NCERT Exemplar Solutions Class 11 Physics Waves — Very Short, Short and Long Answer Questions Q18–36

NCERT Exemplar Problems, Class 11 Physics — Waves, Very Short Answer Questions 18 to 24, Short Answer Questions 25 to 31 and Long Answer Questions 32 to 36. All 19 questions solved. The book numbers them 15.18 to 15.36.

Three ideas carry this page. The speed of a wave is set by the medium: $\sqrt{T/\mu}$ on a string and $\sqrt{\gamma RT/M}$ for sound in a gas, so tension, thickness and temperature change the speed, and through $v = f\lambda$ the frequency of a fixed length or the wavelength of a fixed source. Standing waves fit the boundaries: a string fixed at both ends or an open pipe holds a whole number of half-wavelengths, $f_n = nv/2L$, while a pipe closed at one end holds odd quarter-wavelengths, $f = (2n-1)v/4L$. The form of an equation names the wave: $f(\omega t \pm kx)$ travels, $g(x)\,h(t)$ stands, and a product of two close frequencies in time is beats.

Key insight. Before calculating, write down what is held fixed (the length, the tension, the source frequency) and what is allowed to change. In Questions 18, 19, 22 and 30 that one decision does all the work.

Very Short Answer Questions

Question 18

A sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same, the length of the wire is doubled. Under what conditions would the tuning fork still be in resonance with the wire?

Solution. With the tension and the wire unchanged, the wave speed $v = \sqrt{T/\mu}$ is unchanged. The $n$th mode of a wire of length $L$ has frequency $f_n = \dfrac{nv}{2L}$. Suppose the wire of length $L$ resonates with the fork in its fundamental, $f = v/2L$. For the doubled length,

$$f_n = \frac{nv}{2(2L)} = \frac{n}{2}\cdot\frac{v}{2L},$$

which equals the fork frequency when $n = 2$. So the longer wire stays in resonance if it vibrates in its second harmonic, with two loops and a node at its midpoint. In general, whatever mode the original wire was in, the doubled wire must vibrate with twice as many loops.

Question 19

An organ pipe of length $L$ open at both ends is found to vibrate in its first harmonic when sounded with a tuning fork of $480$ Hz. What should be the length of a pipe closed at one end, so that it also vibrates in its first harmonic with the same tuning fork?

Solution. The fork fixes the frequency and the air fixes the speed, so the wavelength is the same in both pipes. An open pipe in its fundamental holds half a wavelength, $L = \lambda/2$. A closed pipe in its fundamental holds a quarter, $L’ = \lambda/4$. Hence $L’ = L/2$.

Question 20

A tuning fork A, marked $512$ Hz, produces $5$ beats per second, where sounded with another unmarked tuning fork B. If B is loaded with wax the number of beats is again $5$ per second. What is the frequency of the tuning fork B when not loaded?

Solution. Five beats per second mean B is either $507$ Hz or $517$ Hz. Loading a fork with wax adds mass to its prongs and lowers its frequency.

• If B were $507$ Hz, loading would lower it further and the beat rate would rise above $5$.
• If B is $517$ Hz, loading can lower it to $507$ Hz, which again gives $5$ beats per second with A.

Only the second case fits.

Question 21

The displacement of an elastic wave is given by the function $y = 3\sin\omega t + 4\cos\omega t$, where $y$ is in cm and $t$ is in second. Calculate the resultant amplitude.

Solution. Write $3 = A\cos\phi$ and $4 = A\sin\phi$. Then $y = A\sin(\omega t + \phi)$ with

$$A = \sqrt{3^2 + 4^2} = 5 \text{ cm}, \qquad \tan\phi = \frac{4}{3}.$$

The two terms are a quarter-period out of phase, so their amplitudes add like perpendicular vectors.

Question 22

A sitar wire is replaced by another wire of same length and material but of three times the earlier radius. If the tension in the wire remains the same, by what factor will the frequency change?

Solution. For a wire of the same material, the mass per unit length is $\mu = \pi r^2\rho$, so tripling the radius makes $\mu$ nine times as large. The frequency of a given mode is

$$f = \frac{n}{2L}\sqrt{\frac{T}{\mu}} \propto \frac{1}{\sqrt{\mu}} \propto \frac{1}{r},$$

so with $L$ and $T$ unchanged the frequency falls to one-third.

