $x = 0$ and $x = 1.5\ \text{m}$, the clamped ends — never move at all. A fixed pattern with fixed nodes is a stationary wave.

Question 14.11 (b)

Solution. The identity $2\sin A\cos B = \sin(A – B) + \sin(A + B)$ splits the product into two travelling waves:

$$y = 0.03\sin\left(\frac{2\pi}{3}x – 120\pi t\right) + 0.03\sin\left(\frac{2\pi}{3}x + 120\pi t\right)$$

The first travels towards $+x$ and the second towards $-x$, each with amplitude $0.03\ \text{m}$. For each, $k = 2\pi/3\ \text{m}^{-1}$ and $\omega = 120\pi\ \text{rad s}^{-1}$, so

$$\lambda = \frac{2\pi}{k} = 3\ \text{m}, \qquad \nu = \frac{\omega}{2\pi} = 60\ \text{Hz}, \qquad v = \nu\lambda = 180\ \text{m s}^{-1}$$

The string is $1.5\ \text{m} = \lambda/2$ long, so it is vibrating in its fundamental mode, a single loop, as the figure shows.

Question 14.11 (c)

Solution. The speed of the component waves is fixed by the tension, $v = \sqrt{T/\mu}$. The mass per unit length is $\mu = 3.0\times10^{-2}/1.5 = 2.0\times10^{-2}\ \text{kg m}^{-1}$, so

$$T = \mu v^2 = (2.0\times10^{-2})(180)^2 = 648\ \text{N}$$

Question 14.12

(i) For the wave on a string described in Exercise 14.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point $0.375\ \text{m}$ away from one end?

Note on the question. The book refers to “Exercise 15.11” here. There is no such exercise in this chapter; the reference is left over from earlier editions, in which Waves was Chapter 15. The wave meant is the one in Exercise 14.11.

Question 14.12 (i)

Solution. Every point moves as $y = \left[0.06\sin(2\pi x/3)\right]\cos(120\pi t)$: an amplitude that depends on where the point is, times one common time factor.

(a) The time factor $\cos(120\pi t)$ is the same for every point, so all points (apart from the nodes at the ends, which do not move) oscillate at $60\ \text{Hz}$.

(b) Points pass through their maxima together wherever the amplitude factor $\sin(2\pi x/3)$ has the same sign. In general, points in neighbouring loops of a stationary wave move in opposite phase. Here, though, the string is exactly one loop long: $\sin(2\pi x/3)$ is positive all the way from $x = 0$ to $x = 1.5\ \text{m}$, so every point moves in the same phase.

(c) The amplitude $0.06\sin(2\pi x/3)$ varies along the string, from zero at the ends to $0.06\ \text{m}$ at the middle.

Question 14.12 (ii)

Solution. The amplitude of the point at $x$ is $0.06\sin(2\pi x/3)$. At $x = 0.375\ \text{m}$:

$$0.06\sin\left(\frac{2\pi}{3}\times0.375\right) = 0.06\sin\frac\pi4 = 0.06\times0.707 = 0.042\ \text{m}$$

By symmetry the point $0.375\ \text{m}$ from the other end has the same amplitude.

Question 14.13

Given below are some functions of $x$ and $t$ to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all: (a) $y = 2\cos(3x)\sin(10t)$, (b) $y = 2\sqrt{x – vt}$, (c) $y = 3\sin(5x – 0.5t) + 4\cos(5x – 0.5t)$, (d) $y = \cos x\sin t + \cos2x\sin2t$.

Solution. (a) $x$ and $t$ are in separate factors: a fixed shape $\cos3x$ scaled up and down by $\sin10t$, with fixed nodes where $\cos3x = 0$. It is a stationary wave.

(b) Although it is a function of $x – vt$, it has no real value where $x < vt$ and it grows without limit as $x – vt$ grows. As in Question 14.5, a displacement must be finite and defined everywhere, so this represents no wave at all.

(c) Both terms contain the same combination $5x – 0.5t$, and a sine plus a cosine of the same argument is a single sine: $3\sin\theta + 4\cos\theta = 5\sin(\theta + \alpha)$ with $\tan\alpha = 4/3$. So $y = 5\sin(5x – 0.5t + \alpha)$, a harmonic wave of amplitude $5$ travelling towards $+x$ at speed $0.5/5 = 0.1$ (in the units of $x$ per unit of $t$).

(d) Each term on its own is a stationary wave, the first with $k = 1$, $\omega = 1$ and the second with $k = 2$, $\omega = 2$. Nothing travels; it is what a string does when it vibrates in two of its modes at once — the superposition of two stationary waves.

