CBSE · NCERT · Class 11 Physics · Chapter 14

NCERT Solutions: Class 11 Physics Chapter 14 - Waves

17 textbook Q&A17 verifiedFree Content

Chapter-wise NCERT intext questions and exercise answers for Waves, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Exercises 17
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1Exercises17 questions
Q.14.1A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?v
Solution

The linear mass density is $\mu=m/L=2.50/20.0=0.125\,\text{kg m}^{-1}$. The speed of a transverse wave on a stretched string is $v=\sqrt{T/\mu}=\sqrt{200/0.125}=40\,\text{m s}^{-1}$. Time to travel $20.0\,\text{m}$ is $t=L/v=20.0/40=0.50\,\text{s}$.

Answer:

$0.50\,\text{s}$.

Q.14.2A stone dropped from the top of a tower of height 300 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 m s–1 ? (g = 9.8 m s–2)v
Solution

The stone takes $t_1=\sqrt{2h/g}=\sqrt{600/9.8}=7.82\,\text{s}$ to reach the water. The splash sound then takes $t_2=h/v=300/340=0.882\,\text{s}$ to reach the top. Total time $t=t_1+t_2=8.70\,\text{s}$.

Answer:

$8.7\,\text{s}$ after the stone is dropped.

Q.14.3A steel wire has a length of 12.0 m and a mass of 2.10 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 °C = 343 m s–1.v
Solution

The linear mass density is $\mu=2.10/12.0=0.175\,\text{kg m}^{-1}$. For a string, $v=\sqrt{T/\mu}$, so $T=\mu v^2=0.175(343)^2=2.06\times10^4\,\text{N}$.

Answer:

$2.1\times10^4\,\text{N}$ approximately.

Q.14.4Use the formula v = √(γP/ρ) to explain why the speed of sound in air (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity.v
Solution

For an ideal gas, $P/\rho=RT/M$, where $M$ is molar mass. Substituting in $v=\sqrt{\gamma P/\rho}$ gives $v=\sqrt{\gamma RT/M}$. Thus pressure cancels out at fixed temperature. Increasing temperature increases $T$, so $v$ increases. Moist air contains water vapour whose molar mass is smaller than that of dry air; this reduces the effective molar mass and density for the same pressure and temperature, so the speed increases.

Answer:

Using the ideal-gas relation, $v=\sqrt{\gamma RT/M}$; hence it is independent of pressure at fixed temperature, increases as $\sqrt{T}$, and increases when humidity lowers the effective molar mass of air.

Q.14.5You 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 – v t or x + v t, i.e. y = f (x ± v t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave : (a) (x – vt )2 (b) log [(x + vt)/x0] (c) 1/(x + vt)v
Solution

Each expression is a function of only one travelling combination. In (a), $y=(x-vt)^2$ has the form $f(x-vt)$ and travels in the positive x-direction. In (b), $y=\log[(x+vt)/x_0]$ has the form $f(x+vt)$ and travels in the negative x-direction where the logarithm is defined. In (c), $y=1/(x+vt)$ also has the form $f(x+vt)$, except at its singular point. Thus all can possibly represent travelling-wave shapes.

Answer:

Yes. All three can represent travelling waves mathematically, subject to their domains.

Q.14.6A bat emits ultrasonic sound of frequency 1000 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 m s –1 and in water 1486 m s–1.v
Solution

The frequency remains the same on reflection and transmission: $f=1000\,\text{kHz}=1.0\times10^6\,\text{Hz}$. For reflected sound in air, $\lambda_a=v_a/f=340/(1.0\times10^6)=3.40\times10^{-4}\,\text{m}$. For transmitted sound in water, $\lambda_w=1486/(1.0\times10^6)=1.486\times10^{-3}\,\text{m}$.

Answer:

(a) $3.40\times10^{-4}\,\text{m}$; (b) $1.486\times10^{-3}\,\text{m}$.

