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CBSE NCERT Solutions for Class 11 Physics Chapter 15
Back of Chapter Questions
15.1. 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?
Solution:
Given:
Mass of the string, π = 2.50 kg
Tension in the string, π = 200 N
Length of the string, π = 20.0 m
Then, mass per unit length (linear mass density) of the string, π =π
π=
2.50
20=
0.125 kg mβ1
The transverse wave speed (π£) in the string is determined by the relation:
π£ = βπ
π
= β200
0.125= β1600 = 40 m sβ1
Hence, time taken by the disturbance to reach the other end, π‘ =π
π£=
20
40= 0.5 s
15.2. A stone dropped from the top of a tower of height 300 m high 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 sβ1? (π = 9.8 m sβ2)
Solution:
Given:
Height of the tower, β = 300 m
Initial velocity of the stone, π’ = 0
Acceleration, π = π = 9.8 m sβ2
Speed of sound in air = π£ = 340 m sβ1
Let π‘1 be the time taken by stone to reach the pondβs surface.
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Then we use, π = π’π‘ +1
2ππ‘2 to calculate π‘1 by substituting π = β, π‘ = π‘1, π =
π and π’ = 0
β = 0 Γ π‘1 +1
2ππ‘1
2
Therefore, π‘1 = β2β
π= β
2Γ300
9.8= 7.82 s
Also, let π‘2 be the time taken by sound to reach top of the tower of height 300 m.
Then, π‘2 =β
π£=
300
340= 0.88 s
Hence, the total time after which the sound of splash is heard = π‘1 + π‘2 = 7.82 +
0.88 = 8.7 s
15.3. 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 oC = 343 m sβ1.
Solution:
Given:
Mass of the steel wire, π = 2.10 kg
Speed of the transverse wave, π£ = 343 msβ1
Length of the steel wire, π = 12.0 m
Then, mass per unit length (linear mass density) of the steel wire, π =π
π=
2.10
12.0=
0.175 kg mβ1
The transverse wave speed (π£) in the steel wire is determined by the relation,
π£ = βπ
π, where π is the tension in the wire
Squaring on both the sides, we get
π£2 =π
πβ π = ππ£2
Hence, π = 0.175 Γ 3432 = 20,588.575 β 2.06 Γ 104 N
15.4. Use the formula π£ = βπΎπ
π to explain why the speed of sound in air
(a) is independent of pressure,
(b) increases with temperature,
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(c) increases with humidity.
Solution:
(a) Given: π£ = βπΎπ
π β¦ (i)
Where, π = π·πππ ππ‘π¦ =πππ π
ππππ’ππ=
π
π
π = Molecular weight of gas
π = Volume of gas
So, equation (i) reduces to:
π£ = βπΎππ
π β¦ (iib
Using the ideal gas equation for π = 1:
ππ = π π
For constant π, ππ = πΆπππ π‘πππ‘
Since both π and πΎ are constants, π£ = πΆπππ π‘πππ‘
Hence, the speed of sound in air is not dependent on pressure at a constant
temperature.
(b) Given: π£ = βπΎπ
π β¦ (i)
Where, π = π·πππ ππ‘π¦ =πππ π
ππππ’ππ=
π
π βππ = π β¦ (ii)
π = Molecular weight of gas
π = Volume of gas
So, equation (i) reduces to:
π£ = βπΎππ
π
Using the ideal gas equation for π = 1:
ππ = π π
π =π π
π β¦ (iii)
Putting (iii) in (i), we get
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π£ = βπΎπ π
ππ= β
πΎπ π
π β¦ (iv)
Since π, πΎ and π are all constants, π£ β βπ.
Hence, the speed of sound in air increases with increase in temperature.
(c) Let π£π and π£π be the speeds of sound in dry air and moist air respectively.
Let ππ and ππ be the densities of dry air and moist air respectively.