Question 23

At what temperatures (in °C) will the speed of sound in air be $3$ times its value at $0$°C?

Solution. The speed of sound in a gas is $v = \sqrt{\gamma RT/M}$, so $v \propto \sqrt{T}$ with $T$ in kelvin. Tripling the speed needs nine times the absolute temperature:

$$T = 9 \times 273 = 2457 \text{ K} = 2457 – 273 = 2184\ ^\circ\text{C}.$$

Question 24

When two waves of almost equal frequencies $n_1$ and $n_2$ reach at a point simultaneously, what is the time interval between successive maxima?

Solution. Superposing the two waves at a point gives a resultant oscillating at the mean frequency, with an amplitude that rises and falls at the beat frequency $|n_1 – n_2|$. There are $|n_1 – n_2|$ maxima of loudness per second, so successive maxima are separated by the reciprocal.

Short Answer Questions

Question 25

A steel wire has a length of $12$ m and a mass of $2.10$ kg. What will be the speed of a transverse wave on this wire when a tension of $2.06\times10^4$ N is applied?

Solution. The mass per unit length is $\mu = 2.10/12 = 0.175$ kg/m. Then

$$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{2.06\times10^4}{0.175}} = \sqrt{1.177\times10^5} \approx 343 \text{ m/s}.$$

Note that this happens to be close to the speed of sound in air.

Question 26

A pipe $20$ cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a source of $1237.5$ Hz? (sound velocity in air $= 330$ m s$^{-1}$)

Solution. A pipe closed at one end has a node at the closed end and an antinode at the open end, so its fundamental has $L = \lambda/4$:

$$f_1 = \frac{v}{4L} = \frac{330}{4\times0.20} = 412.5 \text{ Hz}.$$

Only odd multiples of $f_1$ are allowed. Now $1237.5/412.5 = 3$, so the source excites the third harmonic (the first overtone of the closed pipe).

Question 27

A train standing at the outer signal of a railway station blows a whistle of frequency $400$ Hz in still air. The train begins to move with a speed of $10$ m s$^{-1}$ towards the platform. What is the frequency of the sound for an observer standing on the platform? (sound velocity in air $= 330$ m s$^{-1}$)

Solution. The source moves towards a stationary observer, so each successive wave is emitted a little closer to the observer and the wavelength in front of the train shrinks. The heard frequency is

$$n’ = n\,\frac{v}{v – v_s} = 400\times\frac{330}{330 – 10} = 400\times\frac{330}{320} = 412.5 \text{ Hz}.$$

Question 28

The wave pattern on a stretched string is shown in Fig. 15.2. Interpret what kind of wave this is and find its wavelength.

Figure redrawn; the printed figure labels the node positions 10, 20, 30, 50 with 40 missing. The label “50” there sits on the fourth node, which is at 40.

Solution. Look at how the shape changes in time. The crests and troughs stay at the same positions (the dashed lines) and only grow, shrink and invert: the string is flat everywhere at $T/4$ and $3T/4$, inverted at $T/2$, and restored at $T$. The points where the string crosses the axis never move. A pattern whose nodes and antinodes stay in place is a stationary (standing) wave, not a travelling one.

Adjacent nodes are $10$ units apart, and the distance between adjacent nodes is $\lambda/2$. So $\lambda = 2\times10 = 20$ cm (equivalently, crest to next crest is $20$ cm).

Question 29

The pattern of standing waves formed on a stretched string at two instants of time are shown in Fig. 15.3. The velocity of two waves superimposing to form stationary waves is $360$ m s$^{-1}$ and their frequencies are $256$ Hz.

(a) Calculate the time at which the second curve is plotted. (b) Mark nodes and antinodes on the curve. (c) Calculate the distance between $A’$ and $C’$.

Solution. (a) At $t = 0$ the string has its maximum displacement everywhere. It is next flat a quarter of a period later, when every point passes through its mean position together:

$$t = \frac{T}{4} = \frac{1}{4\times256} = 9.8\times10^{-4} \text{ s}.$$

(It is flat again at $3T/4$, $5T/4$, and so on; the first instant is $T/4$.)

