Students should refer to Previous Year Questions ICSE Class 10 Physics Sound with solutions that have been prepared by expert teachers of ICSE Class 10 Physics. These questions and solutions are based on past year papers that have come in examinations of ICSE Class 10. Students should understand the type of questions asked and the solutions provided. Also refer to ICSE Class 10 Physics Solutions
ICSE Class 10 Physics Sound Last Year Questions
Students should learn the important questions and answers given below for Chapter Sound in Physics for ICSE Class 10. These board questions are expected to come in the upcoming exams. Students of ICSE Class 10th should go through the board exams questions and answers for ICSE Class 10 Physics which will help them to get more marks in exams.
Board Exam Questions Sound ICSE Class 10 Physics
Previous Year Questions ICSE Class 10 Physics Sound
Short Answer Type Questions I
Question: The human ear can detect continuous sounds in the frequency range from 20 Hz to 20,000 Hz. Assuming that the speed of sound in air is 330 ms-1 for all frequencies, calculate the wavelengths corresponding to the given extreme frequencies of the audible range.
Answer: We know that,
⇒ wavelength, λ = Speed (v) / frequency (v)
Case I : if ν = 20 hz and v = 330 m/s
⇒ λ1, = 330 / 20 = 16.5 m
Case II, if ν = 20,000 Hz and v = 330 m/s
⇒ λ2 = 330 / 20000 = 0.0165 m
Question: An enemy plane is at a distance of 300 km from a radar. In how much time the radar will be able to
detect the plane? Take velocity of radio waves as 3 × 108 ms-1.
Answer: Given, velocity (speed) = 3 × 108 m/s
Distance, (d) = 300 km
= 300 × 1000 m
Total Distance = 2d = 2 × 300 × 1000 m
Time = Total Distance / Speed
= 2 X 300 X 1000 / 3 108 sec
= 0.002 sec
= 2 ms
Question: How does a trawler man catch fish in deep water?
Answer: A trawler man sends ultrasonic waves in deep water, which on striking the shoal of fish, return back to the trawler man. He note the time t taken by the waves in going and coming back, he calculates the depth of fish d by using the relation
d = vt / 2
where, v is the velocity of ultrasonic waves in the water.
Question: (i) Name the phenomenon involved in tuning a radio set to a particular station.
(ii) Define the phenomenon named by you in part (i) above.
Answer: (i) Resonance.
(ii) Resonance is a particular case of forced vibration in which the frequency of forced vibrations is equal to the natural frequency of the body and the body begins to vibrate with increased amplitude.
Question: (i) Draw a graph between displacement and the time for a body executing free vibrations.
(ii) Where can a body execute free vibrations?
Answer:
Question: State two ways by which the frequency of transverse vibrations of a stretched string can be increased.
Answer: (i) By increasing the tension in the string.
(ii) By decreasing the length of the string.
(iii) By decreasing the mass per unit length of the string. (Any two correct points)
Question: (i) What are mechanical waves?
(ii) Name one property of waves that do not change when the wave passes from one medium to another.
Answer: (i) The waves that travel in the medium through the vibrations of the medium particles about their mean positions are known as mechanical waves.
(ii) It is the frequency of the waves that do not change when the wave passes from one medium to another.
Question: A bucket kept under a running tap is getting filled with water. A person sitting at a distance is able to get an idea when the bucket is about to be filled.
(i) What change takes place in the sound to give this idea?
(ii) What causes the change in the sound?
Answer: (i) The frequency of the sound appears to be increased as the air column of bucket is filled with water.
(ii) The changes in the sound is caused, because air column is decreased in filling the water in bucket.
Question: A sound made on the surface of a lake takes 3 s to reach a boatman. How much time will it take to reach a diver inside the water at the same depth? Velocity of sound in air = 300 ms-1; Velocity of sound in water = 1440 ms-1
Answer: Given, Time of sound to reach at boatman t = 3 s
Velocity of sound in air va = 300 m/s
Velocity of sound in water vw = 1400 m/s
Time taken by diver td = ?