We know the formula:
π£ = βπΎπ
π
Therefore, the speed of sound in moist air is:
π£π = βπΎπ
ππ β¦β¦ (i)
And the speed of sound in dry air is:
π£π = βπΎπ
ππ β¦β¦ (ii)
On dividing equations (i) and (ii) we get:
π£π
π£π= β
πΎπ
ππΓ
ππ
πΎπ β¦β¦. (ii)
However, the presence of water vapour reduces the density of air, i.e.,
ππ < ππ
β΄ π£π > π£π
Hence, the speed of sound in moist air is greater than it is in dry air. Thus,
in a gaseous medium, the speed of sound increases with humidity.
15.5. You have learnt that a travelling wave in one dimension is represented by a function
π¦ = π(π₯, π‘) where π₯ and π‘ must appear in the combination π₯ β π£π‘ or π₯ + π£π‘, i.e.,
π¦ = π(π₯ Β± π£π‘). Is the converse true? Examine if the following functions for π¦ can
possibly represent a travelling wave:
(a) (π₯ β π£π‘)2
(b) πππ [π₯+π£π‘
π₯π]
(c) 1
(π₯+π£π‘)
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Solution:
The converse is not true. An obvious requirement for an acceptable function for a
travelling wave is that it should be finite everywhere and at all times.
(a) No, the function does not represent a travelling wave as it is not having
finite value every where and at all times.
Example: For π₯ = β and π‘ = β, function has infinite value.
(b) By substituting π₯ = 0 and π‘ = 0, the given function tends to infinity. So,
the function cannot possibly represent a travelling wave.
(c) By substituting π₯ = 0 and π‘ = 0, the given function tends to infinity. So,
the function cannot possibly represent a travelling wave.
15.6. 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 msβ1 and in water 1486 msβ1.
Solution:
(a) Given:
Frequency of the ultrasonic sound, π = 1000 kHz = 106 Hz
Speed of sound in air, π£π = 340 msβ1
The wavelength (ππ) of the reflected sound is given by:
ππ =π£π
π=
340
106= 3.4 Γ 10β4 m
(b) Given:
Frequency of the ultrasonic sound, π = 1000 kHz = 106 Hz
Speed of sound in water, π£π€ = 1486 msβ1
The wavelength (ππ‘) of the transmitted sound is given by:
ππ‘ =π£π€
π=
1486
106= 1.49 Γ 10β3 m
15.7. 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 sβ1? The
operating frequency of the scanner is 4.2 MHz.
Solution:
Given: Speed of sound in the tissue, π£ = 1.7 km sβ1 = 1.7 Γ 103 m sβ1
Operating frequency of an ultrasonic scanner, π = 4.2 MHz = 4.2 Γ 106Hz
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The wavelength of sound in the tissue= π =π£
π=
1.7Γ103
4.2Γ106= 4.1 Γ 10β4 m
15.8. A transverse harmonic wave on a string is described by
π¦(π₯, π‘) = 3.0 sin (36π‘ + 0.018π₯ +π
4)
where π₯ and π¦ are in cm and π‘ in s. The positive direction of π₯ 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?
Solution:
The equation of a progressive wave travelling from right to left is given by the
displacement equation:
π¦(π₯, π‘) = π sin (ππ‘ + ππ₯ + π) β¦ (i)
The given equation is:
π¦(π₯, π‘) = 3.0 sin (36π‘ + 0.018π₯ +π
4) β¦ (ii)
By comparing equations (i) and (ii), we observe that equation (ii) represents a
travelling wave, propagating from right to left.
So, π = 3.0 cm, π = 36 rad sβ1, π = 0.018 cmβ1 and π =π
4 .
We know that,
π =π
2π and π =
2π
π
But, π£ = ππ
Therefore, π£ = (π
2π) Γ (
2π
π) =
π
π=
36
0.018= 2000 cm sβ1 = 20 m sβ1
(a) Thus, the wave is a travelling wave with speed equal to 20 m sβ1 and
travelling from right to left.
