9 Sound

Question Bank
1. State any two characteristics of wave motion.
Ans. (a) A wave is caused due to periodic disturbance of particles of medium, and the wave by itself is periodic
in nature.
(b) It is the disturbance which travels outward and not the particles of medium. The particles of medium,
simply vibrate either to and fro or up and down about their mean positions.
2. State three characteristics of medium necessary for the propagation of sound waves.
Ans. (a) Medium must be elastic in nature.
(b) Medium must have least frictional resistance between its molecules.
(c) Medium must possess inertia of rest.
3. Explain, why lightning flash is seen before the crack of thunder.
Ans. Light travels at a speed of 3 × 108 ms–1 and hence, reaches instantly to an observer. However, sound travels
at a very slow speed of 330 ms–1 and hence, reaches observer after sometime. Thus, flash of lightning is seen
at once, but crack of thunder is heard later on.
4. Derive a relation between frequency and time period.
Ans. Consider a vibrating body, having a time period ‘T’ and frequency ‘f’.
f vibrations are produced in = 1 s.
1
\ 1 vibration is produced in s.
f
But, time required to produce one vibration is called time period ‘T’
1
\ T = .
f
5. Derive a relation between wave velocity; frequency and wavelength.
Ans. Consider a wave, moving with a velocity ‘v,’ such that ‘f’ is its frequency, ‘T’ the time period and ‘λ’ the wavelength.
In time T, the distance covered by wave = λ.
λ
\ In time one second the distance covered by wave = .
1 T
But, f = ∴ v = f λ
T cm
6. The diagram shows a snapshot of a waveform of frequency 4
50 Hz in a string. The numbers in diagram represent distance in
Displacement

2
centimeters.
For this wave motion find, 0 cm
5 15 25 35
(i) Wavelength (ii) Amplitude (iii) Wave velocity.
2
Ans. (i) Wavelength = 20 cm. (ii) Amplitude = 4 cm.
(iii) Wave velocity = f λ = 50 × 20 = 1000 cms–1 = 10 ms–1. 4

7. Draw a diagram representing a wave of (a) amplitude 4 cm (b) wavelength 2 m. If the frequency of wave is
150 Hz, calculate its velocity.
Ans.
4
(+)
Displacement

2
4 cm
1 2 3 4
0 Distance
in metres
2
(–)
4
2m
f = 150 Hz; λ = 2 m
v = f λ = 150 s–1 × 2 m = 300 ms–1.
Questions Bank 53
 8. The diagram alongside shows a vibrating metal blade clamped at one end. P and R are the extreme positions
occupied by the blade during its course of vibration, Q, being its position of rest. The vibrating blade produces
a note of 480 Hz.
(a) Mark on the diagram amplitude of vibration.
(b) If the velocity of sound in air is 320 ms–1, what is the wavelength of sound produced?
P

Clamp Metal blade

Q

Ans. (a) P
Amplitude

(b) f = 480 Hz.

v 320
λ = f = 480 = 0.66 m.

9. A sound wave of frequency 640 Hz travels 800 m in 2.5 s. Calculate : (a) speed of sound (b) wavelength
of sound wave.
Distance 800 m
Ans. (a) Speed of sound
Time = 2.5 s = 320 ms .
= –1

320
(b) Speed of sound (v) = f λ. ⇒ ∴ 320 = 640 × λ ⇒ ∴ λ = 640 = 0.5 m.
10. A television station broadcasts at a frequency of 4500 MHz. If the speed of television waves is 3 × 108 ms–1.
Calculate the wavelength of television waves.
Ans. v = 3 × 108 ms–1, f = 4500 MHz = 4.5 × 109 Hz.
v 3 × 108
4.5 × 109 = 0.67 × 10 m = 0.067 m.
∴ Wavelength, λ = = –1
f
11. A longitudinal wave of wavelength 0.03 cm travels in air with a speed of 330 ms–1 .Calculate the frequency
of the wave. Can this wave be heard by normal human ear. Give a reason for your answer.
Ans. v = 330 ms–1; λ = 0.03 cm = 0.0003 m.
v 330 330 × 104
∴ Frequency, f = λ = 0.0003 =
= 1,100,000 Hz.
3
The above wave cannot be heard. It is because, the maximum range to which human ear can hear is 20,000
Hz. As the frequency of 1,100,000 Hz is far in excess therefore it will not produce any sound effect in ear.
12. A sound wave has a frequency of 2000 Hz and wavelength 17 cm. If the wavelength increases to 51 cm, what
is the frequency, the nature of material through which sound is propagating remains same.
Ans. Initial wavelength (λ1) = 17 cm.
Initial frequency (f1) = 2000 Hz.
Final wavelength (λ2) = 51 cm.
Final frequency (f2) = ?
For a given material velocity of sound is a constant quantity.
∴ f2 λ2 = f1 λ1
f2 × 51 cm = 2000 Hz × 17 cm.
2000 × 17

