Sound : Previous Year's Questions



1 Mark Questions

 

Q.1  Name two animals which can produce infrasonic waves.                               

[CBSE, 2012]

Hippopotamus and whale


Q.2  What is reverberation?                                                      

[CBSE (CCE), 2011]

Reverberation is defined as persistence of sound after the source has stopped emitting sound.

This is due to multiple reflections of sound waves.


Q.3  What is the frequency of wave with time period 0.025 s?   

[CBSE (CCE), 2010]

Frequency (f) = 1/ time period (T)

f = {1 \over T} = {1 \over {0.025}} = 40\,\,Hz

Therefore, frequency of the wave = 40 Hz.


Q.4  A baby recognizes her mother by her voice. Name the characteristic of sound involved.

[CBSE (CCE), 2010]

The characteristic of sound involved in uniqueness of the sound is quality of sound or timber.


Q.5  What are infrasonic and ultrasonic sounds ?

[CBSE (CCE) 2010]

Infrasonic sound has frequency less than 20 Hz and ultrasonic sound has frequency higher than 20 kHz.


Q.6  What is the audible range of human ear?                                        

[Board, 2010]

The audible range of human ear is 20 Hz to 20 kHz.


Q.7  Why do we hear sound of an approaching car before the car reaches us?          

[Board, 2010]

This is because velocity of sound is much greater than the velocity of car.

 

2 Marks Questions

 

Q.8  What is SONAR? For what it is used?                                    

[CBSE 2012]

SONAR is SOund Navigation And Ranging. It is a technique used to measure the depth of the sea, locate the sunken ships or icebergs and submarines.


Q.9  Write two points of difference between sound wave and light wave.

[CBSE, 2012]

Differences between sound and light waves:


Q.10  An echo is returned in 6 seconds. What is the distance of reflecting surface from source? [Given that speed of sound is 342 m/s.]                   

[CBSE (CCE), 2011]

Given,

Time in which echo returned, t = 6 s,

Speed of sound, v = 342 m/s

Distance = Speed × Time = 342 × 6 = 2052m

As this distance is twice the distance of reflecting surface from source.

So,

The distance of reflecting surface from source = 2052 / 2 = 1026 m.


Q.11  (i) Define the time period of a wave.                                             

(ii)  Give the relation among speed of sound v, wavelength λ and its frequency ν.

(iii)  A sound wave travels at a speed of 339 ms–1. If its wavelength is 1.5 cm, what is the frequency of the wave?

[Board, 2011]

(i)      Time period (T) - It is defined as the time required to complete one wave.

(ii)      Speed of sound (v) = Wavelength (λ) × Frequency (ν)

(iii)    Given,

Speed of sound, v = 339 ms–1

Wavelength, λ = 1.5 cm= 0.015m

Since, Velocity = Wavelength × Frequency

Therefore, Frequency = {{Velocity} \over {Wavelength}}

 = 339m/s \times 0.015m

= 22600 Hz

The frequency of wave = 22600 Hz


Q.12  A body is vibrating 6000 times in one minute. If the velocity of sound in air is 360 m/s, find:

(a) Frequency of vibration in hertz,

(b) Wavelength of the wave produced.                      

[Board, 2011]

(a) Frequency of vibration in hertz

Given,

Number of vibration in one minute = 6000

Number of vibrations in one sec = {{6000\,\,vib} \over {60\,\,s}} = 100Hz

Therefore, Frequency, ν = 100 Hz

(b) Wavelength of the wave produced

Given,

Velocity of speed in air, v = 360 ms–2

Frequency, ν= 100 Hz

Wavelength, λ = ?

v = \nu \lambda \Rightarrow \lambda = {{360m{s^{ - 1}}} \over {100{s^{ - 1}}}} = 3.6\,m

The wavelength of the wave is 3.6 m


Q.13  Describe two uses of multiple reflections of sound.              

[CBSE (CCE), 2010]

The phenomenon of multiple reflections of sound is used in Stethoscope and Megaphones.