Depth of lake d1 = va × t
= 300 × 3 = 900 m
(as given in question)
Time taken by sound in water to reach diver
Question: Name one factor which affects the frequency of sound emitted due to vibrations in an air column.
Answer: The frequency of sound emitted due to vibrations in air column is affected by the length of air column.
Question: State two differences between light waves and sound waves.
Answer: The differences between sound waves and light waves are
Question: An ultrasonic wave is sent from a ship towards the bottom of the sea. It is found that the time interval
between the sending and the receiving of the wave is 1.5s. Calculate the depth of the sea, if the velocity of sound in sea water is 1400 ms–1.
Answer: Given, Time interval t = 1.5s
D = Distance travelled by ultrasonic waves during
1.5 s.
Velocity v = 1400 m/s
So. D = v × t = 1400 × 1.5 = 2100 m
As, the depth of the sea = d / 2 = 2100 / 2 = 1050 m
Question: A radar sends a signal to an aeroplane at a distance 4.5 km away with a speed of 3x 108 ms -1.
After how long is the signal received back from the aeroplane?
Answer: Given, Distance, d = 45 km = 45000 m
Total distance travelled by radar signal is
2d = 45000 × 2 = 90000 m
Speed, v = 3 × 108 m/s
So time taken t = ?
Question: Explain why musical instruments like the guitar are provided with a hollow box?
Answer: The musical instruments like guitar are provided with a hollow box, because vibrating strings of the
instruments produce a very weak sound which cannot be heard. But string set into vibration produce forced vibrations in the large volume of air filled in the hollow box and thus, loud sound of the frequency matching with the frequency of the vibrating string is produced.
Question: State two ways by which the frequency of transverse vibrations of a stretched string can be decreased.
Answer: The two ways by which the frequency of transverse vibrations of a stretched string can be decreased are
(i) By increasing the length (l), when tension (T) and mass (m) are kept constant.
(ii) By decreasing the tension (T), when length (l) and mass (m) are kept constant.
Question: What is SONAR? State the principle in which it is based.
Answer: SONAR is an abbreviation for Sound Navigation and Ranging. It uses ultrasonic waves to detect and locate objects under water. It is based on the principle of echo and state that the sound heard after reflection from a distant obstacle after the original sound has ceased.
Question: 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 ms–1 in air?
Answer: Given, distance between sound source and cliff,
d = 850 m
Total distance travelled by the sound = 2d
Speed tv = 350 m /s; time t = ?
Question: Radio waves of speed 3 × 108 m/s are reflected from the moon and received back on the earth, the
time elapsed between the sending of the signal and receiving it back at the earth’ s surface is 2.5 s,
what is the distance of the moon from the earth ?
Answer: Given speed, v = 3 × 108 m/s
Time taken by the signal to travel twice the distance
between moon and earth, t = 2.5 s
So, distance between the earth and the moon is given by
Short Answer Type Questions II
Question: (i) What are damped vibrations?
(ii) Give one example of damped vibrations.
(iii) Name the phenomenon that cause a loud sound when the stem of a vibration tuning fork is kept pressed on the surface of a table.
answer: (i) The vibrations of which the amplitude goes on decreasing with time and ultimately vibration stops are called damped vibrations.
(ii) Vibrating string of guitar.
(iii) Resonance
Question: John fires a gun towards a hill and hears its echo after 5 seconds. He then moves 320 m towards the
hill and fires his gun again. This time he hears the echo after 3 seconds. Calculate the velocity of sound.’
Answer: Let the distance of the hill from john be x m. Let the velocity of sound be V.
Question: (i) Name the waves used for echo depth sounding.
(ii) Give one reason their use for the above purpose.
(iii) Why are the waves mentioned by you not audible to us?
Answer: (i) Ultrasonic
(ii) They can travel undeviated through long distance.
(iii) Frequency of ultrasonic waves is above 20000 Hz and audible range of frequency is 20 Hz to 20000 Hz.
Question: (i) What is an echo?
(ii) State two conditions for an echo to take place.
Answer: (i) Clear and distinct sound heard after its reflection from a rigid surface is called an echo.