(b) Amplitude of wave, π = 3.0 cm = 0.03 m
Frequency of wave, π =π
2π=
36
2π= 5.7 Hz
(c) On comparing equations (i) and (ii), we get initial phase angle, π =π
4
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(d) The wavelength is the distance between two consecutive troughs or crests.
So, π =2π
π=
2π
0.018= 348.89 cm = 349 cm = 3.5 m
15.9. For the wave described in Exercise 15.8, plot the displacement (π¦) versus (π‘)
graphs for π₯ = 0, 2 and 4 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:
For the given wave,
π¦(π₯, π‘) = 3.0 sin (36π‘ + 0.018π₯ +π
4)
For π₯ = 0, π¦(0, π‘) = 3.0 sin (36π‘ +π
4) β¦ (i)
β Amplitude, π = 3 cm
β Initial phase angle, π =π
4
β Angular frequency, π = 36 rad sβ1
β Time period, π =2π
π=
2π
36=
π
18 s
For π₯ = 2, π¦(2, π‘) = 3.0 sin (36π‘ + 0.036 +π
4)
For π₯ = 4, π¦(4, π‘) = 3.0 sin (36π‘ + 0.072 +π
4)
For π₯ = 0, substituting various values of βπ‘β in equation (i), we get
π‘(s) 0 π
8
2π
8
3π
8
4π
8
5π
8
6π
8
7π
8
π
π¦(cm) 3β2
2
3 3β2
2
0 β3β2
2
β3 β3β2
2
0 3β2
2
Which can be plotted as below:
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Similarly, the graphs for π₯ = 2 and π₯ = 4 can be plotted.
These graphs have same amplitude and frequency, but they differ in phase given by
π
4,
π
4+ 0.036 and
π
4+ 0.072.
15.10. For the travelling harmonic wave
π¦(π₯, π‘) = 2.0 cos 2π(10π‘ β 0.0080π₯ + 0.35)
where π₯ and π¦ are in cm and π‘ in s. Calculate the phase difference between
oscillatory motion of two points separated by a distance of
(a) 4 π,
(b) 0.5 m,
(c) π
2,
(d) 3π
4
Solution:
The given equation can be rewritten as follows:
π¦(π₯, π‘) = 2.0 cos (20ππ‘ β 0.016ππ₯ + 0.70π)
Where,
Amplitude, π = 2.0 cm
Angular frequency, π = 20π rad sβ1
Angular wave number (π) = 0.016π cmβ1
In an oscillatory motion, the phase difference between two points separated by a
distance of βπ₯ is given by,
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βπ =2π
πβπ₯
βπ = πβπ₯
For βπ₯ = 4 m = 400 cm
βπ = 0.016π Γ 400 = 6.4π rad
For βπ₯ = 0.5 m = 50 cm
βπ = 0.016π Γ 50 = 0.8π rad
(a) For βπ₯ =π
2
βπ =2π
πβπ₯
βπ =2π
πΓ
π
2= π rad
For βπ₯ =3π
4
βπ =2π
πβπ₯
βπ =2π
πΓ
3π
4=
3π
2 rad
15.11. The transverse displacement of a string (clamped at its both ends) is given by
π¦(π₯, π‘) = 0.06 sin (2ππ₯
3) cos(120ππ‘)
where π₯ and π¦ are in m and π‘ in s. The length of the string is 1.5 m and its mass is
3.0 Γ 10β2kg.
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.
Solution:
(a) The general equation representing a stationary wave is given by:
π¦(π₯, π‘) = 2π sin (ππ₯)cos (π€π‘) β¦ (i)
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Given equation:
π¦(π₯, π‘) = 0.06 sin (2ππ₯
3) cos(120ππ‘) β¦ (ii)
Equations (i) and (ii) are similar to each other.
So, the given function represents a stationary wave.
