WBBSE Solutions For Class 9 Physical Science And Environment Chapter 7 Sound

Chapter 7 Sound Very Short Answer Type :

Question 1. Define wave motion.
Answer: A wave is a disturbance by which energy is transferred through vibrations.

Question 2. Define periodic motion.
Answer: The type of motion that repeats after a definite interval of time is called periodic motion.

Question 3. What sound will you hear earlier — the sound of an approaching car or the sound of its horn?
Answer: The sound of the horn will be heard earlier as the velocity of sound is higher than that of any fastest car.

Question 4. What is the safe limit of sound level in dB for our ear?
Answer: According to WHO, the safe limit of sound level is 45 dB. (Internationally acceptable safe limit of sound is 65 dB).

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Question 5. What is the sound level of normal conversation?
Answer: Sound level of normal conversation is 50 decibel (dB).

Question 6. About 1m long thin wire is stretched tightly between two horizontally fixed bodies, when the wire is plucked a sound is heard. Now a V-shaped paper rider is kept on the wire and the wire is again plucked when the paper rider flies away. What does the experiment prove?
Answer: The experiment proves that sound is produced by a vibrating body.

Question 7. is it possible to make any discussion by two persons on the moon?
Answer: No, as there is no material medium on the moon, any sound produced by anyone will not propagate to the other.

Wbbse Physical Science And Environment Class 9 Solutions

Question 8. Why does a person stationed at a higher position hear a sound better than when he is on the ground as wind blows opposite to the direction of sound?
Answer: This happens because when wind blows against the direction of sound, sound waves deviate above the ground.

Question 9. How is sound produced when we speak?
Answer: Sound is produced due to vibration of our vocal chords.

Question 10. How does intensity of sound depend on the density of a medium?
Answer: Intensity of sound increases in a denser medium than in a rarer medium.

Question 11. What is a note of sound?
Answer: It is a sound of mixed frequencies.

Wbbse Physical Science And Environment Class 9 Solutions

Question 12. What is Hertz?
Answer: It is the unit used at present to express frequency. Hertz (Hz) is named after the celebrated German physicist Heinrich Hertz.

Question 13. What is decibel?
Answer: Decibel (dB) is the unit of expressing the intensity of sound.

Question 14. The oscillation of a pendulum cannot be heard. Why?
Answer: The oscillation of a pendulum has a frequency less than 20/sec. So, we cannot hear the sound.

Question 15. What is the unit of frequency?
Answer: Unit of frequency is cycles/second or CPS or Hz.

Question 16. How does the velocity of sound depend on pressure ?
Answer: Velocity of sound is independent of the change in pressure.

West Bengal Board Class 9 Physical Science Solutions Chapter 7 

Question 17. What is musical sound ?
Answer: It is a pleasing sound produced by a periodically vibrating body.

Question 18. What are the characteristics of a musical sound ?
Answer:
Characteristics of musical sound are :
(1) Intensity
(2) Pitch
(3) Quality or Timbre.

Question 19. Which characteristic of musical sound depends on frequency?
Answer: Pitch depends on frequency.

Question 20. Find the optical analogs of note and tone.
Answer: Note is analogous to white or polychromatic light and tone is analogous to colour.

Question 21. How many frequencies are there in a tone?
Answer: A tone is a sound of a single frequency.

Question 22. What is fundamental tone?
Answer: In a note, the sound of least frequency is known as the fundamental tone.

West Bengal Board Class 9 Physical Science Solutions Chapter 7 

Question 23. Can sound travel through a vacuum?
Answer: Sound cannot travel through vaccum.

Question 24. What should be the frequency of audible sounds?
Answer: We can hear sound only if the number of vibrations per second, called frequency of a vibrating body, lies within the limit 20 to 20,000.

Question 25. How does pitch depend on frequency?
Answer: Greater the frequency, higher will be the pitch.

Question 26. Between a plane mirror and a wall of house, which one will reflect more sound?
Answer: A wall of house will reflect sound.

Question 27. What is responsible for the generation of sound?
Answer: Vibration is responsible for the generation of sound.

Question 28. What is the minimum distance of the reflector to produce an echo?
Answer: The minimum distance of the reflector to produce an echo is 16.6 m.

Question 29. What is the unit of wavelength in SI system?
Answer: The unit of wavelength in SI system is metre.

Question 30. What is the velocity of sound at 0°C?
Answer: 332 m/sec.

Question 31. Is medium required for propagation of sound?
Answer: Medium is required for propagation of sound.

Chapter 7 Sound 2 Marks Questions And Answers:

Question 1. What is the limit of audibility of sound ?
Answer:

Limit of audibility of sound

When we hear a sound, the sensation of the sound remains in the ear for some time. This is known as persistence of heating. But all sounds are not audible to the human ear, In order that sound may be audible to the human ear, the frequency of the vibration of the source must be between 20 Hz to 20,000 Hz. This is called the audible range or the limit of audibility for normal human ear.

Wbbse Physical Science And Environment Class 9 Solutions 

Question 2. What is ultrasonic sound ?
Answer:

Ultrasonic sound

Sounds of frequency higher than 20,000 Hz are called ultrasonic. We cannot hear the ultrasonic sounds. The speed of ultrasonic sounds is same as that of audible sound. Some animals like dogs, bats, monkeys, deer, leopards, etc. can hear ultrasonic sounds. Dogs can hear sound of frequency 50,000 Hz and bats can hear sounds upto 100,000 Hz frequency.

Question 3. What is infrasonic or subsonic sound ? State its uses.
Answer:

Infrasonic or subsonic sound

Sounds of frequency lower than 20 Hz are called infrasonic or subsonic. We cannot hear the infrasonic sound. The speed of such sound is same as that of audible sound.

Uses : The infrasonic sound is used in drilling deepest wells because the low frequency vibrations break up hard rock more easily than the ordinary drills.

Question 4. How is the reflection of sound different from that of light?
Answer: The main difference between the reflection of sound and that of light is that sound requires a large reflecting surface which may not be smooth, while for the reflection of light, the reflecting surface must be smooth and may be small.

Question 5. What is called an echo?
Answer:

Echo : When a sound is reflected back from a reflecting surface to the source of sound in-such a way that a repetition of the original sound is heard distinctly, then. this phenomenon is called echo.

Wbbse Physical Science And Environment Class 9 Solutions 

Question 6. Explain why the echo of our’sound is not heard in a small room, but is heard distinctly in a big hall.
Answer: To hear the echo of a sound distinctly, the reflecting surface should be at a minimum distance of 16.6 m from the observer. In a small room the distance is less than 16.6 m, the reflected sound will reach the ear before the original dies out and, therefore, it will not be heard distinctly. In a big hall the distance is generally more than 16.6 m.

Question 7. How is the speed of sound determined by the method of echo ?
Answer: To determine the speed of sound in air, the echo method is used. For this, sound is produced from a known distance, say, d (d must be greater than 16.6), from the reflect- ing surface. The time interval, say, t, in which the echo reaches the place from where the sound was produced, is noted by a stop watch. The speed of sound, say, V is calculated by the formula

V\(=\frac{\text { Total dis tance Travelled}}{\text { Time interval }}\)\(=\frac{2 d}{t}\)

Question 8. What is echo depth sounding ?
Answer:

Echo depth sounding

The ships locate the position of rocks and depth of sea by the reflection of sound from it. This process is called echo depth sounding.

Question 9. What is sonar (sound navigation and ranging) ?
Answer:

Sonar (sound navigation and ranging)

Sonar (sound navigation and ranging) method is used to determine the position of iceberg, enemy’s submarine, sunken ship or depth of sea with the help of ultrasonic waves. Ultrasonic waves are sent in all directions from the ship at sea and are then received on their return after reflection from an obstacle. The position of the obstacle can be measured by measuring the time interval t between the source of sound and waves after reflection. Then by using the formula d = a dis calculated. V is the speed of ultrasonic waves in water.

Question 10. The time interval between a lightning flash and the first sound of thunder was found to be 5 sec. If the speed of sound in air is 330 m/sec, find the distance of the flash from the observer.
Answer:

Given

The time interval between a lightning flash and the first sound of thunder was found to be 5 sec. If the speed of sound in air is 330 m/sec

The lightning flash and thunder occur simultaneously, but light has much larger speed than sound. Hence, the time taken by the lightning flash is negligible. The speed of sound 330 m/sec and time taken = 5 sec.

Therefore, distance of flash from the observer = 330 x 5 m =1650 m.

West Bengal Board Class 9 Physical Science Solutions Chapter 7

Question 11. The smoke from the gun barrel is seen 2 seconds before the explosion is heard. If the speed of sound in air is 330 m/sec, calculate the distance of the observer from the gun.
Answer:

Given

The smoke from the gun barrel is seen 2 seconds before the explosion is heard. If the speed of sound in air is 330 m/sec,

Since the light has a much larger speed than sound, hence we can assume that light takes negligible time and the sound takes 2 seconds to reach the observer.

Distance = Speed x time = 330 x 2 = 660 m.

Wbbse Physical Science And Environment Class 9 Solutions 

Question 12. Show by an example of a natural phenomenon that the velocity of light is much greater than the velocity of sound.
Answer: In a natural phenomenon, the lightning flash and thunder occur simultaneously. We see the lightning flash and after some time hear the sound of thunder.
Question 13. Find the relation between wave velocity (V), frequency (f) and time period (T).
Answer:

Relation between wave velocity (V), frequency (f) and time period (T)

Since the distance travelled by the wave in one time period T is equal to wave length (λ) of the wave,

wave velocity (V)\(=\frac{\lambda}{T}\), but f\(=\frac{1}{T}\)

Hence,V=f×λ

or, wave velocity = frequency x wavelength.

Wbbse Physical Science And Environment Class 9 Solutions 

Question 14. What do you mean by damped vibrations ?
Answer:

Damped vibrations

The amplitude of free vibration of a body is seen to diminish gradually and finally die away altogether after sometime. Thus, the periodic vibrations of decreasing amplitude are called the damped vibrations.

