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Question (1):
What is meant by echo? What are the conditions necessary for its formation?
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Answer:
The sound heard after reflection from a rigid obstacle (a cliff, hill or wall) is called an echo.
Echo is heard if the sound reaches our ear 0.1 s after the original sound dies. To hear echo distinctively the distance between the reflecting surface and the listener should be approximately 17 m.
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Question (2):
Why do echoes produced in an empty auditorium usually decrease when it is full of audience?
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Answer:
When the hall is empty there is no obstacles in between to reflect the sound other than the walls. When the hall is full of audiences the sound produced undergoes multiple reflection from the people and so it overlaps with the sound produced. Hence the listener is not able to distinguish between the original sound and the echo.
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Question (3):
A boy fires a gun and hears the echo 2 seconds later. If he is 480 m away from a wall, calculate the velocity of sound in air.
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Answer:
v = 
= 
= 480 m s-1
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Question (4):
A girl claps and hears the echo after reflection from cliff which is 660 m away. If the velocity of sound is 330 m s-1, calculate the time taken for hearing the echo.
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Answer:
v x t = 2 d
= 
= 4 s
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Question (5):
A rifle shot is fired in a valley between two parallel walls. The echo from one wall is heard 3 s later and the echo from the other wall is heard 8.3 s later. If the velocity of sound at 0oC is 330 m s-1 and the temperature in the valley is 10oC. For every 1 oC. Calculate the width of the valley. For every 1 oC rise in temperature the velocity of sound increases by 0.61 m s-1.
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Answer:
Velocity of sound in valley at 10 oC = 330 + 10 0.61
= 330 + 6.1
= 336.1 m s-1
Distance of the nearest wall,
d = 
= 
= 504.15 m
Distance of the farthest wall,
d = 
= 
= 1008.3 m
 The total distance = 504.15 + 1008.3
= 1512.45 m
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Question (6):
Explain how is the principle of echo used by
(a) the bat during its flight at night.
(b) the dolphin to locate small fish as its prey.
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Answer:
(a) Bats during the flight send a series of twittering sound waves which strike against the object in their path of light and send back echoes to the bat's ear. Hearing the echoes the bats know how they must turn to avoid colliding with the objects at night.
(b) Dolphins are aquatic animals which send out ultrasonic sound to communicate with each other. They have a sound sensing system which enables them to find animals under water with great accuracy due to the echo of the ultrasonic sound produced by them.
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Question (7):
Define the terms velocity and wavelength applied to sound wave and state the relation between velocity, frequency and wavelength.
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Answer:
The distance traveled by a wave in one second is its velocity.
The distance traveled by the wave when it completes one oscillation. It is the distance between two consecutive particles executing the motion of the wave in same phase.
v = f where v is the velocity,  is the wave length and f is the frequency.
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Question (8):
What is the wavelength of sound waves produced in air by a vibrating tuning fork whose frequency is 256 Hz when the velocity of sound in air is
330 m s-1?
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Answer:
v = f
330 = 256 
 = 
= 1.3 m
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Question (9):
State the characteristics of a musical note on which its pitch, loudness and quality depend.
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Answer:
The pitch of a musical sound depends on the frequency of the wave. The higher the frequency higher the pitch.
The loudness of a musical note depends on its amplitude, distance of the observer from the source.
Quality or timbre of a musical note depends on the wave form produced by it.
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Question (10):
Figure shows a snap shot of a wave form of frequency 50 Hz in a string. The number in the string represent distance in centimeter. For this wave motion, find:
(a) wavelength
(b) amplitude, and
(c) velocity
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Answer:
(a) 20 cm
(b) 4 cm
(c) v = f
= 50 20
= 1000 cm s-1
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Question (11):
Frequency of a wave motion is 250 Hz. What is its time period?
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Answer:
Frequency ( ) = 250 Hz
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Question (12):
Define frequency. What is the frequency of a wave with a time period of 0.05 seconds?
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Answer:
Frequency is defined as the waves passing through a point in one second.
Frequency= 20 Hertz
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Question (13):
A sound wave travelling in air has a wavelength of 1.6x10-2 m. If the velocity of sound is 320 m/s, calculate the frequency of sound.
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Answer:
Wavelength of sound (l) = 1.6x10-2 m
Velocity of sound (v) = 320 m/s
Frequency = 20 x 104 hertz
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Question (14):
Give the physical explanation for the difference between musical sounds and noises.