(ii) The reflecting surface should be present at least 16 to 18 m from the source of sound OR the sound be reflected back after 0.1 s. The size of the reflecting surface should be greater than the wavelength of sound wave.
Question: A person standing between two vertical cliffs and 480 m from the nearest cliff shouts. He hears the
first echo after 3 s and the second echo 2s later.
Calculate :
(i) The speed of sound.
(ii) The distance of the other cliff from the person.
Answer:
Question: (i) Draw a displacement-distance graph for two waves A and B such that the amplitude of wave A is three times that of wave B.
(ii) Give one example of forced vibration.
(iii) State two necessary condition for hearing a distance echo.
Answer: (i) The displacement-distance graph is give below
(ii) The vibrations produced in air of the sound box when strings of a sitar are plucked to vibrate.
(iii) (1) The echo must reach the ear at least 0-1 s after the original sound is produced.
(2) The size of reflector must be large enough as compared to the wavelength of the sound wave.
Question: (i) What is the principle on which SONAR is based?
(ii) An observer stand at a certain distance away from a cliff and produced a loud sound. He hears
the echo of the sound after 1.8 s. Calculate the distance between the cliff and the observer, if the
velocity of sound in air is 340 ms –1.
Answer: (i) The SONAR is based on the principle of echo.
(ii) Given, time, t = 1.8 s
Distance travelled, d = ?
Distance = speed × time
= 2d = va × t
Question: A vibrating tuning fork is placed over the mouth of a burette filled with water. The tap of the burette is opened and the water level gradually starts falling. It is found that the sound from the tuning fork becomes very loud for a particular length of the water column.
(i) Name the phenomenon taking place when this happens.
(ii) Why does the sound become very loud for this length of the water column?
Answer: (i) In the above process, the phenomenon taking place is resonance.
(ii) The sound becomes very loud for this length of the water column, because the frequency of the air column becomes equal to the frequency of the tuning fork i.e., fa= ft
Question: (i) What is meant by resonance?
(ii) State two ways in which resonance differs from forced vibrations.
Answer: (i) The phenomenon in which body vibrating under the influence of periodic force, where frequency of the applied periodic force is equal to the natural frequency of the vibrating body, begins to vibrate with increased amplitude.
(ii) The two ways in which resonance differs from forced vibrations are as follows :
Question: (i) A man standing between two cliffs produces a sound and hears two successive echoes at intervals of 3 s and 4 s respectively. Calculate the distance between two cliffs. The speed of sound in the air is 330 ms–1.
(ii) Why will an echo not be heard, when the distance between the source of sound and the reflecting
surface is 10 m?
Answer: (i) Given : time taken from nearer cliff t1=3 s
Time taken from farther cliff, t2 = 4 s
Velocity of sound in air, va = 330 m/s
Total distance, d = ?
Distance of man from nearer cliff,
Total distance between two cliffs
d = d1 + d2 = 495 + 660 = 1155 m
(ii) An echo can not be heard, when the distance between the source of sound and the reflecting surface is 10 m, because to hear the echo distinctly, the minimum distance of the obstacle from the observer should be 17 m.
Question:. (i) Name the type of wave which are used for sound ranging.
(ii) Why are these waves mentioned in (i) above, not audible to us?
(iii) Give one use of sound ranging.
Answer: (i) Ultrasonic waves are used for sound ranging as the frequency is greater than 20 kHz.
(ii) As the frequency of ultrasonic is above human is below 20 kHz.
(iii) Sound ranging is used to find the depth.
Question: A man standing 25 m away from a well produces a sound and receives the reflected sound.
(i) Calculate the time after which he receives the reflected sound, if the speed of sound in air is 350 ms–1
(ii) Will the man be able to hear a distinct echo? Give a reason for your answer.
Answer: Given : distance between man and wall
D = 25m
(i) t = ?
Velocity of sound, v = 350 m/s
V – 2d / t
⇒ t – 2d / v – 2 X 25 / 350 – 0.14s
(ii) Yes, the man will be able to hear a distinct echo, because the least time for distinct echo to be 0.1 s as 0.14 s > 0.1 s.