(b) A wave pulse travelling in positive direction of x-axis is given by,
π¦1(π₯, π‘) = π sin (ππ₯ β ππ‘)
Similarly, a wave pulse travelling in negative direction of x-axis is given
by,
π¦2(π₯, π‘) = π sin (ππ₯ + ππ‘)
The superposition of these two waves gives:
π¦ = π¦1 + π¦2 = π sin (ππ₯ β ππ‘) + π sin (ππ₯ + ππ‘)
π¦ = 2π sin (ππ₯)cos (ππ‘) {πππππ, sin(π΄ + π΅) + sin(π΄ β π΅) = 2π πππ΄πππ π΅}
π¦ = 2π sin (2π
ππ₯)cos (2πππ‘) β¦(iii)
The transverse displacement is given by:
π¦(π₯, π‘) = 0.06 sin (2ππ₯
3) cos(120ππ‘) β¦ (ii)
Comparing equations (ii) and (iii), we get
2π
π=
2π
3
Therefore, the wavelength is 3 m.
By comparing, we also get
120π = 2ππ
So, the frequency of the wave is 60 Hz.
Speed of the wave, π£ = ππ
π£ = ππ = 60 Γ 3 = 180 m sβ1
(c) Given:
Mass of the string, π = 3.0 Γ 10β2 kg
Speed of the transverse wave, π£ = 180 msβ1
Length of the string, π = 1.5 m
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Then, mass per unit length (linear mass density) of the string, π =π
π=
3.0Γ10β2
1.5= 2 Γ 10β2 kg mβ1
The transverse wave speed (π£) in the string is determined by the relation,
π£ = βπ
π, where π is the tension in the string
Squaring on both the sides, we get
π£2 =π
πβ π = ππ£2
β π = 2 Γ 10β2 Γ 1802 = 648 N
15.12. (i) For the wave on a string described in Exercise 15.11, 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 m away from one end?
Solution:
Given:
(i) The transverse displacement of the string is given as:
π¦(π₯, π‘) = 0.06 sin (2ππ₯
3) cos(120ππ‘)
(a) Yes. The frequency of oscillation of all points on the string is same. This is
because, frequency is represented by the time dependent harmonic function
cos(120ππ‘) of the stationary wave. It is clear from this function that
frequency is independent of π₯.
(b) Yes. The phase of all points on the string is same. This is because, phase is
represented by the time dependent harmonic function cos(120ππ‘) of the
stationary wave. It is clear from this function that phase is independent of
π₯.
(c) No. All the points on the string have different amplitudes at different points.
This is because, the amplitude of a stationary wave is given as:
π = 0.06 sin (2ππ₯
3)
Here, as π depends on π₯, amplitude of all the points on the string is not
same.
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(ii) Now, amplitude at a point 0.375 m away from one end is given by,
π = 0.06 sin (2π Γ 0.375
3) = 0.06 sin (
π
4) = 0.06 Γ 0.707 = 0.042 m
15.13. Given below are some functions of π₯ and π‘ 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) π¦ = 2 cos(3π₯) sin (10π‘)
(b) π¦ = 2βπ₯ β π£π‘
(c) π¦ = 3 sin (5π₯ β 0.5π‘) + 4 cos(5π₯ β 0.5π‘)
(d) π¦ = cos π₯ sin π‘ + cos 2π₯ sin 2π‘
Solution:
(a) The general equation representing a stationary wave is given by:
π¦(π₯, π‘) = 2π sin (ππ₯)cos (π€π‘)
Basically, equation of a stationary wave is product of harmonic functions of
π₯ and π‘ separately. Hence, the given function represents a stationary wave.
(b) It does not represent any wave.
(c) It represents a travelling harmonic wave.
(d) Here, the equation is sum of two functions, each representing a stationary
wave. Hence, it represents superposition of two stationary waves.
15.14. 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 Γ 10β2 kg and its linear mass
density is 4.0 Γ 10β2kg mβ1. What is (a) the speed of a transverse wave on the
string, and (b) the tension in the string?
Solution:
Given:
Wire is vibrating with a frequency = 45 Hz
Mass of the wire, π = 3.5 Γ 10β2 kg
Linear mass density, π =π
π= 4.0 Γ 10β2kg mβ1
So, the length of the wire, π =π
π=
3.5Γ10β2
4.0Γ10β2 =7
8= 0.875 m
Standing waves formed on a string of length π have wavelength given by,
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π =2π
π, where π = Number of nodes in the wire
i.e., π = 1, 2, 3, β¦ ππ‘π.