∴ f2 = = 666.67 Hz.
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54 A New Approach to I.C.S.E. Physics - IX
 13. A disturbance in air has wavelength of 22 m and speed 330 ms–1. Calculate the frequency of disturbance.
State whether above disturbance is audible to normal human ear. Give one reason for your answer.
Ans. v = 330 ms–1 λ = 22 m.
v 330 ms–1
22 m = 15 Hz.
∴ Frequency, f = =
λ
The above disturbance is not audible. It is because human ear is sensitive to frequencies between 20 Hz to
20,000 Hz.
14. An ultraviolet radiation has wavelength 150 Å. If the speed of ultraviolet radiation is 3 × 108 ms–1, calculate
(a) frequency of radiation in MHz (b) Time period. [1 Å = 10–10m].
Ans. (a) λ = 150 Å = 150 × 10–10 m = 15 × 10–9 m; v = 3 × 108 ms–1
v 3 × 108 ms–1

∴ Frequency, f = λ = 15 × 10–9 m = 0.2 × 1017 Hz.
0.2 × 1017

∴ Frequency in MHz = = 2 × 1010 MHz.
106
1 1

f 0.2 × 1017 = 5 × 10 second.
(b) Time period, T = = –17

15. The wavelength of waves produced on the surface of water is 20 cm. If the wave velocity is 24 ms–1, calculate:
(a) number of waves produced in one second (b) time required to produce one wave.
Ans. (a) Wavelength (λ) = 20 cm = 0.2 m
  Wave velocity (v) = 24 ms–1
v 24 ms–1
∴ Number of waves produced in one second = Frequency = λ = 0.2 m = 120 Hz.
1 1
120 s = 8.33 × 10 second.
   (b) Time required to produce one wave = Time period (T) = = –3
f
16. A thin metal plate is placed against the teeth of a cog wheel. If the cog wheel is rotated at a constant speed
of 360 rotations per minute and has 80 teeth, calculate : (a) frequency of note produced (b) speed of sound,
if the wavelength is 0.7 m (c) what will be the effect, if the speed of cog wheel is halved ?
Ans. (a) Number of rotations of cog wheel in 1 minute (60 s) = 360
360
∴ Number of rotations of cog wheel in 1 second =
60 s = 6 s .
–1

∴ Frequency of note produced = No. of rotations per second × Teeth in cog wheel
= 6 s–1 × 80 = 480 s–1 = 480 Hz.
(b) Speed of sound = Frequency × Wavelength = 480 s–1 × 0.7 m = 336.0 ms–1.
(c) The frequency of note produced is halved. Thus, in turn will lower the pitch of sound, i.e., a bass note
is produced.
17. A vibrating tuning fork can produce note of wavelength 0.83 m and has a time period of 2.5 × 10–3 seconds.
Calculate the wave velocity note and the distance covered by sound wave in 0.08 s.
Ans. Time period = 2.5 × 10–3s
1 1
∴ Frequency (f) = =
T 2.5 × 10–3 = 400 Hz.
∴ Wave velocity = f λ = 400 × 0.83 = 332 ms–1.
∴ Distance covered by sound in 0.08 seconds = 332 ms–1 × 0.08 s = 26.56 m.
18. The distance between one crest and one trough produced on the surface of water is 0.04 m. If the waves are
produced at a rate of 180 per minute. Calculate : (a) time period (b) wave velocity.
Ans. (a) Number of waves produced in 1 minute (60 s) = 180
180
∴ Number of waves produced in 1 second =
60 s = 3 s
–1

∴ Frequency of waves = 3 s–1.

1 1 1
Now, Time period (T) = = –1 = s = 0.33 s.
f 3s 3
λ
(b) Distance between one crest and one trough = .
2
λ
∴ = 0.04 m ⇒ ∴ λ = 0.04 × 2 = 0.08 m
2
λ 0.08 m
⅓ s = 0.24 ms .
∴ Wave velocity = = –1
T
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Questions Bank 55