Q.14.7A 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 km s–1 ? The operating frequency of the scanner is 4.2 MHz.v
Solution

Here $v=1.7\,\text{km s}^{-1}=1700\,\text{m s}^{-1}$ and $f=4.2\,\text{MHz}=4.2\times10^6\,\text{Hz}$. Thus $\lambda=v/f=1700/(4.2\times10^6)=4.05\times10^{-4}\,\text{m}=0.405\,\text{mm}$.

Answer:

$4.0\times10^{-4}\,\text{m}$, or about $0.40\,\text{mm}$.

Q.14.8A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x + π/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 ?v
Solution

The argument has the form $\omega t+kx+\phi$, so the wave travels in the negative x-direction. Here $\omega=36\,\text{s}^{-1}$ and $k=0.018\,\text{cm}^{-1}$. Hence $v=\omega/k=36/0.018=2000\,\text{cm s}^{-1}$. The amplitude is $3.0\,\text{cm}$ and $f=\omega/(2\pi)=36/(2\pi)=5.73\,\text{Hz}$. At $x=0,t=0$, the phase is $\pi/4$. The distance between successive crests is $\lambda=2\pi/k=2\pi/0.018=349\,\text{cm}$.

Answer:

(a) It is a travelling wave moving in the negative x-direction with speed $2000\,\text{cm s}^{-1}$, or $20\,\text{m s}^{-1}$. (b) Amplitude $3.0\,\text{cm}$ and frequency $5.7\,\text{Hz}$. (c) Initial phase $\pi/4$. (d) $349\,\text{cm}$.

Q.14.10For the travelling harmonic wave y(x, t) = 2.0 cos 2π (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 m, (b) 0.5 m, (c) λ/2, (d) 3λ/4v
Solution

The phase changes with position as $2\pi(0.0080x)$, where $x$ is in cm. Thus $\Delta\phi=2\pi(0.0080\Delta x)$. For $4\,\text{m}=400\,\text{cm}$, $\Delta\phi=2\pi(0.0080)(400)=6.4\pi$. For $0.5\,\text{m}=50\,\text{cm}$, $\Delta\phi=0.8\pi$. Also, by definition, separation $\lambda/2$ gives phase difference $\pi$, and $3\lambda/4$ gives $3\pi/2$.

Answer:

(a) $6.4\pi\,\text{rad}$; (b) $0.8\pi\,\text{rad}$; (c) $\pi\,\text{rad}$; (d) $3\pi/2\,\text{rad}$.

Q.14.11The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin (2πx/3) cos (120 πt) where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 ×10–2 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.v
Solution

The form $A\sin kx\cos\omega t$ is a standing wave. Since $2a=0.06$, each travelling component has amplitude $a=0.03\,\text{m}$: $y_1=0.03\sin(2\pi x/3-120\pi t)$ and $y_2=0.03\sin(2\pi x/3+120\pi t)$. Here $k=2\pi/3$, so $\lambda=3.0\,\text{m}$, and $\omega=120\pi$, so $f=60\,\text{Hz}$. The speed is $v=f\lambda=180\,\text{m s}^{-1}$. The linear density is $\mu=3.0\times10^{-2}/1.5=2.0\times10^{-2}\,\text{kg m}^{-1}$. Hence tension $T=\mu v^2=0.020(180)^2=648\,\text{N}$.

Answer:

(a) A stationary wave. (b) It is the superposition of two waves of amplitude $0.03\,\text{m}$ travelling in opposite directions, each with wavelength $3.0\,\text{m}$, frequency $60\,\text{Hz}$ and speed $180\,\text{m s}^{-1}$. (c) $648\,\text{N}$.

Q.14.13Given 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√(x – vt) (c) y = 3 sin (5x – 0.5t) + 4 cos (5x – 0.5t) (d) y = cos x sin t + cos 2x sin 2tv
Solution

A stationary wave has separated position and time factors, as in (a), $2\cos3x\sin10t$. Expression (b) is a function of $x-vt$, so it is a travelling wave where the square root is defined. In (c), both terms depend on the same travelling combination $5x-0.5t$, so their sum can be rewritten as a single travelling wave of that argument. Expression (d) is a sum of two stationary-wave terms with different angular frequencies and wave numbers; it is not a single travelling wave and has no fixed standing-wave pattern.