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Answer:
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Question (15):
Name the musical instruments one in each case, which produces its notes by using the following. In each case state the method used to produce notes of different pitches and loudness.
(a) A vibrating string.
(b) A vibrating column of air.
(c) Vibration of any other body.
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Answer:
(a) Violin: The player using his fingers increase or shortens the length of the string to produce different frequency notes. Also different strings of different mass and tension produce different frequency or pitch. The loudness can be increased by the sound box provided.
(b) Flute: The length of the air column is varied by pressing the fingers on different holes which varies the frequency and hence the pitch. The loudness is increased by increasing the amount of air vibrating which increases the intensity.
(c) Drums: The membrane is adjusted to produce different pitch. The enclosed volume of air increases the loudness.
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Question (16):
Give scientific reasons for the following
(a) When a nail is hammered into a piece of wood the noise of hammering appears to get higher and higher in pitch.
(b) Marching soldiers are commanded to break steps when they are crossing a long bridge.
(c) Windows sometimes rattle when the low notes of a pipe organ are sounded.
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Answer:
(a) As the nail is hammered into the wood the length of the nail decreases. The frequency and hence the pitch of vibration of the nail increases as its length decreases.
(b) When soldiers march in unison it produces a sound of same frequency and large energy. If this frequency matches with the natural frequency of the bridge then it will cause resonance and the bridge will vibrate with a large amplitude. This may result in the collapse of the bridge.
(c) The frequency of the note produced by the pipe organ matches with the natural frequency of the windows. And the window panes readily take up the vibrations of the organ and start vibrating in unison with large amplitude due to resonance.
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Question (17):
The wavelength and velocity of red light is 7000  and 3x108m/s respectively. Calculate its frequency and time period.
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Answer:
Wavelength ( ) = 7000 
= 7000 x 10-10m (1  =10-10 m)
= 7 x 10-7m
velocity (v) = 3 x 108 m/s
= 0.428 1015 = 4.28 1014 Hz
= 0.002 x 10-12 s
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Question (18):
Sound made in front of a tall building 18 m away, is repeated. Name the phenomenon and briefly explain it.
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Answer:
The phenomenon is echo. When sound gets reflected from an obstacle and reaches our ear after 0.1 s. we are able to distinguish between the original sound and the reflected sound which is called echo.
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Question (19):
An observer stands at a distance of 850 m from a cliff and fires a gun. After what time gap will he hear the echo, if sound travels at a speed of 350 m s-1 in air?
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Answer:
v = 
= 
= 4.86 s
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Question (20):
(a) A vibrating tuning fork is placed over the mouth of a burette with water. The tap is opened and the water level gradually falls. It is observed that the sound becomes the loudest for a particular length of air column.
(i) What is the name of the phenomenon taking place when this happens.
(ii) Why does the sound become the loudest.
(iii) What is the name of the phenomenon taking place when sound is produced for another length of air column and is not the loudest.
(b) What change if any would you expect in the characteristics of a musical sound when we increase
(i) frequency
(ii) amplitude
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Answer:
(a) (i) The phenomenon is resonance.
(ii) When the frequency of the air column at that particular length is equal to or an integer multiple of the frequency of the tuning fork then it starts vibrating with a large amplitude which produces a loud sound.
(iii) This phenomenon is forced vibration. Here the air column vibrates but the amplitude is not as large as for resonance, hence the sound is not loudest.
(b) (i) When the frequency increases its pitch also increases.
(ii) The loudness of the sound increases when its amplitude increases.
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Question (21):
Calculate a) the wavelength b) the time period of a tuning fork of frequency 512 Hz which is set to vibrate. Velocity of sound in air is 320 m/s.
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Answer:
Frequency of the tuning fork (u) = 512 Hz.
Velocity of sound (v) = 320 m/s
= 0.00195 s
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Question (22):
Sound waves travel with a speed of 330 m/s. What is the wavelength of sound, whose frequency is 550 Hz?
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Answer:
Speed of the sound wave (v) = 330 m/s
Frequency of the sound wave ( ) = 550 Hz
Wavelength = 0.6 m
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Question (23):
Derive a relation between wave-velocity, frequency and wavelength.