Question 14.14

A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45\ \text{Hz}$. The mass of the wire is $3.5\times10^{-2}\ \text{kg}$ and its linear mass density is $4.0\times10^{-2}\ \text{kg m}^{-1}$. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

Solution. The length is not given directly, but it follows from the mass and the mass per unit length: $L = 3.5\times10^{-2}/4.0\times10^{-2} = 0.875\ \text{m}$. Both ends are fixed, so both are nodes; in the fundamental mode there are no other nodes, and the wire is half a wavelength long. So $\lambda = 2L = 1.75\ \text{m}$.

(a) $v = \nu\lambda = 45\times1.75 = 78.75 \approx 79\ \text{m s}^{-1}$.

(b) $T = \mu v^2 = (4.0\times10^{-2})(78.75)^2 = 248\ \text{N}$.

Question 14.15

A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency $340\ \text{Hz}$) when the tube length is $25.5\ \text{cm}$ or $79.3\ \text{cm}$. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.

Solution. With the piston in place the air column is closed at one end. The closed end must be a displacement node, because the air cannot move into the piston, and the open end an antinode, because the air there is free to move. The column resonates when its length fits an odd number of quarter-wavelengths, $\ell = (2n – 1)\lambda/4$, which is the hint’s $\nu_n = (2n-1)v/4\ell$.

The shorter length is the first resonance, $\ell_1 = \lambda/4$:

$$\lambda = 4\times25.5\ \text{cm} = 102\ \text{cm}, \qquad v = \nu\lambda = 340\times1.02 = 347\ \text{m s}^{-1}$$

The second length is the next resonance, $3\lambda/4$: $79.3/25.5 = 3.1$, close to $3$. (The next, $5\lambda/4 \approx 128\ \text{cm}$, would not fit in a metre-long tube.) The estimate is only as good as the neglect of edge effects: $79.3\ \text{cm}$ taken alone as $3\lambda/4$ gives $359\ \text{m s}^{-1}$, and the difference of the two lengths, which is exactly $\lambda/2$ whatever the end correction, gives $366\ \text{m s}^{-1}$.

Question 14.16

A steel rod $100\ \text{cm}$ long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be $2.53\ \text{kHz}$. What is the speed of sound in steel?

Solution. The clamp holds the middle of the rod still, so the middle is a displacement node. The ends are free, so they are antinodes. In the fundamental mode there are no other nodes, so the rod holds one antinode-to-antinode stretch, which is half a wavelength: $L = \lambda/2$, giving $\lambda = 2\times1.00 = 2.00\ \text{m}$.

$$v = \nu\lambda = (2.53\times10^3)(2.00) = 5.06\times10^3\ \text{m s}^{-1}$$

Question 14.17

A pipe $20\ \text{cm}$ long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a $430\ \text{Hz}$ source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is $340\ \text{m s}^{-1}$).

Solution. A pipe closed at one end has $\nu_n = (2n-1)\dfrac{v}{4L}$. Its fundamental is

$$\nu_1 = \frac{340}{4\times0.20} = 425\ \text{Hz}$$

and the next mode is $3\times425 = 1275\ \text{Hz}$. The $430\ \text{Hz}$ source is within about $1\%$ of $425\ \text{Hz}$ — as close as the data allow, since the speed of sound and the end effects are not known that precisely — so it excites the fundamental, the first harmonic.

With both ends open, $\nu_n = n\dfrac{v}{2L}$, and the lowest mode is $\nu_1 = 340/0.40 = 850\ \text{Hz}$. Every mode is $850\ \text{Hz}$ or higher, so none is anywhere near $430\ \text{Hz}$ and there is no resonance.

Question 14.18

Two sitar strings A and B playing the note ‘Ga‘ are slightly out of tune and produce beats of frequency $6\ \text{Hz}$. The tension in the string A is slightly reduced and the beat frequency is found to reduce to $3\ \text{Hz}$. If the original frequency of A is $324\ \text{Hz}$, what is the frequency of B?

Solution. Six beats per second means the frequencies differ by $6\ \text{Hz}$, so B is either $324 – 6 = 318\ \text{Hz}$ or $324 + 6 = 330\ \text{Hz}$. The change in tension decides which. A string’s frequency is proportional to $\sqrt{T}$, so reducing A’s tension lowers A’s frequency.

If B were $330\ \text{Hz}$, lowering A from $324\ \text{Hz}$ would move it further from B and the beats would become faster than $6\ \text{Hz}$. If B is $318\ \text{Hz}$, lowering A moves it towards B, and the beats slow down — to $3\ \text{Hz}$ when A reaches $321\ \text{Hz}$. That is what was observed.