Question 15. What are forced vibrations ?
Answer:

Forced vibrations: The vibrations which take place under the influence of an external periodic force are called forced vibrations.

For example :
(1) If a vibrating tuning fork is held in air, the sound emitted by it is very feeble. When the fork is placed on the table, the intensity of sound increases. This is because the air in contact with the table begins to vibrate with the tuning fork. As greater amount of air is under vibration, the intensity of sound increases.

(2) The vibration produced in the board of a guitar when its string is made to vibrate.

(3) All stringed instruments owe their loud sound due to the forced vibrations produced in the air of the sounding box.

Question 16. What is Resonance ?
Answer:

Resonance

Resonance is a particular case of the forced vibrations. When the frequency of the applied periodic force is equal to the natural frequency of the body, it readily takes up the vibrations and begins to vibrate with an increased amplitude. This phenomenon is known as resonance.

To demonstrate the phenomenon of resonance, we mount two identical tuning forks, say, A and B, of the same frequency upon two separate sound boxes with open ends; the boxes are kept to face each other. Now, a tuning fork say A is set into vibration, the other fork B also starts vibrating and a loud sound is heard. The vibrations produced in the second fork B are due to resonance.

Question 17. Explain the transverse vibrations in a stretched string.
Answer:

Transverse vibrations in a stretched string

If a string stretched between two fixed points is pulled and then released, then its transverse vibration gives rise to waves. These waves are reflected at the fixed ends of the string or wire. As a result stationary waves are formed.

The frequency of the note emitted depends on the following three factors :
(1) Length of the string
(2) The tension of string and
(3) The mass per unit length of the string.

Wbbse Physical Science And Environment Class 9 Solutions 

Question 18. What will be the velocity of sound wave when its wavelength and frequency are respectively 3m and 120 Hz? Establish the relation you used.
Answer: Relation between velocity (V), wavelength (λ) and frequency (n) of sound wave is.V =nλ). For the relation V= nλ

Given n = 120 cycle / sec, λ=3m.

So, V = 120/s x 3m = 360 ms.

Question 19. State three possible remedial measures that may be taken for minimising the harmful effects of noise pollution.
Answer: Measure to control noise pollution. To control noise pollution the following measures may be taken:

(1) Reduction of intensity level at the source of noise by setting up sound-absorptive arrangements between the source and receipient of noise

(2) Proper installation of sound-absorbing devices should be done in factories, generators, water pumps, etc.

(3) Enclosures of factories, hospitals, schools may be made of densely planted trees and creepers.

Question 20. Which characteristic of musical sound is a measure of shrillness of sound? How do we measure this characteristic in terms of some physical quantity?
Answer: Pitch is a measure of shrillness of sound. We measure pitch in terms of frequency. A source vibrating with higher frequency emits sound of higher pitch, i.e., the sound is sharp or shrill. If the source vibrates with less frequency, it emits a sound of lower pitch, i.e., the sound is flat. Pitch is a kind of feeling while frequency is a measurable physical quantity.

Question 21. Why is the sound of a sitar more pleasant to hear than the sound of a tuning fork, even they have same pitch and loudness?
Answer: Presence of overtones makes the sound of sitar pleasant. Sound emitted from a tuning fork does not contain any overtone. For this reason, the sound of a sitar is more pleasant to hear than the sound of tuning fork.

Question 22. The period of vibration of a body is 0.02 second. Will this produce audible sound?
Answer: In 0.02 second the body makes 1 vibration.

∴ In 1 second the body makes \(\frac{1}{0.02}\)vibrations.

∴ Frequency of the body is 50 Hz, so it will produce audible sound.

Wbbse Madhyamik Class 9 Physical Science And Environment 

Question 23. How is pleasantness of a note affected by the number of overtones present in it?
Answer: It is found that a note having a fewer number of overtones, not exceeding six, along with its fundamental tone is pleasing to hear, a note that has more than six overtones is harsh to the ear.

Question 24. State how(1) Traffic and transport in cities,(2) Factory activities are responsible for causing noise pollution.
Answer:

(1) Traffic and transport in cities, large number vehicles, many of which are with defective engine, ply speedily over bumpy roads to create noise pollution.

(2) Factories undeterred running machines in factories create noise pollution to workers and peoples of adjoining areas.

Question 25. How do domestic activity and activity of neighbours stand as sources of noise pollution?
Answer: Domestic and neighbouring activities-sound of generators, water pumps, television, stereo, CD players, etc. at high level of sound create noise pollution. Blaring loud speakers and microphones used in places of worship, politicians meetings are also sources of noise pollution.

Question 26. Why doesn’t the sound of explosion occurring in the sun reach the earth?
Answer: The continuous explosive sounds occurring on the sun because of conversion of hydrogen to helium by nuclear fusion is not heard from the earth. This is due to the presence of a large space beyond the earth’s atmosphere without any material medium; the sound of explosion cannot transmit to the earth since sound cannot propagate without the help of a material medium.

Wbbse Madhyamik Class 9 Physical Science And Environment 

Question 27. Tapping sound made from inside on the glass pane of a window of a closed room is audible to persons outside the room but not the voice of a person talking inside the same room. Explain.
Answer: The glass of a window is set into vibration due to tapping, this vibration sets in vibration the air layers outside, so the sound of tapping made on glass is heard from outside. On the other hand, the sound of a person’s voice inside the room is mostly reflectedfrom the walls, doors, windows of the room,

So major part of this sound remains inside the room, a very small part may go outside but that does not produce a fairly audible sound. Moreover, for a large difference of densities of air and glass, a sound made inside the room hardly travels outside.

Question 28. Why is it not advisable to regulate a watch on hearing the sound of a distantsiren?
Answer: The sound from the siren at a distance takes some time to reach a place. If the person adjusts his watch with the idea that the sound of the siren reaches him just at the time of its production, his watch will be slow by a few seconds that are taken by the sound to reach him.

Question 29. To hear a feeble sound coming from a distance clearly, we occasionally hold a palm in curved form behind the ear. Why ?
Answer: Although we do this by a reflex action, the sound waves coming from distance get reflected from the palm and concentrate at the ear, so a clearer sound of more intensity is received.

Question 30. Why is it that a speaker can deliver speech with more comfort in an auditorium than in an open space ?
Answer: This is because in an auditorium, the sound of speech gets reflected from ceilings, walls, etc. and reaches till the far end of the room. In an open space, there is no such reflector, the speaker feels discomfort, for, he has to shout louder in an attempt to make his speech reach the far end of a group of assembled listeners.

Question 31. Why are voices of females or babies more sharp but.of less intensity than the sound of that of a grown up male person ?
Answer: Vocal cords of females or babies are tender, so while they speak or sing, the cords vibrate with high frequency, that is why their voices are very sharp. Vocal cords of male persons are usually stiff, so their frequency of vibration is low and hence male voices are generally flat.

Question 32. Why is the humming sound of mosquitoes more sharp but of less loudness than the sound of roar of a tiger ?
Answer: Frequency of vibration of the wings of a flying mosquito is very high so it produces a sharp sound while they fly near our ears. Loudness of the sound is very low because of small sized vibrating wings. When a tiger roars, the frequency of vibration of its vocal cord is low so the sound produced is flat but due to large size of the tiger’s vocal cord the emitted sound is of greater loudness.

Wbbse Madhyamik Class 9 Physical Science And Environment 

Question 33. If the velocity of sound in air at 0°C be 332 m/s, what is the velocity of sound in air at 27°C?
Answer: If the velocity of sound in air at 27°C be V /ms, then,

V\(=332 \sqrt{\frac{273+27}{273}}\)= 348m/s.

Question 34. A brass pot produces sound on striking but it stops on touching – Explain why.
Answer: Brass pot vibrates on striking. So it produces sound. The vibration stops when touched by hand, and hence, there will be no sound.

Question 35. What is persistence of hearing?
Answer: Time period for which the sensation of sound lasts in the ear is called persistence of hearing.

Question 36. Can sound travel through vacuum ?
Answer:

Explanation : For the propagation of sound, a material medium (solid, liquid gas) is necessary. Sound cannot travel through vacuum.

Question 37.If a person talks or even screams on the surface of the moon, the sound will not be audible. Explain why.
Answer:

Explanation : The moon has no atmosphere or gas surrounding. There is absolutely a vacuum all around the moon. That is why if a person talks or even screams on the surface of the moon, the sound will not be audible.

Question 38. What do you mean by wavelength?
Answer:

Wavelength :
(1) It is the distance a wave or an energy travels through during one complete oscillation of its source.
(2) It is the distance between two consecutive crests or two consecutive troughs in a wave. It is normally represented by λ . Its SI unit is metre (m).

Question 39. What is transverse wave?
Answer:

Transverse wave: Transverse wave motion is that wave motion in which the individual particles of the medium execute vibrations about their mean. positions in a direction perpendicular to the direction of motion of the mean wave.

Wbbse Madhyamik Class 9 Physical Science And Environment 

Question 40. What is the definition of sound?
Answer:

Sound: Sound is the energy, which is generated from a vibrating body, travels through an elastic medium and finally reaches our ear to create a special sensation in the brain.

Question 41. How is sound produced?
Answer: The material bodies capable of producing sound are called the sources of the sound or sounding bodies. Sound is produced due to vibrations of the sounding bodies.

Question 42. What is longitudinal wave?
Answer:

Longitudinal wave: Longitudinal wave motion is that wave motion in which the individual particles of the medium execute vibrations about their mean positions along the direction of motion of the wave.

Question 43. What is called frequency?
Answer:

Frequency: The number of complete oscillations of a source of sound per second is known as its frequency. Unit of frequency is Hertz, its symbol is Hz.

Question 44. What is called oscillation?
Answer:

Oscillation: An oscillation of a particle is its trajectory in which the particle starting from a certain position moves through some distance, and retracing the path it returns to its starting point with the same direction of motion with which it started initially.

Question 45. What do you mean by amplitude? 
Answer:

Amplitude: The maximum displacement of a particle in a medium on either side oi its mean position is called amplitude.

