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NCERT Class 11 Physics Chapter 14 Waves
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Waves
Chapter: 14
| Part – II |
EXERCISE
1. A 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?
Ans: 𝓋 = √T/μ
Calculate the mass per unit:
μ = m/L
μ = 2.50 kg/ 20.0m
= 0.125 kg/m
Calculate the wave speed:
𝓋 = √T/μ
= √200 N/ 0.125 kg/m
𝓋 = √1600 m2/s2
= 40 m/s
Determine the time t,
t = L/𝓋 = 20.0m / 40 m/s
= 0.50 s.
2. A 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).
Ans: Here:
s = ut + ½ gt2
Given that:
s = 300 m
u = 0
g = 9.8 m/s2
t =?
300 = 0 x t + ½ x 9.8 x t2
300 = 4.9 t2
t2 = 300/4.9
t2 = 61.22
t2 = √61.22
= 7.8 seconds.
Time taken for sound to travel:
tsound = distance/speed
= 300/340
= 0.88 seconds
Total time:
ttotal = tfall + tsound = 7.82 + 0.88 = 8.7 seconds.
3. A 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.
Ans: u = √T/μ
μ = mass / length
= 2.10 kg / 12.m
= 0.175 kg/m
Wave speed formula to find the tension (T)
343 = √T/ 0.175
3432 = T/0.175
117649 = T/0.175
T = 117649 x 0.175
= 20586.575 N.
4. Use the formula 𝓋 = √γP/𝘱 to explain why the speed of sound in air:
Ans: u = √γp /𝘱
We know that, pV = nRT
= pV,
Where m is the total mass and M is the molecular mass of the gas.
p = m/M × RT/V = pRT/ M
𝘱/p = RT/M
(a) Is independent of pressure.
Ans: For gas at constant temperature:
As 𝘱 increase, p also increased and vice versa.
This implies that:
v = √γp/p = const.
i.e., velocity is independent of pressure of the gas.
(b) Increases with temperature.
Ans: Since: 𝘱/p = RT/M
Therefore, v = √γp/p = √γRT/M.
Clearly, v, α √T , i.e., speed of sound in air increases with increase in temperature.
(c) Increases with humidity.
Ans: Increase in humidity decreases the effective density of air. Therefore, the velocity
increases.
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 – 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).
Ans: No, the converse is not true. The basic requirement for a wave function to represent a travelling wave is that it must have finite values for all 𝓍 and t. Only the function (a) satisfies this condition and can represent a travelling wave.
6. A 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.
Ans: Here:
λ = 𝓋/𝑓
Given:
𝑓 = 1000 kHz = 1 MHz = 1 × 106 Hz.
𝓋air = 340 m/s
𝓋water = 1486 m/s
Sound in air:
λreflected = 𝓋air/𝑓
= (340 m/s) / 1 × 106 Hz
= 3.4 × 10-4m
= 0.34mm
Sound in air:
λtransmitted = 𝓋water / 𝑓
= 1486 m/s / 1 × 106 Hz
= 1.486 × 10-3m
= 1.486 mm.
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 km s–1? The operating frequency of the scanner is 4.2 MHz.
Ans: λ = 𝓋 / 𝑓
𝓋 = 1.7 km/s = 1700 m/s
𝑓 = 4.2 MHz = 4.2 × 106 Hz.
λ = 𝓋/ 𝑓 = (1700 m/s) / 4.2 × 106 Hz
λ = 4.05 × 10-4 m
= 0.405 mm.
8. A 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 is the speed and direction of its propagation?
Ans: y (𝓍, t) = A sin (k𝓍 – wt + Ø)
For a wave described by y (𝓍, t) = A sin (wt + k𝓍 + Ø), it can be rewritten as:
y (𝓍, t) = A sin (k𝓍 – wt + Ø)
Wave function = 3.0 sin (0.018𝓍 + 36t + 𝜋/4)
K = 0.018 rad/cm
ω = 36 rad/s
y (𝓍, t) = 3.0 sin (0.018𝓍 + 36t + 𝜋/4) = 3.0 sin (0.018𝓍 + 36t + 𝜋/4)
Speed of propagation v:
v = ω / k = (36 rad/s) / (0.018 rad/cm)
= 2000 cm/s
= 20 m/s.
(b) What is its amplitude and frequency?
Ans: ω = 2𝜋𝑓 = 𝑓 = ω/2𝜋
= 36/ 2𝜋
= 5.72 Hz.
(c) What is the initial phase at the origin?
Ans: At 𝓍 = 0 and t = 0 the phase of the waye is:
Initial phase = 𝜋/4.
(d) What is the least distance between two successive crests in the wave?
Ans: λ = 2𝜋 / k
= 2𝜋 / 0.018
= 349.1 cm.
= 3.49m.
9. For the wave described in Exercise 14.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in a travelling wave differ from one point to another: amplitude, frequency or phase?
Ans:
| t | 0 | T/8 | 2/T/8 | 3/T/8 | 4/T/8 | 5/T/8 | 6/T/8 | 7/T/8 | T |
| y | 3/√2 | 3 | 3/√2 | 0 | -3/√2 | -3 | -3/√2 | 0 | 3/√2 |
The oscillatory motion of a travelling wave differs from one point to another only in terms of phase. This means that the waves at different points are identical in shape, but they are shifted relative to each other. The amplitude and frequency of the oscillatory motion remain the same at all points.
10. For 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:
Ans:
Given:
Wave equation: y (x t) = 2. 0 cos 2𝜋 (10 – 0.0080𝓍 + 0.35)
Wave number (k): 0.0080𝜋 rad/cm
Angular frequency (ω): 20𝜋 rad/s
Phase difference (ΔØ) between two points:
ΔØ = kΔ𝓍.