(b) The points where the string crosses the axis at $t = 0$ and stays there at every instant are the nodes: A, B, C, D, E. The points of maximum displacement, under the crests at $A’$ and $C’$, are antinodes: $A’$ and $C’$ (the troughs midway between B and C and between D and E are antinodes too, though the figure does not label them).

(c) The wavelength is $\lambda = v/f = 360/256 = 1.40625$ m. $A’$ and $C’$ are the centres of two successive crests, which are one full wavelength apart (crest, trough, crest), so

$$A’C’ = \lambda \approx 1.41 \text{ m}.$$

Question 30

A tuning fork vibrating with a frequency of $512$ Hz is kept close to the open end of a tube filled with water (Fig. 15.4). The water level in the tube is gradually lowered. When the water level is $17$ cm below the open end, maximum intensity of sound is heard. If the room temperature is $20$°C, calculate

(a) speed of sound in air at room temperature (b) speed of sound in air at $0$°C (c) if the water in the tube is replaced with mercury, will there be any difference in your observations?

Fig. 15.4 shows a vertical glass tube joined at the bottom by a flexible tube to a water reservoir; moving the reservoir changes the height of the water column, and the fork is held over the open top of the tube.

Solution. (a) The water surface is a closed end (a displacement node) and the open top is an antinode. The first maximum, as the level is lowered from the top, is the fundamental of the closed air column, with $l = \lambda/4$ (end correction neglected). So $\lambda = 4\times0.17 = 0.68$ m and

$$v_{20} = f\lambda = 512\times0.68 = 348.16 \text{ m/s}.$$

(b) Since $v \propto \sqrt{T}$,

$$v_0 = v_{20}\sqrt{\frac{273}{293}} = 348.16\times0.9653 \approx 336 \text{ m/s}.$$

(c) The resonance length depends only on the frequency and the speed of sound in the air column, not on the liquid that closes it. With mercury the resonance is still heard at $17$ cm. Mercury, being far denser, reflects sound more completely than water, so the resonance may be somewhat louder, but its position does not change.

Question 31

Show that when a string fixed at its two ends vibrates in $1$ loop, $2$ loops, $3$ loops and $4$ loops, the frequencies are in the ratio $1 : 2 : 3 : 4$.

Solution. Each loop of a standing wave is the segment between two adjacent nodes, which is half a wavelength. If the string of length $L$ vibrates in $n$ loops,

$$L = n\frac{\lambda_n}{2} \quad\Rightarrow\quad \lambda_n = \frac{2L}{n}, \qquad \nu_n = \frac{v}{\lambda_n} = \frac{nv}{2L}.$$

The speed $v = \sqrt{T/\mu}$ is the same for every mode, because the tension and the string are unchanged. So $\nu_n \propto n$, and for $n = 1, 2, 3, 4$,

$$\nu_1 : \nu_2 : \nu_3 : \nu_4 = 1 : 2 : 3 : 4.$$

Long Answer Questions

Question 32

The earth has a radius of $6400$ km. The inner core of $1000$ km radius is solid. Outside it, there is a region from $1000$ km to a radius of $3500$ km which is in molten state. Then again from $3500$ km to $6400$ km the earth is solid. Only longitudinal (P) waves can travel inside a liquid. Assume that the P wave has a speed of $8$ km s$^{-1}$ in solid parts and of $5$ km s$^{-1}$ in liquid parts of the earth. An earthquake occurs at some place close to the surface of the earth. Calculate the time after which it will be recorded in a seismometer at a diametrically opposite point on the earth if wave travels along diameter?