But given that the wire vibrates in its fundamental node, so π = 1
Then, π = 2π = 2 Γ 0.875 = 1.75 m
The speed of a transverse wave on the string is given by,
π£ = ππ = 45 Γ 1.75 = 78.75 m sβ1 β 79 m sβ1
The transverse wave speed (π£) in the string is determined by the relation,
π£ = βπ
π, where π is the tension in the string
Squaring on both the sides, we get
π£2 =π
πβ π = ππ£2 = 4.0 Γ 10β2 Γ (78.75)2 = 248.06 N β 248 N
15.15. 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.
Solution:
Given:
Frequency of the tuning fork, π = 340 Hz
One end of the tube is open and the other end is closed by the piston, so it behaves
as a closed organ pipe, which produces only odd harmonics.
Therefore, the pipe is in resonance with the fundamental mode and the third
harmonic (Since, 79.3 β 3 Γ 25.5)
Below is the figure for a fundamental mode formed due to pipe open at one end and
closed at other end.
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π1 =π
4, where length of the pipe, π1 = 25.5 cm = 0.255 m
Therefore, π = 4π1 = 4 Γ 0.255 = 1.02 m
Speed of sound in air = ππ = 340 Γ 1.02 = 347 m sβ1
15.16. 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?
Solution:
Given:
Length of the steel rod, π = 100 cm = 1 m
The fundamental frequency of vibrations of the rod, π = 2.53 kHz = 2.53 Γ
103Hz
When the rod is clamped at its middle, then in its fundamental mode of vibration,
antinodes (A) are formed at its two ends, and a node (N) is formed at the middle,
as shown in the given figure
It is obvious that, πΏ =π
4+
π
4=
π
2
π = 2πΏ = 2 m
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We know that, π£ = ππ = 2.53 Γ 103 Γ 2 = 5.06 Γ 103 m sβ1
15.17. 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 sβ1).
Solution:
Given:
Length of the pipe, π = 20 cm = 0.2 m
Frequency of the source = ππ‘β normal mode of frequency, ππ = 430 Hz
Speed of sound in air = 340 m sβ1
For a pipe closed at one end:
ππ = (2π β 1)π£
4π, for π = 1, 2, 3, β¦
430 = (2π β 1)340
4 Γ 0.2
2π β 1 =430 Γ 4 Γ 0.2
340= 1.01
π β 1
Thus, the first mode of frequency of vibration is resonantly excited by a 430 Hz
source. For a pipe open at both ends:
ππ =ππ£
2π, for π = 1, 2, 3, β¦
π =2πππ
π£=
2 Γ 0.2 Γ 430
340= 0.506
As the number of the mode of vibration (π) has to be an integer, the same frequency
source of 430 Hz does not produce a resonant vibration in the pipe, whose both
ends are open.
15.18. Two sitar strings A and B playing the note βGaβ 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?
Solution:
Given:
Original frequency of string A, ππ΄ = 324 Hz
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Let the original frequency of string B = ππ΅
Decrease in the tension of a string decreases its frequency.
If the original frequency of B (ππ΅) were greater than that of A (ππ΄), a further
increase in ππ΅ should have resulted in an increase in the beat frequency.
But, the beat frequency is found to decrease from 6 Hz to 3 Hz.
This shows that ππ΅ < ππ΄.
Since ππ΄ β ππ΅ = 6 Hz, and ππ΄ = 324 Hz, we get
324 β ππ΅ = 6
βππ΅ = 318 Hz
Hence, the original frequency of string B is 318 Hz.
15.19. 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 βeyesβ,
(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.
Solution:
(a) Since the displacement node is a point where displacement is zero, the
variation of pressure is maximum at this point. So, it is a pressure antinode.