 

3 Marks Questions

 

Q.14  (a)  What is one vibration in a second called as?                                 

(b) A tuning fork produces 256 waves in four seconds. Calculate the frequency of the tuning fork.

[Board, 2013]

(a) One Hertz (Hz) is defined as one vibration in one second.

(b) Frequency = {{Number\,\,of\,\,waves} \over {Time}}  = {{256} \over 4}  = 64\,\,Hz.

Therefore, the frequency of the tuning fork is 64 Hz.


Q.15  (a) Suppose a person whistles standing on the moon. Will the person standing nearby hear the sound? Explain giving reasons.

(b) What kind of wave needs a material medium to propagate?       

[CBSE 2013]

(a) No, the person standing nearby would not be able to hear the whistle as there is no atmosphere on the moon. Sound waves needs medium for travelling, but the moon has no medium for the propagation of sound. Therefore, no sound could be heard.

(b) Mechanical waves need material medium to propagate. For example: sound waves.


Q.16  A nail was gently touched by the hammer and then was hit harder.

(a) When will be the sound created louder?

(b) Which characteristic of sound here is responsible for change in sound?

(c) Give the SI unit of loudness.                                                

[CBSE 2013]

(a) Sound is produced only when the nail is hit harder not when it is gently touched.

(b) Amplitude of the pulsating body.

(c) The SI unit of loudness is decibel (dB)


Q.17  What are ultrasonic waves? Name one animal which emits ultrasonic waves and explain how it uses the waves? What is the role of ultrasonic waves in medical science?

[Board, 2012]

Ultrasonic waves are those waves which have frequency more than 20 kHz.

Bats use ultrasonic waves to fly and search for their prey at night without colliding.

Bat produces ultrasound waves in the direction of its way and hear back the echoes produced by reflection of the waves while striking the object in the path. Through this process they judge the position of the object and change their path.

Role of ultrasound waves in Medical science:

  • Imaging of internal body structure and functioning of heart in humans.
  • Breaking of stones produced in kidney into smaller ones to flush out through urination.
  • Observing the foetus development in the mother’s womb for any abnormality or defect.

Q.18  Give reasons for the following:                                                          

(a) The reverberation time of a hall used for speeches should be very short.

(b) A vibrating body produces sound. However, no sound is heard when a simple pendulum oscillates in air.

(c) Sounds of same loudness and pitch, but produced by different musical instruments such as violin and flute are distinguishable.

[Board, 2012]

(a) The time for which the sound persists in the atmosphere after the sound has stopped producing from the source is known as Reverberation time. If reverberation time of a hall is long, then the multiple echoes produced would interfere with the original sound, and the speeches would not be heard clearly and distinctly.

(b) Sound is produced only when the frequency of the wave is greater than 20 Hz. As a simple pendulum produces waves less than 20Hz they cannot be heard.

(c) The quality or timbre is the characteristic which helps in distinguishing two sounds of same pitch and loudness. So, the musical instruments like violin and flute produces distinguishable sound.


Q.19  Define the term ‘tone’. A person is listening to a sound of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions reaching his ears from the source?

[CBSE (CCE), 2012]

Tone is the sound of single frequency.

Given,

Frequency, ν = 500 Hz;

Distance, s = 450 m

Time period, T=?

T = {1 \over \nu } = {1 \over {500}} = 0.002\,\,s

Therefore, the time interval will be 0.002 s


Q.20  (a)  State a condition for an echo to be heard.

(b)  Bats cannot see, then how do they catch their Prey.

[CBSE (CCE), 2011]

(a) For echo to be heard the minimum distance between the source and the obstacle should be 17.2m.

(b) Bats produce ultrasonic waves in the direction of their way. They hear back the echoes produced by the reflection of waves while striking their prey. By this the position of prey is judged and is being eaten by the bat.