Solution. Only P waves can cross the molten layer, so the signal is a P wave all the way. Along the diameter it crosses each layer twice, once going in and once coming out, except the inner core, which it crosses along a full diameter of $2000$ km:

Region	Distance along the diameter	Speed	Time
Outer solid shell	$2\times(6400 – 3500) = 5800$ km	$8$ km/s	$725$ s
Molten layer	$2\times(3500 – 1000) = 5000$ km	$5$ km/s	$1000$ s
Inner solid core	$2\times1000 = 2000$ km	$8$ km/s	$250$ s

$$t = 725 + 1000 + 250 = 1975 \text{ s} = 32 \text{ min } 55 \text{ s}.$$

Question 33

If $c$ is r.m.s. speed of molecules in a gas and $v$ is the speed of sound waves in the gas, show that $c/v$ is constant and independent of temperature for all diatomic gases.

Solution. From kinetic theory, $P = \tfrac{1}{3}\rho c^2$, so

$$c = \sqrt{\frac{3P}{\rho}} = \sqrt{\frac{3RT}{M}}.$$

By Laplace’s formula the speed of sound is

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

Dividing, $T$ and $M$ cancel:

$$\frac{c}{v} = \sqrt{\frac{3}{\gamma}}.$$

For a diatomic gas (at ordinary temperatures, with the vibrational modes not excited) $\gamma = 7/5$, so

$$\frac{c}{v} = \sqrt{\frac{15}{7}} \approx 1.46,$$

the same for every diatomic gas and at every temperature.

Question 34

Given below are some functions of $x$ and $t$ to represent the displacement of an elastic wave.

(i) $y = 5\cos(4x)\sin(20t)$ (ii) $y = 4\sin(5x – t/2) + 3\cos(5x – t/2)$ (iii) $y = 10\cos[(252 – 250)\pi t]\cos[(252 + 250)\pi t]$ (iv) $y = 100\cos(100\pi t + 0.5x)$

State which of these represent (a) a travelling wave along $-x$ direction, (b) a stationary wave, (c) beats, (d) a travelling wave along $+x$ direction. Give reasons for your answers.

(The book labels the four functions (a) to (d) as well; they are numbered (i) to (iv) here to keep them apart from the four descriptions.)

Solution.

• (i) $5\cos4x\sin20t$ is a product of a function of $x$ alone and a function of $t$ alone, so every point oscillates in step with an amplitude $5|\cos4x|$ fixed by its position: a stationary wave.
• (ii) Both terms depend on $x$ and $t$ only through $5x – t/2$. A constant phase $5x – t/2$ needs $x$ to increase as $t$ increases, so it is a travelling wave along $+x$ (speed $\tfrac{1/2}{5} = 0.1$ in the given units). Combining the terms gives a single sinusoid of amplitude $5$.
• (iii) There is no $x$ at all; at a point the displacement is a fast oscillation at $\tfrac{(252+250)\pi}{2\pi} = 251$ Hz whose amplitude $10\cos(2\pi t)$ varies slowly. This is the form of the sum of two oscillations of $250$ Hz and $252$ Hz: beats, at $2$ per second.
• (iv) $100\cos(100\pi t + 0.5x)$ has phase $\omega t + kx$, constant when $x$ decreases as $t$ increases: a travelling wave along $-x$.

Question 35

In the given progressive wave $y = 5\sin(100\pi t – 0.4\pi x)$, where $y$ and $x$ are in m, $t$ is in s, what is the (a) amplitude (b) wave length (c) frequency (d) wave velocity (e) particle velocity amplitude?

Solution. Compare with $y = A\sin(\omega t – kx)$: $A = 5$ m, $\omega = 100\pi$ rad/s, $k = 0.4\pi$ rad/m.

• (a) $A = 5$ m.
• (b) $\lambda = 2\pi/k = 2\pi/0.4\pi = 5$ m.
• (c) $f = \omega/2\pi = 50$ Hz.
• (d) $v = \omega/k = 100\pi/0.4\pi = 250$ m/s (check: $f\lambda = 50\times5 = 250$).
• (e) The particle velocity is $\partial y/\partial t = A\omega\cos(\omega t – kx)$, whose amplitude is $A\omega = 5\times100\pi = 500\pi \approx 1571$ m/s.

Question 36

For the harmonic travelling wave $y = 2\cos2\pi(10t – 0.0080x + 3.5)$, where $x$ and $y$ are in cm and $t$ is in second, what is the phase difference between the oscillatory motion at two points separated by a distance of (a) $4$ m (b) $0.5$ m (c) $\lambda/2$ (d) $\dfrac{3\lambda}{4}$ (at a given instant of time)? (e) What is the phase difference between the oscillation of a particle located at $x = 100$ cm, at $t = T$ s and $t = 5$ s?