Question 46. Define wave.
Answer:

Wave: It is defined as a disturbance that moves through a medium transferring energy from a point to another without any physical transportation of the material between the points.

Question 47. What is called wave velocity?
Answer:

Wave velocity: In a medium, the distance travelled by the wave in one second is defined as the velocity of propagation of the wave or wave velocity. It is normally represented by V.

Wbbse Madhyamik Class 9 Physical Science And Environment 

Question 48. What is meant by reflection of sound?
Answer:

Reflection of sound : When sound waves hit on the boundary separating twe homogeneous media, a portion of sound changes it direction from the surface of sepaeration and returns to the first medium. This is known as reflection of sound.

Question 49. Extablish the relation connecting wave velocity, wavelength and frequency.
Answer:

Relation among wave velocity, wavelength and frequency : Let the frequency and wavelength of a wave be n and J respectively.

According to the definition :
‘n’ numbers of complete waves are generated in 1 sec. Since wavelength is A, so in 1 sec, wave advances by n x λ = nλ distance.

Now, the distance traversed by the wave in 1 sec is its velocity (V).

So,V=n

Wave velocity = Frequency x Wavelength.

Question 50. What do you mean by Free or Natural vibration?
Answer:

Free or Natural vibration: When a body vibrates with its own natural frequency, is said to execute natural or free vibrations. The frequency of free vibration depends on the density, shape and elasticity of the body.

Example: The frequency of a tuning fork depends on the length and cross-section of its arms as well as the density and elasticity of the material of the tuning fork.

Question 51. What is called forced vibration?
Answer:

Forced. vibration: When a body is maintained in a state of vibration by a strong periodic force, the vibrations are called forced vibrations.

Example: Intensity of sound generated from a tuning fork is not very high, and hence, can not be heard from a distance. Now, when this tuning fork is touched on a table, it causes forced vibration on the table and thereby sound is generated from the table also. Intensity of this sound is relatively higher and hence can be heard from a distant location.

Question 52. What is called resonant vibration?
Answer:

Resonant vibration: It is a particular case of forced vibration when the frequency of the driving force equals the natural frequency of the body. When such condition is fulfilled, amplitude of oscillation becomes very high.

Example: A vibration is generated in the air column when a vibrating tuning fork is placed on the mouth of a pipe filled with air. Air being forced vibrated creates this sound.

Question 53. State the laws of reflection of sound.
Answer:

Laws of reflection of sound :
(1) The incident sound, reflected sound and the normal to the surface of separation through the point of incidence like on the same plane.

(2) The angle of incidence is equal to the angle of reflection.

Question 54. Is pitch synonimous with frequency? Or, Mention the differences between frequency and pitch of a sound wave.
Answer: Pitch is not the same as frequency. The pitch refers to sensation as perceived by the listener, and hence, is subjective. Frequency, on the other hand, is a measurable quantity, and hence, is an objective quantity. Actually, pitch depends on frequency. If frequency is the cause, pitch is its effect.

Question 55. In what state does a body generate sound? In which of the media, solid or liquid, sound travels faster?
Answer: A body in a state of vibration generates sound. Sound travels faster in solid than in liquid.

Wbbse Madhyamik Class 9 Physical Science And Environment 

Question 56. Why are two sounds heard at short intervals at one end of a fong hollow iron pipe when its other end is struck lightly with a solid body ?
Answer: When a long hollow iron pipe is struck at one end, the produced sound travels through air as well as the material medium of the pipe. Since, sound travels faster through solid than through air, two sounds are heard at the other end of the pipe.

Question 57. Deduce the minimum distance of the reflector that can produce echo of a transient sound.
Answer:

Minimum distance of the reflector for listening the echo of transient sound :

The feel of a transient sound that remains in our. brain is\(\frac{1}{10}\)see after it reaches our ear.

This period is known as persistence of hearing. If a second sound reaches our brain within

This \(\frac{1}{10}\)see of time, the latter one can not be registered by the brain separately. Obviously

A gap of at least \(\frac{1}{10}\) sec has to be there between the original sound and reflected sound so that the brain can recognize them separately.

Now, if the velocity of sound in air be 332 m/sec, in \(\frac{1}{10}\) see it travels through 33.2 m. For producing an echo, the total length of the path traversed by a sound from the source to the reflector and back should be at least 33.2 m. Hence for hearing echo of a transient sound, the minimum distance of the reflector from the source of sound is half of 33.2 m, i.e., 16.6.m.

Question 58. Give two applications of echo.
Answer:

Application of echo :
(1) Determination of velocity of sound in air with the help of a stopwatch and a pistol.
(2) Depth of sea, height of an aeroplane above ground may also be determined with the help of echo.

Question 59. What is called musical sound?
Answer:

Musical sound : A continuous sound pleasing to the ear and produced by regular and periodic vibrations is known as musical sound.

Example: Sounds generated from different musical instruments like violin, piano, flute, etc. are musical sounds.

Question 60. What is noise?
Answer:

Noise: A noise is a sound which is discontinuous, unpleasant to ear and produced by irregular and non-periodic vibrations

Example: The sounds of moving cars, trains, machines, sound of explosions, etc. are the examples of noise. a

Wb Class 9 Physical Science Question and Answers

Question 61. What do you mean by tone?
Answer:

Tone : A musical sound generally consists of number of components having different frequencies. Such a sound is known as note.

Question 62. What is fundamental tone and what is overtone?
Answer:

Fundamental tone: In a note the tone of the lowest frequency is known as fundamental tone.

Overtone: All tones, except fundamental tone, present in a note are known as overtones.

Question 63. What is the intensity of a musical sound?
Answer:

Intensity : It means how forcefully or how strongly the sound reaches a listenerIntensity or loudness of a sound depends on the energy contained per unit volume of themedium through which sound passes.

Question 64. What are the factors which affect the intensity of a musical sound?
Answer:

The intensity of a musical sound depends upon the following factors :
(1) Amplitude of the vibration of the source,
(2) Size of the sounding body,
(3 Distance of the observer from the source,
(4) Presence of other bodies
(5) Density of the medium.

Question 65. What is pitch?
Answer:

Pitch: The characteristic of a musical sound by which a sharp or a shrill sound is distinguished from a dull, flat sound of the same intensity is known as its pitch. Pitch of a note depends Pitch of a note depends on its frequency. Greater the frequency higher will be the pitch. Wavelength being inversely proportional to frequency, the sound of low wavelength has higher pitch and vice-versa.

Question 66. What is quality of a musical sound?
Answer:

Quality of a musical sound: It is that characteristic which enables us to distinguish one sound from another even though the two sounds may have the same intensity or pitch. Richer the quality, more pleasant is the sound.

Example: If harmonium and violin are played simultaneously, one can easily distinguish between these two even without seeing. This distinction is felt out of the property called quality of sound.

Wb Class 9 Physical Science Question and Answers

Question 67. What are the sources of sound pollution?
Answer:

Sources of sound pollution :
(1) The sound of vehicle causes sound pollution.
(2) The machinery used in industry creates sound of intensity more than tolerance limit and thereby causes sound pollution.
(3) Loudspeakers, amplifiers, air horn played at top volume, are major sources of sound pollution.

Question 68. What are the harmful effects of noise pollution?
Answer:

Harmful effects of noise pollution :
(1) Sound pollution may cause hindrance in the growth of nervous system of a child in the womb.
(2) One may become deaf due to sound pollution.
(3) The sound of high intensity and pitch may cause blood pressure, nervous break down, heart disease, loss of memory, loss of concentration, etc.

Question 69. Why are the voices of females or babies more sharp than that of a grown up male person?
Answer:

Explanation: Vocal cords of females or babies are tender, so, while they speak or sing the cords vibrate with high frequency, for this reason their voices are very sharp. Vocal cords of male persons are usually stiff, so their frequency of vibration is low and hence male voices are generally flat.

Question 70. Why doesn’t the sound of explosion occuring in the sun reach the earth?
Answer:

Explanation : The continuous explosive sounds occuring on the sun because of conversion of hydrogen to helium by nuclear fusion are not heard from the earth. This is due to presence of a large space beyond the earth’s atmosphere without any material
medium. The sound of explosion cannot transmit to the earth since sound cannot propa- gate without the help of a material medium.

Question 71. What are called match numbers?
Answer:

Mach number: The ratio of speed of a body to the speed of sound is called the match number of the body. If the match number of a body be greater than 1, then the velocity of the body is supersonic velocity.

Question 72. How does speaking tube operate?
Answer:

Speaking tube : If we talk at one end of a long tube, it can be heard clearly on the other end. The sound generated at one end does not get scattered. On the contrary, it undergoes several reflections within the tube and finally reaches the other end.

Wb Class 9 Physical Science Question and Answers

Question 73. How does stethoscope function?
Answer:

Stethoscope: Doctors use this gadget for chest examination. This is basically a reflecting tube. This gadget consists of two rubber tubes with a metallic plate at the juncture of the tube. The sound of the heart beat reaches from the metallic plate to the ears of the doctor through multiple reflection of the sound within the rubber tubes.

Question 74. What are harmonics and what is an octave?
Answer:

Harmonics: Those overtones whose frequencies are integral multiples of the frequency of the fundamental tone are called Harmonics.

Octave: The particular harmonic whose frequency is double the fundamental frequency is known as the second harmonic or octave of the fundamental tone.

Question 75. Velocity of sound in air is 350 m/sec. Frequency of a tunning fork is 700/sec. Find the wavelength of the sound generated by the tunning fork.
Answer:

We know, n = 700 sec

v=nλ ,  λ  =?

or,λ  =\(\frac{v}{n}\)      v=350m/sec

⇒ \(=\frac{350}{700}\)

=0.5m.

Question 76. The frequency of a tuning fork is 440. If the wavelength of the wave is 0.73 m then find the velocity of sound.
Answer:

We know,  n= 440

v=nλ,  λ=0.73m

or,v=440×0.73 ,  v=?

= 321.2 m/sec.