Q.21  In the figure given here, a displacement-distance graph for a wave is shown. The wave velocity is 320 m/s. Find                                                              

(a) Wavelength 

(b) Frequency

(c) Amplitude 

sd                                                                      

[CBSE (CCE), 2011]

(a)  Wavelength,  λ = Distance between two consecutive crests = 0.4 m

(b)  Frequency,  \nu = {v \over \lambda } = {{320} \over {0.4}} = 800\,\,Hz

(c)  Amplitude = During wave propagation, the maximum displacement of particles in a medium from mean position = 2 cm


Q.22  Explain a technique to find fault in a block of metal using reflection of sound.

[CBSE, 2010]

Ultrasound waves are used to detect a fault in a block of metal using reflection of sound. These waves are made to pass through one side of the metal block. The detectors placed on the other side detect the transmitted waves. If there is any flaw or defect like crack or hole, etc. then ultrasound gets reflected back and does not reach the detector.

sd1

 

5 Mark Questions

 

Q.23  (a) Give three medical uses of ultrasound.                        

(b) A ship which is stationary is at a distance of 2800 m from the sea-bed. The Ship sends an ultrasound signal to the sea-bed and its echo is heard after 4 s. Find the speed of sound in water.      

[Board, 2014]

(a) Ultrasound can be used in:

(i) Monitoring foetus during pregnancy.

(ii) Medical imaging of internal organs to check their size, structure and physiological functioning in humans.

(iii) Breaking kidney stones into smaller sizes to flush out in the urine. 

(b) Given,

Distance, d = 2800 m

Time, t = 40 sec

Velocity, v =?

Total distance travelled by sound = speed of sound × time

So, 2d = v × t

⇒  v=2d / t

⇒  v = 2 × 2800 / 4

= 1400 m/sec


Q.24  State reason for the following statements:                                      

(a) Ceiling of good conference halls and concert halls are curved.

(b) We hear sound produced by the humming bees while that of moving pendulum is not heard.

(c) Sound waves are mechanical waves.

(d) Sometimes we hear echo of sound.

(e) People in their old age suffering from hearing loss, wear hearing aids.

[Board, 2012]

(a)  Ceiling of good conference halls and concert halls are curved so that the sound waves after reflecting from these walls reaches every part of the hall and can be easily heard by the listeners.

(b)  We hear sound produced by the humming bees while that of moving pendulum is not heard. This is because the frequency of vibration of wings of bees is in the audible range of Humans (20Hz to 20 kHz) whereas frequency of vibrations of pendulum is less than 20 Hz so cannot be heard.

(c)  Sound waves are mechanical waves as they need material medium for propagation which is the characteristic of the mechanical waves.

(d) We sometimes hear the echo of a sound produced because the distance between the source of the sound and the obstacle is at least 17.2.

(e) People in their old age suffering from hearing loss, wear hearing aids because hearing aid amplifies the electrical signal and converts it into sound which can then be heard much clearly by the old people.

(f) People in their old age suffering from hearing loss, wear hearing aids. This is because hearing aid amplifies the electrical signal and converts it into sound which can then be heard much clearly by the old people.


Q.25  (a) The stone is dropped from a tower of 500 m height into a pond of water at the base of the tower. When is the splash heard at the top? (Given g = 10 ms–2 and speed of sound = 340 ms–1).

(b) How do the sound waves cause vibrations in the eardrum of human ear?

[CBSE (CCE), 2010]

(a) Given,

Height of the tower, h = 500m

Time taken by stone to reach water surface from the top of the tower:

{t_1} = \sqrt {{{2h} \over g}} = \sqrt {{{2 \times 500} \over {10}}} = \sqrt {100} = 10\,\,s

Time taken by the sound to reach the object:

{t_2} = {h \over v} = {{500} \over {340}} = 1.47\,\,s

Time at which splash is heard at the top:

{t_1} + {t_2} = 10 + 1.47 = 11.47\,\,s

(b) Sound waves are longitudinal waves which vibrate back and forth in the direction of its movement. It consists of compressions and rarefactions which results in pressure variation in the medium. This causes vibrations in the eardrum of human ear with required frequency and hence the sound becomes audible.



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