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Answer:
Wave velocity is defined as the distance travelled by the wave in one second.
or
Wave velocity = Wavelength x frequency
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Question (24):
Radio waves of speed 3 108 m s-1 are reflected off the moon and received back on earth. The time elapsed between the sending of the signal and receiving it back at the earth station is 2.5 s. What is the distance of the moon from the earth?
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Answer:
= 
= 3.75 108 m
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Question (25):
(a) A sound wave of wave length 0.332 m has a time period of 10-3 s. If the time period is decreased to 10-4 s. Calculate the wave length and frequency of the new wave.
(b) Name the subjective property of sound related to its frequency and of light related to its wavelength.
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Answer:
time taken to travel 
= 
= 0.33 103
= 330 m s-1
Time period of 2nd wave = 10-4 s
Therefore wavelength  = V T
= 330 x 10-4
= 0.033 m
Frequency = 
= 
= 103 Hz
(b) Pitch is related to the frequency of sound and colour is related to the wavelength of light.
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Question (26):
(a) State two characteristics of wave motion.
(b) What is the relation between frequency, wavelength and speed of a wave?
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Answer:
(a)
 A wave motion is periodic in nature.
 The particles of the medium do not move from their mean position but execute vibration but only the energy is transmitted from one point to another.
(b) Speed = wavelength frequency
v = 
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Question (27):
A longitudinal wave of wavelength 1 cm travels in air with a speed of 330 ms-1. Calculate the frequency of the wave. Can this wave be heard by a normal human being?
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Answer:
= 33,000 Hz
The sound is not audible to human ear because frequency 20 to 20,000 Hz is the audible range of human ear.
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Question (28):
A person standing between two vertical cliffs and 640 m away from the nearest cliff shouted. He heard the 1st echo after 4 seconds and the second echo 3 seconds later. Calculate (i) the velocity of sound in air and (ii) the distance between the cliff.
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Answer:
Velocity of sound v = 
= 
= 320 m s-1
Distance the farthest cliff, d = 
= 
= 1120 m
Therefore the distance between the cliffs,
D = 640 + 1120
= 1760 m
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Question (29):
A ship on the surface of water sends a signal and receives it back after 4 seconds from a submarine inside the water. Calculate the distance of the submarine from the ship (The speed of sound in water is 1450 m s-1).
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Answer:
= 2900 m or 2.9 km
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Question (30):
A man fires a gun and hears its echo after 5 seconds. The man then moves 310 m towards the hill and fires his gun again. This time he hears the echo after 3 seconds. Calculate the speed of sound.
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Answer:
Let d be the distance between the man and the hill in the beginning.
=  ?. (1)
He moves 310 m towards the hill. Therefore the distance will be (d - 310)m.
 . (2)
Since velocity of sound is same, equating (1) and (2), we get
3d = 5d - 1550
2d = 1550
d = 775 m
= 310 m s-1
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Question (31):
The figure below shows a tuning fork with one of its prongs fastened to one end of a spring whose other end is fastened to a rigid support, when the tuning fork is made to vibrate, how does the pattern of the coils of the spring change.
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Answer:
The spring will execute vibrations having compressions and rarefaction which is how the tuning fork vibrates.
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Question (32):
Give the range of frequencies of infrasonic, audible and ultrasonic and mention one use of each of them.
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Answer:
Infrasonic  frequency is below 20 Hz
Use: to drill holes in rocks.
Audible sound  frequency range 20 Hz to 20,000 Hz.
Use: means of communication.
Ultrasonic  frequency above 20,000 Hz.
Use: Sonar in navigation and finding depth of sea.
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Question (33):
State the unit of loudness. Give the value for a soothing sound and unbearable sound.
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Answer:
Unit of loudness is decibel.
For a soothing sound the loudness is between 20-40 decibels.
Any sound louder than 120 decibels is unbearable.
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Question (34):
Echo is not heard in a small room. Explain.
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Answer:
We can distinguish the sound of echo if it is heard 0.1 s after it is made. If the echo reaches our ears before 0.1 s it overlaps with the sound produced due to persistence of hearing. The sound has to travel at least 17 m before it strikes an obstacle for it to reach our ear after 0.1 s. This does not happen in a small room.
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Question (35):
What types of waves are produced when
(1) Bell rings in air,
(2) The string of a guitar is plucked,
(3) By a vibrating tuning fork?
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Answer:
(1) Longitudinal waves are produced in air.
(2) Transverse waves are produced in the string.
(3) In the tuning fork longitudinal waves are provided.
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