Wb Class 9 Physical Science Question and Answers

Question 77. A boat approaches a high vertical cliff. When the anchor is dropped, an echo is heard 2 seconds later. How far is the boat from the cliff? (velocity of sound = 332 m/sec)
Answer:

As the velocity of sound is 332 msec, so total distance travelled by sound in 2 sec = 2 x 332 = 664 m.

For an echo, the sound must go to the cliff and return through the same distance.

So the distance of the cliff \(=\frac{664}{2}\) m  =332 m.

Question 78. In measuring the depth of an ocean, it was found that the sound of an explosion and its echo by the bed had a time interval of 20 sec between them. What is the depth of the ocean? (velocity of sound in water = 1436 m/sec)
Answer:

As the velocity of sound in water = 1436 m/sec, so total distance travelled by the sound in 20 sec = 20 x 1436 m.

∴ The depth of the ocean \(=\frac{20 \times 1436}{2}\)m

=14360m.

Question 79. Find the frequency of a tuning fork whose wavelength is 1.7 m. (velocity of sound in air = 340 m/sec)
Answer:

We know,  v = 340 m/sec

v=nλ,   λ =1.7m

or,n= \(\frac{v}{λ }\)     n=?

⇒\(=\frac{340}{1.7}\)

=200 Hz.

∴ Frequency of the tuning fork = 200 Hz. 

Question 80. A radio station transmits waves of wavelength 100 m. If the speed of the waves is 3 x 108 m/s, find the frequency of the radio station.
Answer:

Here, wavelength of the wave (λ) = 100 m.

Wave velocity (v) = 3 x \(10^8\) m/s

Using v = nλ, we get   n = \(\frac{v}{λ }\) =\(=\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{100 \mathrm{~m}}\)

=3 x \(10^6\)

∴ Frequency of the radio station = 3 x \(10^6\)Hz = 3 MHz.

Question 81. The water ripples are produced at a frequency of 50 Hz. If the wavelength of these ripples is 2.4 cm, calculate the speed of the ripples.
Answer:

Here, frequency (n) = 50 Hz.

Wavelength (λ) = 2.4 cm = 0.024 m.

Using the relation, v= nλ, we get, v= 50 Hz x 0.024m = 1.2 m/s.

∴ The speed of the ripples = 1.2 m/s.

Wb Class 9 Physical Science Question and Answers

Question 82. A wave pulse on a string advances with a speed of 16 m/s. What would be the wavelength of the wave on the string if its frequency is 200 Hz ?
Answer:

Here, wave velocity (v) = 16 m/s and frequency (n) = 200 Hz.

Using the relation, v = nλ, we get, λ\(\frac{v}{n}\)\(=\frac{16 \mathrm{~m} / \mathrm{s}}{200 \mathrm{~Hz}}\)

=0.08m.

∴  Wavelength of the wave = 0.08 m.

Question 83. If the frequency of a tuning fork be 400 Hz and the velocity of sound in air be 320 ms, find how far sound travels when the fork completes 30 vibrations.
Answer:

Here, frequency (n) = 400 Hz.

Velocity of sound (v) = 320 m/s

Using the relation v = nλ, we get, Wavelength, λ =\(\frac{v}{n}\)  =\(=\frac{320}{400}\) m=0.8m.

∴ In 1 vibration of the fork sound travels 0.8 m.

Hence, in 30 complete vibrations sound travels (0.8 x 30)m = 24 m.

Question 84. The period of vibration of a body is 0.02 second. Will this Produce audible sound?
Answer:

In 0.02 second the body makes 1 vibration.

∴ In 1 second the body makes\(\frac{1}{0.02}\) or 50  vibrations.

∴  Frequency of the body is 50 Hz, producing audible sound.

Question 85. A sound wave has frequency 2 kHz and wavelength 40 cm. Calculate the speed of the wave.
Answer:

Given, n = 2kHz =2 4\(10^3\) Hz;  /= 40 cm = 0.40 m.

∴ v=nλ=2 ×\(10^3\) ×0.4.0=800m\(s^{-1}\)

Question 86. Calculate the time taken by a sound wave of frequency 100 Hz and wavelength 50 cm to travel a distance of 500 m.
Answer:

Given, n = 100 Hz;  f= 90 cm = 0.5m; S = 500 m
V=nλ= = 1000 x 0.5 = 500 m\(s^{-1}\)

∴ t\(=\frac{s}{v}\)=\(\frac{500}{500}\)=1s.

Question 87. A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in a tissue in which the speed of sound is 1.7 kms\(s^{-1}\)? The Operating frequency of the scanner is 4.2 MHz. (1 MHz = 10°)
Answer:

Given, v = 1.7 km\(s^{-1}\) = 1700 m\(s^{-1}\)    n= 4.2 MHz = 4.2 x 106 Hz.
Since, v = nλ

∴ λ\(=\frac{v}{n}\)\(=\frac{1700}{4.2 \times 10^6}\)

=4×\(10^{-4}\) m=0.4mm.

Question 88. Calculate the minimum distance between the listener and the reflector for hearing distinctly the echo of an inarticulate sound, if the velocity of the sound be 340 m/s.
Answer:

Let the minimum distance of the reflector from the listener be x meter. Thus sound would travel 2x metres. As the persistence of hearing is 1/10s, so it is the time required to travel 2x metres.

Now,V\(=\frac{2 x}{t}\)

or,x\(=\frac{v t}{2}\)=\(\frac{340 \times 1}{2 \times 10}\)=17m.

Question 89. In an experiment for measuring the depth of sea, sound of explosion reflected from sea bed is heard after 12 s. Find the depth of the sea at that place. Velocity of sound in water = 1500 m/s.
Answer:

Depth of sea = ½ x 1500 x 12 = 9000 m = 9km.

Chapter 7 Sound 3 Marks Questions And Answers:

Question 1. Deseribe the basic mechanics of hearing.
Answer:

Basic mechanics of hearing

Hearing is the process by which humans use their ears to detect and perceive sounds. Ears are important for hearing and for controlling a sense of position and balance.

Each ear is divided into three sections:
1. The outer ear
2. The middle ear and
3. The inner ear.

The middle and inner parts of the ear are located in hollow spaces on either side of the head within the temporal bones of the skull.
To hear sound, the ear has to do three basic things:
(1)Direct the sound waves into the hearing part of the ear.
(2)Sense the fluctuations in air pressure.

Translate these fluctuations into an electrical signal that the brain can understand.

Question 2. Describe the structure and functions of the ear.

Answer:

Each human ear is divided into three sections:

The outer ear or external ear :

The external part ofthe ear consists of the pinna and ear lobe. The pinna or ear shell is the shell-like part of the external ear and it is made of cartilage and skin. The pinna directs sound waves from the outside into the external auditory canal (ear canal), which in turn channels sound waves to the tympanic membrane (known as the eardrum), causing it to vibrate.

The tympanic membrane is a thin, semi-transparent, flexible membrane that separates the outer and middle ear. The outer ear functions to collect sound (acoustic energy) and funnel it to the eardrum.

The middle ear :
The middle ear is an air-filled space that contains three tiny bones known as ossicles which transmit sound. The bones are known individually (according to their shapes) as the
1. Malleus (hammer)
2. Incus (anvil)
3. Stapes (stirrup).

Sound waves that reach the tympanic membrane cause it to vibrate. In turn, the eardrum sets into motion the first ear bone, which transmits the motion to the second bone (the incus). Finally, the third bone (the stapes) works like a piston to amplify and transform the sound energy into mechanical energy.

This mechanical energy is then transmitted from the stapes to the hearing part (cochlea) of the inner ear via the oval window (a thin membrane between the middle and inner ear). The middle ear is connected to the back of the nose (nasopharynx) by the Eustachian tube.The Eustachian tube is a narrow tube that connects the middle ear to the back of the
nose and throat.

The inner ear: The delicate membranous inner ear (labyrinth) is enclosed and protected by a bony chamber that is referred to as the bony labyrinth.

The inner ear contains two main structures :
(1)The cochlea: In the shape of a snail, which is involved in hearing. The round window (fenestra cochlea) is a membrane that connects the cochlea to the middle ear. It helps to dampen the vibrations in the cochlea.

(2)The vestibular system: (consisting of the semicircular canals, saccule and utricle), which is responsible for maintaining balance and a sense of position.

Question 3. What are the sources of Noise Pollution?
Answer:

Sources of Noise Pollution: Noise pollution like other pollutions is also a by product of industrialization, urbanizations and modern civilization.

The main sources of noise pollution are:

1. Road Traffic Noise: The main sources of traffic noise are the motors and exhaust system of autos, smaller trucks, buses and motorcycles. This type of noise can be aug mented by narrow streets and tall buildings, which produce a canyon in which traffic noise reverberates.

2. Air Craft Noise: The low flying military and other air cratts.

3. Noise from railroads: The noise from locomotive engines, horns and whistles, and switching and shunting operation in rail yards can affect neighboring communities and railroad workers.

4. Construction Noise: The noise from the construction of highways, city streets and buildings is a major contributor to the urban scene.
5. Noise in Industry
6. Noise in building
7. Noise from consumer products
8. Fire crackers.
9. Loud speakers in functions

Wb Class 9 Physical Science Question and Answers

Question 4. What are the harmful effects of noise pollution?
Answer: Noise pollution makes men more irritable. The effect of noise pollution is multifaceted & interrelated.

The effects of noise pollution on human beings, animals and property are as follows :
1. It decreases the efficiency of a man: Regarding tne impact of noise on human efficiency there are a number of experiments which print out the fact that human efficiency increases with noise reduction.

2. Lack of concentration: For better quality of work there should be concentration. Noise causes lack of concentration.

3. Fatigue:  Because of noise pollution, people cannot concentrate on their work. Thus they have to give their more time for completing the work and they feel tired.

4. Abortion is caused: There should be cool and calm atmosphere during the pregnancy. Unpleasant sounds make a lady irritative in nature.