Answer:

(a) stationary wave; (b) travelling wave; (c) travelling wave; (d) none of these as a single travelling or stationary wave.

Q.14.14A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10–2 kg and its linear mass density is 4.0 × 10–2 kg m–1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?v
Solution

The wire length is $L=m/\mu=(3.5\times10^{-2})/(4.0\times10^{-2})=0.875\,\text{m}$. In the fundamental mode of a string fixed at both ends, $f=v/(2L)$, so $v=2Lf=2(0.875)(45)=78.75\,\text{m s}^{-1}$. The tension is $T=\mu v^2=4.0\times10^{-2}(78.75)^2=2.48\times10^2\,\text{N}$.

Answer:

(a) $78.8\,\text{m s}^{-1}$; (b) $2.48\times10^2\,\text{N}$.

Q.14.15A 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 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.v
Solution

For a tube closed at one end and open at the other, consecutive resonance lengths differ by $\lambda/2$. Thus $\lambda/2=79.3\,\text{cm}-25.5\,\text{cm}=53.8\,\text{cm}=0.538\,\text{m}$, so $\lambda=1.076\,\text{m}$. With $f=340\,\text{Hz}$, $v=f\lambda=340(1.076)=3.66\times10^2\,\text{m s}^{-1}$.

Answer:

$3.66\times10^2\,\text{m s}^{-1}$.

Q.14.16A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel?v
Solution

The clamped middle is a displacement node and the two free ends are antinodes. In the fundamental mode, each half of the rod is one quarter wavelength, so the full rod length is $L=\lambda/2$. Thus $\lambda=2L=2.00\,\text{m}$. With $f=2.53\,\text{kHz}=2.53\times10^3\,\text{Hz}$, $v=f\lambda=2.53\times10^3\times2.00=5.06\times10^3\,\text{m s}^{-1}$.

Answer:

$5.06\times10^3\,\text{m s}^{-1}$.

Q.14.17A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source ? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s–1).v
Solution

For a pipe closed at one end, allowed frequencies are $(2n-1)v/(4L)$. With $L=0.20\,\text{m}$, the fundamental is $v/(4L)=340/0.80=425\,\text{Hz}$, close to $430\,\text{Hz}$, so the fundamental mode is resonantly excited. For a pipe open at both ends, the allowed frequencies are $nv/(2L)=n(340/0.40)=850n\,\text{Hz}$. Since $430\,\text{Hz}$ is not an allowed open-pipe frequency, resonance will not occur.

Answer:

For the closed pipe, the source excites the fundamental mode approximately. If both ends are open, the source is not in resonance.

Q.14.18Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?v
Solution

Reducing the tension in string A reduces its frequency. Since the beat frequency also decreases, string A must originally have had a frequency greater than string B. Therefore $f_A-f_B=6\,\text{Hz}$. With $f_A=324\,\text{Hz}$, $f_B=324-6=318\,\text{Hz}$.

Answer:

$318\,\text{Hz}$.

Q.14.19Explain 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.v
Solution

(a) In a sound standing wave, pressure variation is related to compression and rarefaction, which are maximum where neighbouring particles move oppositely around a displacement node. At a displacement antinode, neighbouring particles move together and pressure variation is minimum. (b) Bats emit ultrasonic pulses and analyse the reflected echoes; time delay gives distance, direction comes from the receiving geometry, and echo strength and structure give information about size and nature. (c) A violin and sitar may have the same fundamental frequency, but their harmonic content, relative intensities and transients are different, so their quality or timbre differs. (d) Transverse mechanical waves require shear restoring forces. Solids have shear modulus and can support both transverse and longitudinal waves, whereas gases have essentially no shear modulus and support only longitudinal pressure waves. (e) A pulse is a superposition of many frequency components. In a dispersive medium, these components travel with different speeds, so their relative phases change and the pulse shape is distorted.

Answer:

(a) Pressure variation is greatest where displacement is zero and least where displacement is greatest. (b) Bats use echoes of ultrasonic waves. (c) The notes differ in waveform and overtones. (d) Solids have shear elasticity; gases do not. (e) Different frequency components travel at different speeds in a dispersive medium.