(b) Bats can generate and detect ultrasonic waves (π > 20 kHz) so that they
can detect the object's distance by the interval between the waves they
generate and the echo they receive. They identify the nature and size of the
object from the intensity of the echo, and they can determine the direction
of the object from the tiny interval between getting the echo from their two
ears.
(c) The sound quality generated relies on the instruments ' overtones. However,
the Sitar and Violin may have the same fundamental frequency, but they
generate distinct overtones that are integral multiples of fundamental
frequencies and therefore we can distinguish between the two sounds.
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(d) Only longitudinal waves can travel through gases as they do not have shear
elasticity. But solids can support both transverse and longitudinal waves as
they have both shear and volume elasticity.
(e) The property of a dispersive medium tells that waves of various
wavelengths travel with different velocities or at distinct speeds in distinct
directions. Thus, when a sound pulse passes through it, which is a
combination of waves of various wavelengths, it becomes distorted.
15.20. A train, standing at the outer signal of a railway station blows a whistle of frequency
400 Hz in still air. (i) What is the frequency of the whistle for a platform observer
when the train (a) approaches the platform with a speed of 10 m sβ1, (b) recedes
from the platform with a speed of 10 m sβ1? (ii) What is the speed of sound in each
case? The speed of sound in still air can be taken as 340 m sβ1.
Solution:
(a) Given:
Frequency of the whistle, π = 400 Hz
Speed of sound in still air, π£ = 340 m sβ1
Speed of the train, π£π = 10 m sβ1
When the train moves towards the platform, the apparent frequency (π£β²) of
the whistle is given by:
π£β² = (π£
π£ β π£π ) π
= (340
340 β 10) Γ 400 = 412 Hz
(b) When the train recedes the platform, the apparent frequency (π£β²β²) of the
whistle is given by:
π£β²β² = (π£
π£ + π£π ) π
= (340
340 + 10) Γ 400 = 389 Hz
(ii) The speed of sound is not dependent on relative motion between source and
observer. Hence, the speed of sound in air in both cases remains the same.
15.21. A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air.
The wind starts blowing in the direction from the yard to the station with at a speed
of 10 m sβ1. What are the frequency, wavelength, and speed of sound for an
observer standing on the stationβs platform? Is the situation exactly identical to the
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case when the air is still and the observer runs towards the yard at a speed of
10 m sβ1? The speed of sound in still air can be taken as 340 m sβ1.
Solution:
Case 1:
For the stationary observer:
Source frequency in still air, π = 400 Hz
Speed of sound, π£ = 340 m sβ1
Speed of the wind, π£π = 10 m sβ1
When the wind starts blowing in the direction from the yard to the station
with a speed of 10 m sβ1.
Therefore, for an observer standing on the stationβs platform, the effective
speed of sound, π£πππ = 340 + 10 = 350 m sβ1
As observer and source both are at rest, so the frequency remains the same,
i.e., 400 Hz.
The wavelength (π) of the sound heard by the stationary observer is given
by,
π =π£πππ
π=
350
400= 0.875 m
Case 2:
For the running observer:
Velocity of the observer, π£π = 10 m sβ1
Since there is a relative motion between the sound source (train) and the
observer, when the observer is moving towards the source, the frequency of
the sound heard by the observer will not be equal to that of the source due
to Doppler effect.
So, the observed frequency is given by,
π£β² = (π£ + π£π
π£) π
= (340 + 10
340) Γ 400
= 411.8 Hz
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Also, as wavelength of sound waves is not affected by motion of the
observer, it remains unchanged.
Thus, the first situation is not identical to the case when the air is still and
observer runs towards the station-yard at a speed of 10 m sβ1. In the latter
situation, medium is at rest. So, the effective speed of sound = 340 m sβ1
Additional Exercises
15.22. A travelling harmonic wave on a string is described by
π¦(π₯, π‘) = 7.5βsinβ (0.0050π₯ + 12π‘ +π
4)
(a) what are the displacement and velocity of oscillation of a point at π₯ = 1 cm,
and π‘ = 1 s? Is this velocity equal to the velocity of wave propagation?