Question: (i) A man stand at a distance of 68 m from a cliff and fires a gun. After, what time interval will he hear the echo. If the speed of sound in air is 340 ms–1?
(ii) If the man had been standing at a distance of 12 m from the cliff would he have heard a clear echo?
Answer: (i) Given : distance between man and cliff d= 68 m
Total distance travelled by the sound
= 2d
Velocity of sound v = 340 m/s
Time, t = ?
T- 2d / v – 2X2 68 / 340 – 0.4 s
(ii) No he would not have heard a clear echo, because to hear echo distinctly, the reflecting surface in air should be at a minimum distance of 17 m from the listener.
Question: (i) What is the principle on which SONAR is based?
(ii) Calculate the minimum, distance at which a person should stand infront of a reflecting surface so that he can hear a distinct echo.(take speed of sound in air = 350 ms -1).
Answer: (i) SONAR is based on the principle of echo
(ii) Speed of sound, v = 350 m/s
Distance, d = ? Time, t = 0.1 s
[∴ persistence of hearing]
d – v X t / 2 – 350 X 0.1 / 2 – 17.5m
Question: (i) What is meant by an echo? Mention one important condition that is necessary for an echo to be heard distinctly.
(ii) Mention one important use of echo.
Answer: (i) Echo : The sound heard after reflection from a distant obstacle after the original sound has ceased.
(a) To hear echo distinctly, the reflecting surface in air should be at a minimum distance of 17 m from the listener.
(ii) The uses of echo are
(a) By SONAR
(b) By bats, dolphin and fisherman. [anyone]
Question: A man standing infront of a vertical cliff fire a gun. He hears the echo after 3 s. On moving closer
to the cliff by 82.5 m, he fires again. This time, he hears the echo after 2.5 s. Calculate by,
(i) The distance of the cliff from the initial position of the man.
(ii) The velocity of sound.
Answer: (i) Let d be the distance of cliff from the initial position of man and v be the velocity of sound.
So, the distance travelled by sound in 3 s = 2d metre
Therefore speed of sound v = 2d / time = 2d / 3
On moving closer to the cliff by a distance of 82.5 m then
(iii) Substituting the value of d in eq (i), we get v = 2 × 495/3=330 m/s
Question: When a tuning fork, struck by a rubber pad is held over a length of air column in a tube, it produces a
loud sound for a fixed length of the air column.
(i) Name the above phenomenon.
(ii) How does the frequency of the loud sound compare with that of the tuning fork ?
(iii) State the unit for measuring loudness,
Answer: (i) The given phenomenon is based on resonance.
(ii) The frequency of the loud sound is equal or integral multiple of the frequency of the tuning fork.
(iii) The unit for measuring loudness is decibel (dB).
Question: A radar is able to detect the reflected waves from an enemy aeroplane, after a time interval of 0.02
ms. If the velocity of the waves is 3 × 108ms–1, Calculate the distance of the plane from the radar.
Answer: Given, time, t = 0.02 × 10–3 s
Velocity, v = 3 × 108 m/s
As the distance travelled by the wave is twice of the distance between energy and source.
v – 2d / t
⇒ d – v X t / 2 – 3 X 108 X 0.02 X 10-1 / 2
– 3000m – 3km
Question: (i) Sound made infront of a tall building 18 m away, is repeated. Name the phenomenon and briefly explain it.
(ii) A tuning fork, held over an air column of a given length, produces a distinct audible sound. What did you call this phenomenon? How does it occur?
Answer: (i) As we know the minimum distance between source and reflector is 17 m and here the distance is 18 m so the phenomena due to which sound is repeated is echo.
(ii) We call the phenomenon as resonance. This occurs when the frequency of forced vibrations produced by the tuning fork equal to the natural frequency of the air column.
Question: A sound wave of wavelength 0.332 m has a time period of 10–3 s, If the time period is decreased to
10–4 s, calculate the wavelength and frequency of the new wave.
Answer: Given, wavelength, λ1 = 0.332 m
Time period, T1=10–3 s
Time period, T2 = 10–4 s
Wavelength, λ2=?