5. It causes Blood Pressure: Noise pollution causes certain diseases in human. It attacks on the persons’ peace of mind. The noises are recognized as major contributing factors in accelerating the already existing tensions of modern living. These tensions result in certain diseases like blood pressure or mental illness, etc.

6. Temporary or Permanent Deafness: The effect of nose on audition is well recognized in mechanics, locomotive drivers, telephone operators, etc. All have their hearing impairment as a result of noise at the place of work.

7. Poor quality of Crops: Now it is well known to all that plants are similar to human beings. They are also as sensitive as man. There should be cool & peaceful environment for their better growth. Noise pollution causes poor quality of crops in a pleasant atmosphere.

8. Effect on Animals: Noise pollution damages the nervous system of animal. Animals fooses the control of its mind. They become dangerous.

9. Effect on Buildings: Loud noise is very dangerous to buildings, bridges and monuments. It creates waves which struck the walls and put the building in danger condition. It weakens the edifice of buildings.

Question 5. Show that sound is produced by vibration.
Answer: Only a vibrating body can produce sound. It can be shown by the following experiment. A tuning fork fitted on a hollow rectangular box is taken. Tuning fork is a U-shaped bar with a stem in the middle. A pith ball is suspended from a stand in such a way that the ball just touches one of the arms (or prong) of the tuning fork. Now struck the tuning fork with a rubber hammer, the vibration in the arm is produced with sound fork and the pith ball jumps to and fro. The pith ball vibrates

As long as the sound is produced by the tuning fork. When the prong of tuning fork stops vibrating, no sound ‘ is heard. The pith ball also stops to vibrate. This shows that sound is produced by vibration.

Wb Class 9 Physical Science Question and Answers

Question 6. What are mechanical waves? Explain the kinds of mechanical waves.
Answer: The waves which require material media for their transmission are the elastic waves or the mechanical waves.

For example vibration in a rope or stretched string, vibration on the surface of water, sound waves in air, etc.
Mechanical waves are of two kinds :
(1) The transverse waves and
(2) The longitudinal waves.

(1) Represents longitudinal waves and
(2) Represents transverse waves.

Question 7. What is transverse wave? Explain with examples.
Answer:

Transverse wave : When each particle of the medium executes vibration in the direction perpendicular to the direction of the propagation of the wave, then it is called Transverse wave.

For example waves on the surface of water, vibrations in a string.

Question 8. What is longitudinal wave? Explain with example.
Answer:

Longitudinal wave: When each particle of the medium executes vibration about its mean position in the direction of propagation of the wave, then it is called transverse wave.

For example Sound wave travelling in air, souind wave inside water, etc. The vibrations of the particles in air medium due to propagation of longitudinal wave form compression and rarefaction in air. In the give figure, C represents the position of  compression and R represents the position of rarefaction by a vibrating body (or tuning fork).

Question 9. State the characteristics of wave motion.
Answer:

The characteristics of a wave motion are the following :
(1) The wave is caused due to periodic disturbance and the wave itself is periodic.
(2) When a wave is transmitted in a medium, energy is transferred without actual physical transfer of matter.
(3) The energy transfer in the medium takes place with a constant speed which depends on the nature of the medium.
(4) If more than one waves are travelling through a medium, then they move independently causing interference of waves, reflection of waves, etc.

Wb Class 9 Physical Science Question and Answers

Question 10. State the uses of ultrasonic sound.
Answer:

The uses of ultrasonic sound are :
(1) Ultrasonic sound is used for detecting flaws in metal casting or automobile tires.
(2) It is used in welding two pieces of metals together with help of an ultrasonic vibrator.
(3) tis used in cleaning crockery with the help of cleansing liquid and ultrasonic vibrations.
(4) It is used in controlling smoke and clearing fog.
(5) It is used in certain diseases and also to kill the bacteria in the milk.

Wb Class 9 Physical Science Question and Answers

Question 11. What is infrasonic or subsonic sound? State its uses.
Answer:

Infrasonic or subsonic sound

Sounds of frequency lower than 20 Hz are called infrasonic or subsonic. We cannot hear the infrasonic sound. The speed of such sound is same as that of audible sound.

Uses : The infrasonic sound is used in drilling deepest wells because the low frequency vibrations break up hard rock more easily than the ordinary drills.

Question 12. What should be the minimum distance for the echo to be heard?
Answer: The sensation of sound persists in our ear for about 0.1 second after the exciting stimulus ceases to act. Hence the echo is heard ‘distinctly if it reaches the ear at least 0.1 second later than the original sound is heard.

If dis the distance between the observer and the obstacle and V is the speed of sound, then the distance covered by the sound to reach the obstacle and then come back is 2d. Thus the time taken is

t=2d/V   or,d=Vt/2.

By putting t = 0.1 second, V = 332 m/sec at ordinary temperature,

d= (332×0.1)/2 = 16.6 m (or nearly 17 m).

Hence, to hear the echo of a sound distinctly, the reflecting surface should be at a minimum distance of 16.6 m (or 17 m nearly) from the observer.

Question 13. A man fired a gun standing between two parallel cliffs. He bate two successive echoes after 3 seconds and 5 seconds respectively. What is the distance between the two hills ? (The velocity of sound in air = 330 m/sec).
Answer: Let the distance between the man and first hill be x m and the distance between the man and second hill be y m.

Hence, distance between two hills = (x + y) m.

Time taken to hear the echo in first case is 5 sec.

Now, v= 330

or, x= m=495m

Similarly, y = m = 825m

Therefore, distance between the hills = (x + y) m

= 495 m + 825 m = 1320 m.

Wb Class 9 Physical Science Question and Answers

Question 14. What do you mean by natural vibration or free vibration?
Answer: Whenever a body capable of vibration is displaced slightly from its equilibrium position and then left to itself, the body begins to vibrate freely in its own natural way called free vibration or natural vibration of the body. The amplitude and frequency remain constant.

Free or natural vibration: The vibration of a body with constant amplitude and constant frequency are called free vibrations or natural vibrations.
For example :
(1) when a tuning fork is struck against a hard rubber pad, it vibrates with natural frequency.
(2) When the bob of a simple pendulum is displaced slightly from its mean position, it starts vibrating with its natural frequency.

Question 15. Explain musical sound and noise.
Answer:
All sounds which produce sensation of hearing can be roughly divided into two kind :
(1) Musical sound and
(2) Noise.

(1) Musical sound: Musical sound is a pleasant, continuous and uniform sound produced by the periodic and regular vibrations.
For example The sound produced by Piano, Flute, Violin, etc.(2) Noise: Noise is an unpleasant sound produced by an irregular succession of disturbance and discontinuous sound.
For example, The sound produced by large crowd, the horn of a bus, the sound of a hammer on an iron sheet, etc.

Question 16. State the characteristics of a musical sound.
Answer:

There are three main characteristics of musical sound :
(1) Pitch or shrillness
(2) Loudness and
(3) Duality or timber.

(1) Pitch or Shrillness: It is that characteristic of sound by which an acute or shrill note can be distinguished from a grave or flat note. If the pitch is higher, the sound is said to be shrill and if the pitch is low, the sound is flat.

Pitch of a note depends on its wavelength or frequency.

(2) Loudness: It is the property by virtue of which a loud sound can be distinguished from a faint one, both having the same pitch.

Loudness depends on the intensity or amplitude of wave. The loudness of sound de- pends on the following factors :
(1) Loudness is proportional to the square of the amplitude.
(2) Loudness varies inversely as the square of the distance.
(3) Loudness depends on the surface area of the vibrating body.
(4) Loudness depends on the density of the medium.
(5) Loudness depends on the presence of resonant.bodies.

(3) Quality or timber: Quality or timber of a musical sound is that characteristic that distinguishes two sounds of the same loudness and same pitch but emitted by two different instruments. The quality of a musical sound depends on the wave form.

Wb Class 9 Physical Science Question and Answers

Question 17. Define fundamental note and overtones.
Answer:

Fundamental note and overtones

When the stretched string of an instrument is plucked or struck, the string vibrates and the transverse waves are produced in the string. When a string is plucked at the middle, the string vibrates in one loop.

When the string is plucked at one-fourth part ns Bo ae of the string it vibrates in two loops then it is 2nd Harmonic called first overtone or first subsidiary vibration. The frequency of this vibration is greater that of fundamental note, when the string is i ae
plucked at a distance \(\frac{1}{6}\) of the length from End, it vibrates in three loops. This is called second overtone or second subsidiary vibration. The frequency of this vibration is three times that of the fundamental tone.

Question 18. Describe a sonometer.
Answer:

Sonometer

A sonometer is a hollow wooden tension. One end of the wire is fixed to a rigid peg, while the other end after passing over a pulley carries a pan. By placing weights on this pan desired tension can be applied to the string. The string rests on two movable wooden triangular bridges fixed with metal plates and placed on the sonometer box.

As the box is hollow when the string vibrates, the air inside the box also vibrates along with it and the intensity of sound produced increases. By changing the positions of the bridges the vibrating length of the wire can be changed.

The frequency of sound produced by the vibrations of the string depends on:
(1) The length of the wire
(2) The tension of the wire and
(3) The mass pér unit length of the wire.

By keeping tension constant of a particular string, sounds of different frequencies can be produced by changing the vibrating length of the wire. It could be seen that lesser the length of the wire (i.e., distance between the bridges), higher is the frequency of the sound produced.

West Bengal Board Class 9 Physical Science Solutions

Question 19. Write down Newton’s formula for velocity of sound in air and then state and explain Laplace’s correction.
Answer:

According to Newton, the formula for the velocity of sound in a gaseous medium,

V\(=\sqrt{\frac{P}{D}}\)

Where V is the velocity of sound in the gaseous medium and P and D are the pressure and density of the medium respectively. In establishing this relation Newton assumed that the process of movement of sound through a gaseous medium was a slow process and in the process there was no change in the temperature of the medium. Thus the process was under isothermal (constant temperature) condition.

Now, at N.T.P.