Solution. The phase is $\phi = 2\pi(10t – 0.0080x + 3.5)$. At a fixed instant, two points $\Delta x$ apart differ in phase by

$$\Delta\phi = 2\pi\times0.0080\,\Delta x \quad (\Delta x \text{ in cm}),$$

and $\lambda = 1/0.0080 = 125$ cm.

• (a) $\Delta x = 400$ cm: $\Delta\phi = 2\pi\times0.0080\times400 = 6.4\pi$ rad.
• (b) $\Delta x = 50$ cm: $\Delta\phi = 2\pi\times0.0080\times50 = 0.8\pi$ rad.
• (c) $\Delta x = \lambda/2$: $\Delta\phi = \dfrac{2\pi}{\lambda}\cdot\dfrac{\lambda}{2} = \pi$ rad.
• (d) $\Delta x = 3\lambda/4$: $\Delta\phi = \dfrac{3\pi}{2}$ rad.
• (e) At a fixed point the phase changes with time as $2\pi\times10\,\Delta t$. The printed question reads “$t = T$ s”; the book’s answer, $80\pi$ rad, corresponds to $t = 1$ s, which is evidently what was meant: $\Delta\phi = 2\pi\times10\times(5 – 1) = 80\pi$ rad. (If $T$ is read as the period, $0.1$ s, the difference is $2\pi\times10\times4.9 = 98\pi$ rad.) Either way it is a whole number of cycles, so the particle is in the same state of oscillation at the two instants.

Common mistakes

• Assuming the wire stays in its fundamental in Question 18. A doubled length resonates with the same fork only in the second harmonic.
• Picking $507$ Hz in Question 20. Wax lowers the frequency; only a fork above $512$ Hz can keep the same beat count after loading.
• Using Celsius in $v \propto \sqrt{T}$ in Questions 23 and 30. The temperature must be absolute; convert back to °C only at the end.
• Allowing even harmonics in a closed pipe in Question 26. A pipe closed at one end has only odd multiples of $v/4L$.
• Calling Fig. 15.2 a travelling wave in Question 28. The crests never move along the string; only their size changes.
• Taking $A’C’$ as $\lambda/2$ in Question 29. $A’$ and $C’$ are both crests, one wavelength apart; a crest and the next trough are $\lambda/2$ apart.
• Forgetting that the inner core is crossed once, along its full diameter, in Question 32. The shells are crossed twice; the core contributes $2000$ km in one pass.
• Mixing centimetres and metres in Question 36. $k$ is per centimetre, so $4$ m must be entered as $400$ cm.

Practise next

• A pipe $25$ cm long is closed at one end. Taking the speed of sound as $340$ m/s, which harmonic is excited by a source of $1020$ Hz?

Show answer

The third harmonic. The fundamental is $f_1 = v/4L = 340/(4\times0.25) = 340$ Hz, and $1020/340 = 3$, an odd multiple, so the source excites the third harmonic.

• A fork A of $256$ Hz gives $4$ beats per second with fork B. When B is filed (which raises its frequency) the beats rise to $6$ per second. What was the frequency of B?

Show answer

$260$ Hz. B is $252$ or $260$ Hz. Filing raises it; from $260$ Hz that widens the gap with A (to $262$ Hz, $6$ beats), while from $252$ Hz it would narrow the gap and reduce the beats.

• At what temperature in °C is the speed of sound in air twice its value at $27$°C?

Show answer

$927$ °C. $v \propto \sqrt{T}$, so doubling $v$ needs $T = 4\times300 = 1200$ K, which is $927$ °C.

• Exemplar Waves: Multiple Choice Questions — the same ideas in objective form, from reflection at a denser boundary to the Doppler effect.
• NCERT textbook Chapter 14 — Waves for the theory of travelling and standing waves, organ pipes and beats.
• Exemplar Oscillations — the previous unit, whose simple harmonic motion every particle of a wave performs.