Frequency, f1 and f2=?
Long Answer Type Questions I
Question; (i) A wire of length 80 cm has a frequency of 256 Hz. Calculate the length of a similar wire
under similar tension, which will have frequency 1024 Hz.
(ii) A certain sound has a frequency of 256 hertz and a wave length of 1.3 m.
(a) Calculate the speed with which this sound travels.
(b) What difference would be felt by a listener between the above sound and another sound travelling at the same speed, but of wavelength 2.6 m?
Answer: (i) Frequency of stretched wire having length l is given by
(ii) (a) Given, f = 256 Hz, λ=1.3 m
Speed = frequency × wavelength
Speed = f × λ
= 256 × 1.3 m/s
= 332.8 m/s
(2) wavelength of another sound = 2.6 m
Therefore frequency of this sound = 128 Hz
as f × 1 / λ)
Therefore sound will be heard less.
Question: (i) What is meant by the terms :
(a) amplitude (b) frequency of a wave?
(ii) Explain, why stringed musical instruments, like the guitar, are provided with a hollow box
Answer: (i) (a) The maximum displacement of the particle of medium on either side of its mean position is known as amplitude of wave. It is denoted by A.
(b) The number of vibrations made by the particle in one second is known as frequency of wave. It is denoted by f or ν.
(ii) The musical instruments like guitar are provided with a hollow box, because vibrating strings of the instrument produce a very weak sound which cannot be heard, but strings set into vibration produce forced vibration, the large volume of air filled in hollow box and the loud sound of the frequency of the vibration string is produced.
Question: The following diagram shows the displacement time graph for a vibrating body.
(i) Name the type of vibration produced by the vibrating body.
(ii) Give one example of a body producing such vibrations.
(iii) Why is the amplitude of the wave gradually decreasing?
(iv) What will happen to the vibrations of the body after sometime?
Answer: (i) Damped vibrations are produced by the vibrating body.
(ii) A simple pendulum oscillating in air, tuning fork vibration in air are examples of a body producing such vibrations.
(iii) The amplitude of the wave gradually decreasing due to friction as energy is continuously losing.
(iv) Its amplitude decrease and finally stops.
Question: (i) A person is tuning his radio set to a particular station. What is the person trying to do to tune it?
(ii) Name the phenomenon involved, in tuning the radio set.
(iii) Define the phenomenon named by you in part (ii).
Answer: (i) The person is trying to change the frequency of his radio set to a particular station.
(ii) The phenomenon of resonance is involved in tuning the radio set.
(iii) Resonance is a special case of forced vibrations. When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body readily begins to vibrate with an increased amplitude. This phenomenon is known as resonance. The vibrations of large amplitude are called the resonant vibrations.
Question; (i) Sometimes when a vehicle is driven at a particular speed, a rattling sound is heard.
Explain briefly, why this happens and give the name of the phenomenon taking place?
(ii) Suggest one way by which the rattling sound could be stopped.
Answer: (i) Sometimes when a vehicle is driven at a particular speed, a rattling sound is heard, because the frequency of movement of piston in the engine becomes equal to the natural frequency of the part which is rattling. The phenomenon taking place is resonance.
(ii) The rattling sound can be stopped by changing the speed of the vehicle.
Question: (i) Differentiate between resonance and forced vibrations.
(ii) The wavelength of waves produced on the surface of water is 20 cm. If the wave velocity is 24 ms–1, calculate
(a) The number of waves produced in one second and
(b) The time required to produce a wave.
Answer: (i) Refer to ans. 7 (i) (2 marks).
(ii) Given, wavelength, y – 20 cm = 0.2 m
Wave velocity v = 24 m/s
(a) Number of wave/s f = ?
(b) Time required t = ?
(a) Number of waves/s
Question: (i) Give one example each of natural vibration, forced vibration and resonance.
(ii) Mention one practical use of echoes.
Answer; (i) Natural vibration : A tuning fork struck against a hard rubber pad.
Forced vibration : Vibration produced in the board of a guitar when its string is made to vibrate.