P= 76 cm of Hg pressure

= 76 x 13.6 x 980 dyne/\(\mathrm{cm}^2\)

D= 0.001293 g/\(\mathrm{cm}^2\)

∴ V\(=\sqrt{\frac{76 \times 13.6 \times 980}{0.001293}}\)

= 28000cm/s= 280 m/s

The calculated value of the velocity of sound in air at NTP as calculated following Newton’s formula was found to be much less than the experimentally obtained value of 332 m/s at NTP. Laplace corrected this formula and state that in gaseous medium its temperature does not remain contant. This corrected formula is

V=V\(=\sqrt{\frac{Y P}{D}}\)
where Y is a constant and its value in the case of air is 1.41. According to this formula,

velocity of sound in air at NTP is 280 x \(\sqrt{1.41}\) = 332.5 m/s. Laplace’s value for velocity of

sound in air is thus in good agreement with the experimental values.

Question 20. What is vibration or oscillation? What is a tuning fork?
Answer:

(1) Vibration: When a body moves back and forth repeatedy about a mean position, its motion is called oscillation or vibration. If a system is disturbed from its mean position and left to itself, it vibrates about its mean position.

(2)Tuning fork: The tuning fork is a U-shaped steel bar free at both the ends with a steam or hand at the middle. The two arms of the V-shaped rod are known as the prongs. If any of the prongs is struck with a hammer wrapped with cloth, the prong starts vibrating in a periodic manner. Vibrations produce longitudinal waves in air which when reach our ears, sound is heard.

Question 21. Demonstrate by a simple experiment that sound waves produced in a medium do not transfer the particles. of the medium from one place to another.
Answer:

That during propagation of sound waves there is no displacement of material medium, can be shown by the following simple experiment. One cylindrical shaped metal tube of about one metre long and 10 cm diameter and tappered at one end is filled with white smoke by burning turpentine-soaked paper or cloth.

The flame of a burning candle is placed at the open mouth of the tappered end of horizontally mounted tube. Now, a sound is produced at the other open end of the tube by striking two pices of wooden blocks. The candle flame is found to move away from the mouth of tube and no smoke comes out. This shows that sound waves during propagation transfer energy but no particles of medium.

West Bengal Board Class 9 Physical Science Solutions

Question 22. State the condition for resonance. How does natural vibration differ from forced vibration?
Answer:

Conditions for Resonance: When the frequency of the applied periodic force is either equal to or is an integral multiple of the natural frequency of the vibrating body, resonance occurs.

Thus, if the periodic force of frequency n is applied on a body vibrating with its natural frequency, resonance would occur only when n =f, 2f, 3f …. and the body starts vibrating with an increased amplitude.

Difference between natural vibration and forced vibration: In absence of any external force, the frequency of natural vibration of a body depends on the mass, shape and elasticity of the body. On the other hand, in forced vibration a vibrating body acauires the frequency of the applied force but the magnitude of amplitude remains constant. Natural vibration dies down after a while due to frictional force. But the forced vibration ontinues so long as the applied external external periodic force remains in force.

Question 23. Mention the differences between musical sound and noise.
Answer:

Differences between musical sound and noise : There is no sharp line of demarcation between a musical sound and a noise. The same sound which seems pleasant to one person may appear to be unpleasant to. another person under different conditions.
The following are the main differences between musical sound and noise :

Musical sound	Noise
1.    It is pleasant and smooth to hear, and produces pleasant sensation in the brain.	1.    It is harsh, discordant and displeasing to hear and produces unpleasant sensation in the brain.
2.    It is produced by the vibrations which are regular, periodic and continuous.	2.    It is produced by an irregular succession of disturbances which are irregular and non-periodic.
3.    There is no sudden change in intensity and loudness of musical sound	3.    There is a sudden change in intensity and loudness of noise.

 

Question 24. What is noise pollution? Mention two of its hazardous effects on public health.
Answer: Noise pollution means creation of discomfort, disturbance and irritation which result to ill effects to mental and physical health.

Harmful effects of noise pollution:
(1) Noise level a little above 85 dB damages our physiological processes. That of 130-140 dB may deafen us and noise of 160 dB may destroy our eardrum permanently.

(2) Noise pollution causes anger, mental depression, loss of concentration that causes school-going pupils unmindful in their studies. Workers in factories and drivers of lorries working in an environment of high noise level are often found to be of rude and aggressive nature.

West Bengal Board Class 9 Physical Science Solutions

Question 25. Describe in brief an experiment which shows that a material medium is necessary for propagation of sound.
Answer:

Experiment to prove the necessity of a material medium for propagation of sound :

Experimental setup: A glass jar is placed in an airtight condition on a platform provided with a suction pump. An electric bell kept suspended inside the jar can be operated from outside with a battery.

Experiment: The electric bell is set into operation. Air from inside the jar is gradually drawn out.

Observation: A sound of the bell is heard when the jar is full of air but the sound gradually fades away on withdrawal of air. The sound is again heard clearly when air is reintroduced.

Inference: Since sound from the electric bell is not heard in absence of air within the jar, it proves that sound from a source cannot propagate in absence of a material medium.

Question 26. A pitcher under a tap of running water when gradually fills with water produces flat sounds at the beginning but in later stages the sound becomes sharper and sharper. Why?
Answer: lt may be remembered that the pitch of a sound increases when wavelength decreases and pitch decreases if wavelength increases. Here, at the beginning the sound wave formed due to vibration of the air contained in the pitcher spreads through almost the whole volume of the pitcher, so wavelength is sufficiently large.

Pitch of the sound is thus low, i.e., the sound is flat. As water gradually fills the pitcher, the air space in it slowly decreases, so the sound wave occupies smaller spaces for which its wavelength gradually shortens. The pitch of sound increases gradually, i.e., the sound becomes more and more sharp.

West Bengal Board Class 9 Physical Science Solutions

Question 27. A bat while flying makes an ultrasonic sound to detect-an obstacle in its path. The bat hears an echo when it has moved ‘d’ cm forward. If the velocity of the bat be \(V_B\)cm/s and that of the ultrasonic sound is \(V_s\) cm/s, how far is the obstacle from the bat’s initial position?
Answer: Let the initial distance of the bat from the reflector be x cm. The time in which the bat moves d cm, the sound travels the total distance (x + x – d) cm when the bat hears echo.

∴\(\frac{x+x-d}{V_S}\)=\(\frac{d}{V_B}\)  or(2x−d) \(V_B\) = \(V_s\)×d.

x=\(=\frac{d}{2 V_B}\)(\(V_s\) +\(V_B\))cm away from the reflector.

Question 28. The driver of an engine produced a whistle sound of the engine from a distance 800 m away a hill to which the engine was approaching. The driver heard the echo of the whistle after 4.5 seconds. If the velocity of sound is 340 m/s, find the speed of the engine.
Answer: Let the distance between the hill and the position of the engine where the driver received the echo be ‘y’ m. The total distance travelled by the sound for producjng echo = (800 + y) m. Since the echo was heard after 4.5 seconds, in 4.5 seconds sound travelled (340 x 4.5) m or 1530 m.

800 + y = 1530 or y = 730 m.

Now, the engine travelled through a distance (800 – 730) m or 70 before the reaching the spot of reception of echo. This distance of 70 m was covered by the engine in 4.5 s.

∴ Speed of the engine\(=\frac{70 m}{4.5}\)= 15.5 m/s = 55.8 km/hr.

West Bengal Board Class 9 Physical Science Solutions

Question 29. A man fired a gun standing somewhere between two hills. He heard one eaho after\(1 \frac{1}{2}\)  seconds and the other after \(2 \frac{1}{2}\)  seconds. Find the distance between the hills if the velocity of sound be 340 m/s.
Answer: The first echo was heard due to reflection of the sound of gunfire from the nearer hill. So, the distance of the man from this hill is half the distance that sound travelled in

\(1 \frac{1}{2}\) seconds and it is equal \( \frac{1}{2}\)\(\left(340 \times 1 \frac{1}{2}\right)\)m or 255 m. Similarly, the distance of the other

hill from the man that caused the second echo after \(2 \frac{1}{2}\)seconds is \( \frac{1}{2}\)\(\left(340 \times 1 \frac{1}{2}\right)\)  m or

425 m. So, the distance between the two hills = (255 + 425) m = 680 m.

Question 30. A man at a certain distance from a cliff receives the echo of a sound after ‘x’ second. He then approaches the cliff by ‘a’ metre and finds the time for echo to be ‘y’ seconds. Show that the velocity of sound is \(\frac{2 z}{x-y}\)m/s.
Answer: Let ‘V’ m/s be the velocity of sound and the initial distance between the man and the cliff be dm.

So, in ‘x’ seconds sound travels through Vx m for making echo and this is equal to 2d m.

∴ VX = 2d. or d=\(\frac{V x}{2}\)
When the man approaches ‘a’ m, the distance of the cliff from him is (d – a) m. So, for
echo the sound.travels Vy m and it is equal to 2 (d – a) m. BY

∴ 2 (d- a) = Vy. or \(2\left(\frac{V x}{2}-a\right)\) (putting the value of ‘d’ shown earlier)

Or, Vx – 2a=Vy  ∴V   \(=\frac{2 a}{x-y}\)So, velocity of sound is \(=\frac{2 a}{x-y}\) m/s.

West Bengal Board Class 9 Physical Science Solutions

Question 31. A body vibrating with a particular frequency generates waves of wavelength 0.2 m in a medium ‘A’ where velocity of the wave is 160 m/s. Calculate the wavelength of the wave it will generate in another medium ‘B’ in which its velocity is 240 m/s.
Answer: From the relation V = ni, where V = velocity of wave in the medium A, n =frequency , λ= wavelength,

160 =nx0.2

∴n= \(=\frac{160}{0.2}\) = 800 Hz.

Again, in the medium, B, V =n λ or 240 = 800 x  λ

∴ λ = \(\frac{240}{800}\) Hz.

∴ The required wavelength is 0.3 m. Answer:

Question 32. Why are sound waves called elastic waves?
Answer:

Reason :
(1) Waves require sometime to travel from one place to another. Same thing happens for sound waves.