(b) Locate the points of the string which have the same transverse
displacements and velocity as the π₯ = 1 cm point at π‘ = 2 s, 5 s and 11 s.
Solution:
(a) The given harmonic wave is:
π¦(π₯, π‘) = 7.5βsinβ (0.0050π₯ + 12π‘ +π
4)
For x = 1 cm and t = 1 s,
π¦(1,1) = 7.5βsinβ (0.0050 + 12 +π
4)
7.5βsinβ (12.0050 +π
4)
= 7.5βsinπ
Where,
π = 12.0050 +π
4= 12.0050 +
3.14
4= 12.78βrad
=180
3.14Γ 12.79 = 732.81Β°
β΄ π¦(1,1) = 7.5βsinβ(732.81Β°)
= 7.5 sin(90 Γ 80 + 12.81Β°) = 7.5sin12.81Β°
= 7.5 Γ 0.2217
= 1.6629 β 1.663βcm
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The velocity of the oscillation at a given point and time is given as:
π£ =π
ππ‘π¦(π₯, π‘) = (
π
ππ‘) [7.5βsin (0.0050π₯ + 12π‘ +
π
4)]
= 7.5 Γ 12 cos (0.0050π₯ + 12π‘ +π
4)
At π₯ = 1 cmβ and π‘ = 1 s
π£ =π
ππ‘π¦(1,1) = 90cos (12.005 +
π
4)
= 90cos(732.81Β°) = 90cos(90 Γ 8 + 12.81Β°)
= 90cos(12.81Β°)
= 90 Γ 0.975 = 87.75βcm/s
Now, the equation of propagating wave is given by:
π¦(π₯, π‘) = πsin(ππ₯ + π€π‘ + π)
Where,
π =2π
π
β΄ π =2π
π
And π = 2ππ
β΄ π =π
2π
Speed, π£ = ππ =π
π
Where,
π = 12βrad/s
π = 0.0050βmβ1
β΄ π£ =12
0.0050= 2400βcm/s
Hence, the velocity of the wave oscillation at π₯ = 1 cm and π‘ = 1 s is not
equal to the velocity of the wave propagation.
(b) The relation between wavelength and propagation constant is given by,
π =2π
π
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Therefore, π =2π
π=
2Γ3.14
0.005= 1256 cm = 12.56 m
All the points situated at distance ππ, (where π is an integer) from the point
π₯ = 1 cm have the same transverse velocity and displacement.
15.23. A narrow sound pulse (for example, a short pip by a whistle) is sent across a
medium.
(a) Does the pulse have a definite
(i) frequency, (ii) wavelength, (iii) speed of propagation?
(b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split
of second after every 20 s), is the frequency of the note produced by the
whistle equal to 1
20 or 0.05 Hz?
Solution:
(a) The narrow sound pulse such as a short pip by a whistle does not have a
definite frequency or wavelength. But, the speed of the sound pulse remains
the same, which is the same as that of the speed of sound in that medium.
(b) No, the frequency of the note produced by the whistle is not equal to 1
20 or
0.05 Hz. As it is only the frequency at which the pulse gets repeated.
15.24. One end of a long string of linear mass density 8.0 Γ 10β3 kg mβ1 is connected to
an electrically driven tuning fork of frequency 256 Hz. The other end passes over
a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all
the incoming energy so that reflected waves at this end have negligible amplitude.
At π‘ = 0, the left end (fork end) of the string π₯ = 0 has zero transverse
displacement (π¦ = 0) and is moving along positive π¦-direction. The amplitude of
the wave is 5.0 cm. Write down the transverse displacement π¦ as function of π₯ and
π‘ that describes the wave on the string.