Resonance : While tuning a wireless receiver, we adjust its natural frequency equal to that of the incoming wireless carries waves, so that a loud sound of incoming signal is heard.
(ii) The practical use of echoes is to determine distance between two objects
v = 2d / t
d = vt / 2
Where, d is the distance (not less than 17 m) between the source of sound and reflecting surface, t is time taken.
Question: (i) How does a stretched string on being set into vibration, produce the audible sound?
(ii) Will the sound be audible, if the string is set into vibration on the surface of the moon? Given reason for your answer.
Answer: (i) On striking, a vibrating string produces a very weak sound which cannot be heard at distance. Thus, all the stringed instruments are provided with a sound box. The vibrating string produces forced vibrations of larger amplitude as now a larger volume of air is set into vibrations. Hence, this sound becomes audible.
(ii) The sound produced on the surface of the moon by a vibrating string cannot be heard, because there is absence of any medium and we know that sound travels only in a medium.
Long Answer Type Questions II
Question: In the diagram below, A, B, C, D are four pendulums suspended from the same elastic string PQ. The length of A and C are equal to each other while the length of pendulum B is smaller than that of D. Pendulum A is set into a mode of vibrations.
(i) Name the type of vibrations taking place in pendulums B and D?
(ii) What is the state of pendulum C?
(iii) State the reason for the type of vibrations in pendulums B and C.
Answer: (i) Pendulum B and D execute forced vibrations
(ii) Pendulum C will be in a state of Resonance
(iii) Natural frequency of B does not match with Natural frequency of A. Natural frequency of C matches that of A
Characteristics of Sound
Short Answer Type Questions I
Question: How is the frequency of a stretched string related to :
(i) Its length?
(ii) Its tension?
Answer:
Question: What are the factors on which the following characteristics of a musical note depends?
1. intensity 2. timbre.
Answer: Factors on which the following characteristics of musical note depends are
(1) Intensity depends upon amplitude.
(2) Timbre depends upon
(i) waveform of sound
(ii) resonance and sound pressure.
Question: The ratio of amplitude of two waves is 3 : 4. What is the ratio of their
(i) Loudness?
(ii) Frequencies?
Answer: (i) 9 : 16
(ii) 1 : 1
Question: (i) What do you understand by loudness of sound?
(ii) In which units is the loudness of sound measured?
Answer: (i) Loudness is magnitude of auditory ensation.
(ii) bel/dB/phon (any one)
Question: What is meant by noise pollution? Name one source of sound causing noise pollution.
Answer: The disturbance produced in the environment due to undesirable loud and harsh sound of level above 120 dB.
Source : Honking of vehicles in traffic jams
Question: (i) State the safe limit of sound level in terms of decibel for human hearing.
(ii) Name the characteristic of sound in relation to its waveform.
Answer: (i) Safe limit of sound level – upto 120 dB.
(ii) Quality of sound waves. Source; Honking of vehicle in traffic jams.
Question: Name the unit used for measuring the sound level.
answer: Decibel (dB) is the unit used for measuring the sound level.
Question: Which characteristics of sound will change, if there is a change in its (i) amplitude (ii) Waveform
Answer: (i) Loudness of sound will change its amplitude.
(ii) Quality of sound will sound will change, if
there is a change in its wavelength.
Question: (i) Three musical instruments give out notes at frequency listed below. Flute; 400 Hz guitar 200
Hz trumpet; 500 Hz which one of these pitch? Of 256 Hz resonate? 288 Hz, 341 Hz, 333 Hz, 512 Hz.
Answer: (i) Trumpet having highest frequency of 500 Hz have highest pitch, because pitch is directly proportional to the frequency.
(ii) The frequency 512 Hz will resonate a tuning fork of 256 Hz as 512 Hz is an integral multiple of the natural frequency of the tuning fork.
Question: When acoustic resonance take place a loud sound is heard. Why does this happen? Explain.
Answer: When acoustic resonance take place a loud sound is heard because the amplitude of the sound is increased. As loudness of the sound is directly proportional to the square of the amplitude i.e. L × A2. So loudness of sound increases.