(2) Vibrations of source are necessary for the production of elastic waves. Sound waves also require vibrations of the source for their production.

(3) The medium through which elastic waves pass is not bodily moved when the waves pass through it. The same thing happens for the sound waves.

(4) Like all other elastic waves, sound waves require a material medium for their propagation.

(5) Sound waves like all other elastic waves undergo reflection, refraction and interference.

Question 33. Show that a material medium is necessary for the propagation of sound.
Answer:

Material medium is necessary for the propagation of sound

A material medium is necessary for the propagation of sound. It cannot travel in vacuum. A material medium necessary for the propagation of sound can be demonstrated by the following experiment.

An electric bell is suspended inside an air tight glass bell-jar connected to a vacuum pump as shown in the given figure. On completing the electric bell circuit we héar the sound of electric bell. Now, air is withdrawn from the jar by starting the vacuum pump, the intensity of sound goes on decreasing. When all the air has been drawn out we do not hear any sound, while the hammer of electric bell is still seen striking repeatedly.
This means that sound is produced due to vibrations of bell, but not transmitted to us. This proves that sound requires a material medium for transmission and it cannot travel through vacuum.

Question 34. Define the following terms related to vibration :
(1) Vibration
(2) Amplitude
(3) Time Period
(4) Frequency.

Answer:

(1) Vibration: One complete to and from motion of a body about its mean position is called one vibration.

(2) Amplitude: The minimum displacement of a vibrating body from its mean position (or rest) is called its amplitude.

(3)Time period: The time (in seconds) taken by a vibrating body to complete one vibration is called its time period.

(4)Frequency: The number of complete vibrations that the vibrating body makes in one second is called its frequency. :

West Bengal Board Class 9 Physical Science Solutions Chapter 7  Question 35.

Explain that sound has wave like motion.
Answer: A material medium and the vibration of a body are essential for the propagation of sound. When sound propagates through a material medium, then there is no bodily movement of that medium. Hence, it is obvious that sound has wave-like motion. That sound is wave

like in nature can be understood by the following facts :
(1) Sound wave undergoes reflection.
(2) Sound wave travels with a definite velocity in a medium.
(3) Sound waves suffer deviation from the straight path in passing through obstacles which come on the way. This phenomenon is called diffraction.
(4) The superposition of sound may take place at a particular instant. This phenomenon is called interference of sound.

Question 36. Explain the mechanism of propagation of sound through air medium.
Answer:

The mechanism of propagation of sound through air medium

Sound propagates in air by longitudinal waves. When the prong of a tuning fork is struck against a hard rubber pad, it starts vibrating. The vibrations form longitudinal waves in air which reach our ear and we perceive the sensation of sound. The represents the formation of longitudinal waves in air due to the vibrations of one prong P of a tuning fork.

When the prong moves to the right from its mean position P to A, it compresses the layers of air close to it and thus the volume of these layers at C decreases and pressure increases. These air layers at high pressure and high density, called compression (c), have a tendency to expand.

When the prong returns from A to its mean position, the compression moves to right and reaches to another compression (c). When the prong reaches from mean position P to the other extreme B (to the left), it leaves behind a region of low pressure producing a rarefaction R. The compression shifts from next position at iC.

Thus, in one complete vibration of the prong, one compression and one rarefaction are formed. These compression and rarefaction are alternately produced and continue to travel forward to make a sound wave in air.

Question 37. On what factors does the speed of sound in air depend?
Answer:

The speed of sound is influenced by the following factors:

(1) Effect of density: The speed of sound is inversely proportional to the square root of the density of the medium. For example, the density of oxygen is 16 times the density of hydrogen, therefore, the speed of sound in hydrogen is four times than in oxygen.

So, if v be the velocity of sound in a gas of.density d, then.\(v \propto \frac{1}{\sqrt{d}}\)

(2) Effect of temperature: The speed of sound increases with an increase in temperature of the gas. For small change in temperature, the speed of sound in air increases by about 0.6 m/sec for each degree Celsius rise in temperature.

(3) Effect of humidity: The speed of sound increases with an increase in humidity. The sound travels faster in moist air than in dry air.

(4) Effect of wind: The speed of sound increases or decreases according to the direction of the wind.

(5) Effect of pressure: The speed of sound is independent of pressure.

(6) Effect of wavelength and amplitude: The speed of sound is independent of wave length and amplitude of sound wave.

Question 38. What are the differences between the light and sound waves?
Answer:

The distinction between the light and sound waves are :

Light Waves	Sound Waves
1.    Light waves are the electromagnetic waves.	1.    Sound waves are the mechanical waves.
2.    They can travel in vacuum.	2.    The required material medium for propagation.
3.    These waves are transverse.	3.    These waves are longitudinal (through air).
4.    The speed of light waves is very high = 3x\(10^8\) m/sec.	4.    The speed of sound waves in air is very low = 300 m/sec.

Question 39. Describe the comparison between the transverse and longitudinal waves.
Answer:

The following points explain the comparison between the transverse and longitudinal waves.

Transverse Wave	Longitudinal Wave
1.    The particles vibrate perpendicular to the direction of the propagation of wave.	1.    The particles vibrate’ in the same direction of the propagation of wave.
2.    It forms a series of crests and troughs.	2.    tt forms a series o! compression and refracting.
3.    It can propagate only in solids.	3.    It can propagate in att types of mediums.

Question 40. What is reflection of sound? Explain the reflection of sound with the help of an experiment.
Answer: Sound is reflected by an obstacle obeying the laws of reflection as of light.

The following experiment demonstrates the reflection of sound obeying the law of reflection.
Experiment: In front of a plane surface or a board (say PQ) two hollow tubes (say A and B) are arranged in such a way that they may be inclined equally with the plane surface. Place a clock near the mouth of one tube (A) and ear at the end of another tube (B) as shown in the given diagram. Place a screen (S) between the two tubes
After adjusting the two tubes, when the ticking of clock is clearly heard to the ear placed at B, angle AOS and angle BOS are measured. It is found that the angle of incidence AOS and angle of reflection BOS are equal.

West Bengal Board Class 9 Physical Science Solutions Chapter 7  Question 41.

Define the following terms related to wave motion :
(1) Wave velocity
(2) Frequency
(3) Periodic time
(4) Wavelength
(5) Amplitude
Answer: In the given wave motion of a wave (1) Wave Velocity: The distance travelled by a wave in one second is called its wave velocity. The wave velocity depends on
(1) The density of medium and
(2) The elasticity of medium.
It is denoted by the letter V. The unit of wave velocity is m/s.

(2) Frequency: The number of waves passing through a point in one second is called the frequency of the wave. It is represented by the letter f. The S.1. unit of frequency is hertz (Hz).

(3) Periodic time or time period: The time taken by a wave to pass through a point is called the time period of the wave.

It is denoted by the letter T.

The relation between time period and wave is given by T\(=\frac{1}{f}\)  or, f\(=\frac{1}{T}\) .

(4) Wavelength: The distance travelled by the wave in one time period is called the wavelength. It is denoted by the letter 4 (lambda). Its unit is metre (m).

(5) Amplitude: The maximum displacement of a vibrating particle about its mean position is called amplitude.

Question 42. What is reflection of sound? State the laws of reflection of sound.
Answer:

Reflection of sound : Like light, sound also suffers reflection. When sound is incident on a hard surface like that of brick, wood, stone or a plaster wall, it bounces back.

The reflection of sound does not require a polished or shining surface like a miror. As the wavelength of sound waves are much longer than that of light waves, so sound waves require a much larger reflector for reflection.

Like light, the reflection of sound also obeys two laws; viz.—

(1) The incident sound wave, the reflected sound wave and the normal at the point of incidence on the reflector, all lie in the same plane.

(2) The angle of reflection of sound wave is always equal to the angle of incidence.

Question 43. Write a short note on reverberation.
Answer:

Reverberation: inside the domes of cathedrals or ordinary large rooms echo is not heard because the reflecting sufraces situated there are less than the minimum distance required for the formation of echo. The reflected sound reaches the ear at such a quick succession that it can not be distinguished from the original sound. So one gets the impression that original sound has been prolonged. This effect of prolonging of original sound is called reverberation.

The quality of sound can be improved by the choice of a particular amount of reverberation. In case of orchestral and choral music excessive reverberation is desirable. Again, in some cases excessive reverberation makes the speech of a person indistinct. With excessive reverberation rock music also becomes less pleasing.

For an architect, reverberation time of an auditorium is the main concern. For a good acoustic building, the ceiling of the auditorium should be made by the soft material, the windows should be covered with heavy curtains, padded seats should be arranged in the hall and some plants in pots should also be arranged in the hall to reduce the reverberation time.

Question 44.Mention some measures to control sound pollution.
Answer:

Control of sound pollution: There are three specific measures that could be taken to control sound pollution. These are :
(1) Reducing the noise at the source :
(1) Automobiles and machines should be provided with silencers.
(2) The loud sound of impact on metal surface in factories can be dampened by covering the impact surface with a rubber material.

(2) Reducing the noise on its path :
(1) By taking the source of noise to an isolated area
(2) by placing a noisy machine in a double-walled room
(3) by coating the walls of the auditorium and cinema halls with sound-absorbing materials.

(3) Protecting the receiver :
(1) Some people such as factory workers, bus and engine drivers and mechanics, who have to work in a noisy atmosphere should use ear plugs.
(2) Many devices like ear muffs, noise helmet, etc. are available in the market which prevent sound entering into the ears.

Moreover, each one of us can contribute a lot towards reducing noise by practising silence, or talking with low pitch, keeping the volumes of stereo-music and radio low. Blowing of horns by automobiles and trucks are prohibited in certain cities to reduce the noise and its adverse effects on the health of the citizens.

West Bengal Board Class 9 Physical Science Solutions Chapter 7 

Question 45. Explain the formation of a wave.
Answer:

Explanation of the formation of wave: The dropped piece of pebble exerts a downward mechnical force on a few nearby water particles. The particles move a bit down ward, forming a temporary depressed region there, which is called Trough. At the same time, the mutual attractive force, known as the cohesive force, acting between these particles and their neighbouring ones, offers an upward pull on them and so they move up again. They move a bit up forming an elevated region which is called Crest. Again, by inertia of motion the particles move down, cross the level and again move up and so on. !n this way an undulatory motion of the particles sets in.