Solution:
Linear mass density, π = 8.0 Γ 10β3 kg mβ1
Frequency of the tuning fork, π = 256 Hz
Mass of the pan, π = 90 kg
Amplitude of the wave, π = 5.0 cm = 0.05 m β¦ (i)
Tension in the string, π = ππ = 90 Γ 9.8 = 882 N
The transverse wave speed (π£) in the string is determined by the relation,
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π£ = βπ
π, where π is the tension in the string
= β882
8.0 Γ 10β3 = 332 m sβ1
Angular frequency, π = 2ππ
= 2 Γ 3.14 Γ 256
= 1608.5 = 1.61 Γ 103 rad sβ1 β¦ (ii)
Wavelength, π =π£
π=
332
256 m
Therefore, propagation constant, π =2π
π
=2Γ3.14
332
256
= 4.84 mβ1 β¦ (iii)
The equation of a travelling wave is given by the displacement function:
π¦(π₯, π‘) = π sin (ππ‘ β ππ₯) β¦ (iv)
Substituting the values from equations (i), (ii), and (iii) in equation (iv), we get the
displacement function:
π¦(π₯, π‘) = 0.05 sin(1.61 Γ 103 π‘ β 4.84 π₯) m, where π₯, π¦ are in m and π‘ in s.
15.25. A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An
enemy submarine moves towards the SONAR with a speed of 360 km/h. What is
the frequency of sound reflected by the submarine? Take the speed of sound in
water to be 1450 m/s.
Solution:
Given:
Frequency of the SONAR system, π = 40.0 kHz = 40 Γ 103Hz
Speed of sound in water, π£ = 1450 m/s
Speed of the enemy submarine (observer), π£π = 360 km/h = 360 Γ5
18=
100 m/s
As the source (SONAR) is at rest and the enemy submarine (observer) moves it,
Therefore,
π£β² = (π£ + π£π
π£) π
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= (1450 + 100
1450) Γ 40 Γ 103 = 42.76 kHz
This frequency (π£β²) is reflected by the enemy submarine and is observed by the
SONAR (which now acts as an observer).
Thus, in this case, π£s = 360 km/h = 360 Γ5
18= 100 m/s
Therefore, the frequency of sound reflected by the submarine is given by:
π£β²β² = (π£
π£ β π£s) πβ² = (
1450
1450 β 100) Γ 42.76 kHz = 45.9 kHz
15.26. Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can
experience both transverse (π) and longitudinal (π) sound waves. Typically the
speed of π wave is about 4.0 km sβ1, and that of π wave is 8.0 km sβ1. A
seismograph records π and π waves from an earthquake. The first π wave arrives
4 min before the first π wave. Assuming the waves travel in straight line, at what
distance does the earthquake occur?
Solution:
Given:
Speed of π wave, π£s = 4.0 km sβ1
Speed of π wave, π£p = 8.0 km sβ1
The time gap between π and π waves reaching the seismograph, π‘ = 4 min =
240 s
Let the point of earthquake occur at a distance of π· km from the seismograph.
Also, let π‘π and π‘π are the respective times taken by π and π waves to reach the
seismograph from the location of the earthquake.
So, π‘ = π‘π β π‘π =π·
π£sβ
π·
π£p=
π·
4β
π·
8=
π·
8= 240 s
β π· = 8 Γ 240 = 1920 km
Therefore, the earthquake occurs at a distance of 1920 km from the seismograph.
15.27. A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the
sound emission frequency of the bat is 40 kHz. During one fast swoop directly
toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air.
What frequency does the bat hear reflected off the wall?
Solution:
Given:
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Frequency of sound emitted by the bat, π = 40 kHz
Let the velocity of sound in the air be π£.
So, the velocity of the bat, π£b = 0.03π£
The apparent frequency striking the wall is given by,
π£β² = (π£
π£ β π£b) π
= (π£
π£ β 0.03π£) Γ 40 kHz
=40
0.97 kHz
This frequency (π£β²) is reflected by the wall and is received by the bat moving
towards the wall.
So, π£s = 0
π£β²β² = (π£ + π£π
π£) πβ²
= (π£ + 0.03π£
π£) Γ
40
0.97
= (1.03 Γ 40
0.97) = 42.47 kHz
⧫ ⧫ ⧫
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