Question 14.19

Explain why (or how): (a) in a sound wave, a displacement node is a pressure antinode and vice versa, (b) bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”, (c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes, (d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and (e) the shape of a pulse gets distorted during propagation in a dispersive medium.

Question 14.19 (a)

Solution. The pressure change in a sound wave depends not on how far a layer of air has moved but on how much it has been squeezed or stretched, that is, on how differently neighbouring layers have moved. At a displacement node the layer itself stays put while the layers on either side move towards it together, then away from it together, so the air there is compressed and rarefied the most: a pressure antinode. At a displacement antinode neighbouring layers move by nearly the same amount in the same direction, so the air is carried back and forth without being squeezed: a pressure node.

Question 14.19 (b)

Solution. Bats emit short pulses of ultrasound and listen for the echoes. The time taken for an echo to return gives the distance ($d = vt/2$, since the sound goes there and back). The difference between the echo’s arrival at the two ears, and its loudness at each, gives the direction. The strength of the echo, and how it changes with frequency, depend on the size and on how hard or soft the reflecting surface is. Because ultrasound has a very short wavelength, even small insects reflect it well.

Question 14.19 (c)

Solution. A note from a real instrument is not a single pure frequency. It is its fundamental plus a set of overtones (harmonics), and which overtones are present, and how strong each is, depends on the instrument and how it is played. Two notes with the same fundamental — the same pitch — can therefore have quite different waveforms. The ear hears that difference as the quality, or timbre, of the sound.

Question 14.19 (d)

Solution. A wave needs a restoring force for the kind of deformation it causes. A transverse wave shears the medium, sliding one layer sideways past the next, so it needs a medium with a shear modulus — one that resists a change of shape. Solids have one. Gases have no shear modulus: their layers slide past each other with no restoring force. Longitudinal waves compress and expand the medium, which needs only a bulk modulus, and every medium, gases included, resists compression.

Question 14.19 (e)

Solution. A pulse is not a single harmonic wave but a superposition of many harmonic waves of different wavelengths. In a dispersive medium the speed of a wave depends on its wavelength, so these components travel at different speeds and drift out of step with one another. Their sum — the shape of the pulse — therefore changes as it travels, usually spreading out.

Common mistakes

• Question 14.2: forgetting one of the two times. The splash is heard after the fall and the sound’s return trip. Using only the fall gives $7.8\ \text{s}$; using only the sound, $0.88\ \text{s}$.
• Questions 14.5 and 14.13(b): assuming any function of $x – vt$ is a wave. Being a function of $x \pm vt$ is necessary but not sufficient; a real displacement must also be finite and defined everywhere. $2\sqrt{x – vt}$ and $(x – vt)^2$ fail.
• Question 14.8(a): getting the direction backwards. With $\omega t$ and $kx$ added, as in $36t + 0.018x$, the wave travels towards $-x$, from right to left. Students often read any wave equation as moving towards $+x$.
• Question 14.10: mixing centimetres and metres. $k$ is in rad per centimetre, so $4\ \text{m}$ must go in as $400\ \text{cm}$. Leaving it as $4$ gives $0.064\pi$ instead of $6.4\pi$.
• Question 14.11(c): using the total mass instead of the mass per unit length. The tension is $\mu v^2$ with $\mu = 0.02\ \text{kg m}^{-1}$; using $3.0\times10^{-2}\ \text{kg}$ as it stands gives $972\ \text{N}$.
• Question 14.12(i)(b): saying the points differ in phase. That is true of a travelling wave. In a stationary wave all points within one loop move in phase, and here the whole string is one loop.
• Question 14.15: treating the tube as open at both ends. Taking $25.5\ \text{cm}$ as $\lambda/2$ gives $173\ \text{m s}^{-1}$. The piston makes a closed end, a node, so the first resonance is $\lambda/4$.
• Question 14.18: choosing $330\ \text{Hz}$. Both $318$ and $330\ \text{Hz}$ give $6$ beats; only the effect of lowering A’s tension, which lowers its frequency, tells them apart.

Practise next

• Chapter 13, Oscillations — every point of a medium carrying a harmonic wave executes the simple harmonic motion of that chapter; revise amplitude, phase and period there.
• Chapter 8, Mechanical Properties of Solids — the elastic moduli behind wave speeds: Young’s modulus sets the speed of sound in the rod of Question 14.16, and the shear modulus is why solids carry transverse waves (14.19(d)).