(a) 4 m.
Ans: ΔØ = 0.0080𝜋 × 400 = 10.05 rad
= 3.18𝜋 rad.
(b) 0.5 m.
Ans: ΔØ = 0.0080𝜋 × 50
= 0.4𝜋 rad.
(c) λ/2.
Ans: v = ω/k
v = (20𝜋) / (0.0080𝜋) = 2500 cm/s
λ = v/f = v/(ω/2𝜋) = 2500 / (20𝜋 / 2 𝜋) = 250 cm
ΔØ = k (λ/2) = 0.0080𝜋 × (250/2) = 𝜋 rad.
(d) 3λ/4.
Ans: ΔØ = k (3λ / 4) = 0.0080𝜋 × (3 × 250 / 4) = 1.5𝜋 rad.
11. The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin,
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?
Ans: The function represents a stationary wave. It does not represent a travelling 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?
Ans: Using the relation,
(c) Determine the tension in the string.
Ans: T = mv2
m = mass / length
= (3 × 10-2) / 1.5
= 2 × 10-2 kg m-1
T = 2 x 10-2 × (180)2
= 648 N.
12. (i) For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same. Explain your answers.
Ans: All the points on the wave have,
(a) Frequency.
Ans: Same frequency everywhere.
(b) Phase.
Ans: Different phases at different points.
(c) Amplitude?
Ans: Varies along the string.
(ii) What is the amplitude of a point 0.375 m away from one end?
Ans: A (𝓍) = A0 sin (k𝓍)
Amplitude calculation:
Here, k = n𝜋 / L
Amplitude 𝓍 = 0.375 m is,
A (0.375) = A0 sin (n𝜋 x 0.375/ L).
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).
Ans: It represents a stationary wave.
(b) y = 2√𝓍 – ut.
Ans: It does not represent either a stationary wave or a travelling wave.
(c) y = 3 sin (5𝓍 – 0.5t) + 4 cos (5𝓍 – 0.5t).
Ans: It represents a travelling wave.
(d) y = cos x sin t + cos 2x sin 2t.
Ans: It is a superposition of two stationary waves.
14. A 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?
Ans: Here,
𝓋 = 𝑓λ
λ = 2L
The linear mass density μ is defined as:
μ = m/L
L = m/μ = 3.5 × 10-2 / 4.0 × 10-2
= 0. 875 m
The wavelength λ is:
λ = 2L = 2 × 0.875 = 1.75 m
The speed 𝓋 of the wave is:
𝓋 = 𝑓λ = 45 × 1.75 = 78. 75 m/s
(b) Tension in the string:
𝓋 = √T / μ
T = μ𝓋2
T = 4.0 × 10-2 × (78.75)2
T = 4.0 × 10-2 × 6202 . 56
= 248.1 N.
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 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.
Ans: 𝑓 = nv / 2L
For the first resonance ( n = 1)
340 Hz = (1)v / (2 × 0.255 m)
v = 340 Hz × 2 × 0.255 m
= 173.4 m/s.
For the second resonance ( n = 2)
= 340 Hz = (2)v / (2 × 0.793 m)
= 340 Hz × 2 × 0.793 m/2
= 174.44 m/s
Average of the value:
= (173.4 m/s + 174 .44 m/s) /2
= 347.84 / 2
= 173 .92 m/s
16. A 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?
Ans: Given:
L = 100 cm = 1.00m
𝑓 = 2.53 kHz = 2530 Hz
Here:
L = λ / 2
λ = 2L
λ = 2 × 1.00 = 2.00 m
v = 𝑓λ
𝑓 = 2530 Hz and λ = 2.00 m
v = 2530 × 2.00
= 5060 m/s.
17. A 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).
Ans: Given:
L = 20cm = 0.20 m
v = 340 m/s
𝑓 = 430 Hz
Wavelength of the first harmonic mode:
λ1 = 4L = 4 × 0.20 m
= 0.80 m.
Frequency of the first harmonic mode:
𝑓1 = V /λ1
= 340 m/s / 0.80
= 425 Hz.
λ2 = 2L = 2 × 0.20 = 0.40 m
𝑓2 = v / λ2
= 340 m/s / 0.40 m
= 850 Hz.
18. Two 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?
Ans: Let v1 and v2 be the frequencies of strings A and B, respectively. Given that, v1 = 324 Hz and the number of beats b = 6 Hz, the frequency of string B can be:
v2 = v1 ± = 324 ± 6.
v2 = 330 Hz or 318 Hz.
19. Explain why (or how):
(a) In a sound wave, a displacement node is a pressure antinode and vice versa.
Ans: In a sound wave, a decrease in displacement (displacement node) corresponds to an increase in pressure (pressure antinode). Conversely, an increase in displacement is associated with a decrease in pressure.
(b) Bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”.
Ans: Bats emit ultrasonic waves of high frequencies from their mouths. When these waves reflect off obstacles and return to the bats, they provide information about the distance, direction, nature, and size of the obstacles.
(c) A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes.
Ans: The timbre of a violin note differs from that of a sitar because each instrument produces distinct harmonics. These differences in harmonics allow the human ear to distinguish between the sounds of the two instruments.
(d) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases. and
Ans: This is because gases exhibit only bulk modulus of elasticity, whereas solids have both shear modulus and bulk modulus of elasticity.
(e) The shape of a pulse gets distorted during propagation in a dispersive medium.
Ans: A sound pulse is composed of waves with varying wavelengths. In a dispersive medium, these waves travel at different velocities, leading to distortion of the sound wave.