Question: Two waves of the same pitch have their amplitude in the ratio 2 : 3. What will be the ratio of their
loudness?
(ii) What will be the ratio of their frequency its?
Answer: (i) As loudness ∝ (amplitude)
(iii) The ratio of frequency is 1 : 1 as pitch depends upon the 24 characteristics of musical sound when we increase
(i) its frequency (ii) its amplitude
Short Answer Type Questions II
Question: Name the factor that determines :
(i) Loudness of the sound heard.
(ii) Quality of the note.
(iii) Pitch of the note.
Answer: (i) Amplitude of vibration.
(ii) Wavelength of sound wave.
(iii) Frequency of vibration.
Question: Following diagram shows three different mode of vibrations P, Q and R of the same string.
(i) Which vibration will produce a louder sound and why?
(ii) The sound of which string will have maximum shrillness?
(iii) State the ratio of wavelength of P and R.
Answer: (i) The mode of vibration R will produce a louder sound, because its amplitude is more than that of P and Q.
(ii) The sound of P string will have maximum shrillness because its frequency is more than that of R and Q.
(iii) Frequency fp= 3 f, frequency fr = f
λ – v / f
The ratio of wavelength of P and R is given by,
Question: (i) Name the characteristics of sound which enables a person to differentiate between two sounds with equal loudness but having different frequencies.
(ii) Define the characteristic named by you in (i).
(iii) Name the characteristic of sound which enables a person to differentiate between two sounds of the same loudness and frequency but produced by different instruments.
Answer: (i) Pitch or shrillness is the characteristic of sound which enable a person to differentiate between two sounds with equal loudness but having different frequencies.
(ii) Pitch is that characteristic of sound by which an acute (or shrill) note can be distinguished from a grave or flat note.(iii) Quality is that characteristic of sound which enables a person to differentiate between two sounds of the same loudness and frequency but produced by different instruments.
Question: (i) The ratio of the amplitude of two waves is 4 : 9. What is the ratio of their intensities?
(ii) A pendulum has a frequency of 5 vibrations per second. An observer starts the pendulum and fires a gun simultaneously. He hears the echo from a cliff after 8 vibrations of the pendulum. If the velocity of sound in air 340 ms–1, w;hat is the distance between the cliff and the observer?
Answer:
(ii) Given, f = 5 vib / s = 5 Hz
Velocity of sound= 340 m/s
Distance between observer and cliff, d =?
∴ 5 vibrations take 1 s.
∴ 8 vibrations will require = 1 / 5 × 8 – 1.6s
Question: A vibrating tuning fork is placed over the mouth of a burette filled 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.
Answer: (i) The phenomenon taking place is resonance in air column.
(ii) The frequency of vibrating tuning fork becomes equal to the frequency of vibration air column. As a result amplitude of the sound produced increases. Since, loudness is proportional to the square of amplitude so sound is loudest.
(iii) Forced vibration i.e. frequency of forced vibration is not equal to the frequency of the vibrating air column.
Question: (i) State three characteristics of musical sound.
(ii) How does the musical sound differ from noise?
Answer: (i) The three characteristic if a musical sound are.
Pitch : It is the frequency of sound perceived by human ear. A high frequency gives rise to high pitch note and a low frequency produce a low pitch note.
Loudness (Intensity) : It is the property by virtue of which a loud sound can be distinguished upon the amplitude of wave (I ∝ A2)
Quality (Timbre) : It is that characteristics of musical sound which distinguish between two sounds of same loudness and same pitch but emitted by two different instruments.
(ii) The musical sound differ from noise in the following ways
Question: (i) Two friends were playing on their identical guitars whose string were adjusted to give
notes of the same pitch. Will the quality of the two notes be the same? Give a reason for your answer.
(ii) Give the relation among wavelength, time period and wave velocity of a wave motion.
Answer: (i) Yes the quality of the two notes will be same. Quality of a wave depend on the wavelength and the number of subsidiary notes. Since both guitars are identical and have same pitch the notes will have an identical quality.
(ii) The relation between wavelength λ, time period (T) and wave velocity (v) of wave motion is given by v – λ / t