Thus, the portions of the medium where the surface is at a maximum distance above the normal level are called crests and those at a maximum distance below the normal level are called troughs. The upward and downward arrows, in the diagram, correspondigly indicate cohesive forces which are in opposition to the downward and upward motions of water particles. But energy does not transmit in such a wavy or zig zag manner.

Energy travels in a straight line inside an elastic continuous medium. lf the displacements of a particle in the medium about its mean position are plotted against time, the curve obtained is of a wavy pattern. This is shown in the diagram. In drawing the dia- gram, displacements of each particle in the upward directions or in the right hand directions are considered as positive, those in reverse directions are taken as negative.

The time required for one complete oscillation of a particle in the medium is called the Time period (T). T is
divided in four equal parts, T/4, T/2, 3T/ 4 and T for the respective displacements OA (+), BO (-), OC (—) and DO (+). ‘O’ is the mean position of the vibrating particle. Thus, the shape of the time-displacement curve is wavy

 

West Bengal Board Class 9 Physical Science Solutions Chapter 7 

Question 46. Write the conditions for the formation of echo.
Answer:

The conditions for the occurrence of echo are :
(1) Reflection of sound must occur from a distant and large body so that the reflected sound reaches the listener.

(2) The minimum distance between the reflector and a listener for hearing echo of a transient sound is 16.6 m and that for a monosyllabic articulate sound should be 33.2 m.

(3) The minimum time interval between a transient sound and its echo and that between a monosyllabic articulate sound and its echo should be respectively 1/10 second and 1/5 second.

(4) A sound and its echo should be similar to hear, only the echo will be a bit fainter than the original sound.

Question 47. Write a short note on the applications of echo.
Answer:

Applications of Echo :

(1) Echo depth sounding : To find depth of an ocean, the device SONAR is attached to a floating ship. Ultrasonic wave from the ship is sent from SONAR to sea bed. The re flected wave from the sea bed is received by an electronic receiver. The time taken by the ultrasonic wave travelling from source and coming back is noted. Knowing the velocity of sound in water, the depth of sea is calculated as

Depth of sea\(=\frac{\text { velocity of sound through water } \times \text { time }}{2}\).

(2) Determination of height of an aeroplane above ground: A sound from SONAR device in the aeroplane is sent from an aeroplane when it a sec B is at the point A on its path along which it flies high above and parallel to the ground. Echo of the sound reaches the aeroplane when it is at the position B after a time t from the instant of gunshot.

Let, velocity of the aircraft be v and that of sound in air V. In the time ‘t’ the aircraft moves. through AB and the sound through AOB.

∴Ac=CB=V.\(\frac{t}{2}\) ;   OA=OB=V.\(\frac{t}{2}\)

∴Height of the aeroplane from the ground,

CO\(=\sqrt{A O^2-A C^2}\)\(=\sqrt{\left(\frac{V t}{2}\right)^2-\left(\frac{v t}{2}\right)^2}\)=\(\frac{t}{2} \sqrt{V^2-v^2}\)

From the knowledge of V, v and t, the height of a flying aeroplane can be determined.

West Bengal Board Class 9 Physical Science Solutions Chapter 7

(3) Determination of velocity of sound in air with the help of a stopwatch and a pistol: In a wide open space where a high wall stands, a person fires a pistol and simultaneously starts the stop-watch. The sound reflects from the wall and the person hears an echo when he stops the stop-watch immediately. The time interval between the initial sound of firing and its echo is noted, which is t, say.

Since the sound of pistol firing is transient, the minimum distance between the wall and the person must be half of 1/10th of the velocity of sound in air at that time there. Dividing twice the distance between the man and the wall with this time t, velocity of sound ini air at the temperature there is determined.

Question 48. What are the characteristics of progressive waves?
Answer:

The characteristics of progressive waves are :
(1) Progressive waves are produced due to periodic motion of the particles of the medium.

(2) Each particle of the medium executes exactly similar vibrations about to that of wave motion, the wave may be transverse or longitudinal.

(3) The medium, the wave moves in the medium with a particular velocity which is dependent on the density and elasticity of the medium.

(4) Wave transports energy through the medium but not the medium itself.

(5) When a wave passes through a medium, every particle of the medium passes through same changes in pressure and density.

West Bengal Board Class 9 Physical Science Solutions Chapter 7 

Question 49. What are the factors on which pitch depends?
Answer:

Factors on which pitch depends :
(1) Pitch increases with increase in frequency: In the musical scale, ‘Sa’ — ‘te’ — ‘ga’ — ma’, etc. the frequency gradually increases from ‘sa’ so pitch also increases gradually from ‘sa’, pitch of ‘re’ is more than that of ‘sa’, pitch of ‘ma’ is more than those of ‘sa’ and ‘re’ and so on.

Humming sound of mosquitoes is more sharp than the roar of a tiger: A mosquito Produces a sharp sound because the frequency of vibration of the wings of a flying mosquito is very high. When a tiger roars, the frequency of vibration of its vocal chords is low, so the sound produced is flat but due to large size of the tiger’s vocal chord the emitted sound is of high intensity.

The sound of a wax-coated tuning fork is flatter than that produced by it in its usual condition. The wax-coating increases its mass, so its frequency decreases, causing the flat emitted sound.

(2) Pitch increases when the distance between the source of sound and a listener decreases and vice versa: The pitch of the sound of whistle of an approaching engine gradually increases since its distance from a listener gradually decreases.

(3) Pitch increases or decreases corresponding to decrease or increase of wavelength: Since, velocity of sound in a medium = frequency x wavelength. For constant velocity, if frequency of a sound increases or decreases, its wavelength correspondingly decreases or increases proportionally. So, pitch of a sound increases when wavelength decreases and pitch decreases if wavelength increases.

Question 50. Two persons A and B standing one km apart in open air. ‘A’ fires a gun and ‘B’ receives sound after 3 second of the flash. Calculate the velocity of sound in air.
Answer:

We know,      s=1      km=1000m

V\(=\frac{s}{t}\)                 t=3 sec

or, V\(=\frac{1000}{3}\)        V=?

= 333.3 m/sec.

∴ Velocity of sound in open air = 333.3 m/sec.

Question 51. What should be the minimum distance of the reflector from the source of sound to hear the echo of the word ‘Beautiful’ ? (velocity of sound = 340 m/sec)
Answer:

The word ‘beautiful’ is a trisyllabic word. So to hear it distinctly,

one would require \(\left(3 \times \frac{1}{5}\right)\) sec \(=\frac{3}{5}\) sec to pronounce it.

In  \(=\frac{3}{5}\) sec sound travels\(\left(340 \times \frac{3}{5}\right)\)m = 204 m.

Thus, the minimum distance of the reflector  \(=\frac{204}{2}\)m= 102 m.

West Bengal Board Class 9 Physical Science Solutions Chapter 7 

Question 52. A vibrating body produces 24 waves in 3 seconds. The distance between a crest and a trough is 0.15 m. Find the
(1) Frequency
(2) Wavelength
(3) Velocity of the wave.
Answer:

Here,(1) Frequency (n) \(=\frac{\text { No. of complete vibrations }}{\text { time taken }}\)=\(\frac{24}{3}\)=  8Hz

(2) Wavelength (λ) = 2 x distance between two consecutive crests and troughs
= (2 x 0.15)m=0.3 m.

and(3) Wave velocity (v) = nλ= (8 x 0.3) ns = 2.4 m/s.

Question 53. The frequency of a tuning fork is 256 Hz and sound travels to a distance of 30 m while the fork completes 24 vibrations. Find the wavelength of the sound emitted and also the velocity of sound.
Answer:

During 24 complete vibrations sound travels 30 m.

∴  During one complete vibration sound travels =\(\frac{30}{24}\)  m = 1.25 m.

∴ Wavelength of the sound (λ) = 1.25 m.

Here, Frequency (n) = 256 Hz.

∴ Velocity of sound wave (v) = nλ = 256 Hz x 1.25 m = 320 m/s.

Question 54. A tunning fork has frequency 256 Hz. Sound produced with it travels 20 m when the fork makes 16 vibrations. Find the wavelength and the velocity of the sound wave.
Answer:

Wavelength is the distance travelled by sound in the time required for one

vibration of the fork. So, wavelength, λ = \(\frac{20}{16}\) m.=1.25Now, from the relation , v = n [v

velocity of sound, nλ = frequency, λ = wavelength = 256 x 1.25 m/s = 320 m/s.

So, the required wavelength is 1.25 m and velocity of sound is 320 m/s.

Question 55. A body vibrating with a particular frequency generates waves of wavelength 0.2 m in a medium ‘A’ where the velocity of wave is 160 m/s. Calculate the wavelength it will generate in another medium ‘B’ in which its velocity is 240 m/s.
Answer:

From the relation V = nλ , where V = velocity of wave in the medium A, n = frequency, λ = wavelength,
160=n×0.2

n\(=\frac{160}{0.2}\)=800 Hz.

Again,in the medium,B,V=nλ    or, 240 × 800×λ

∴ λ\(=\frac{240}{800}\)=0.3

∴  The required wavelength is 0.3 m.

West Bengal Board Class 9 Physical Science Solutions Chapter 7 

Question 56. The velocity of sound in a certain gas is four times that in air at the same temperature. When a tuning fork is sounded in the air, a wave of frequency 480 Hz and wavelength A, m is produced. The same fork is sounded in the gas and if 1, m is the wavelength of the wave, find the ratio of the wavelengths of sound produced in the gas to that in air.
Answer:

Given

The velocity of sound in a certain gas is four times that in air at the same temperature. When a tuning fork is sounded in the air, a wave of frequency 480 Hz and wavelength A, m is produced. The same fork is sounded in the gas and if 1, m is the wavelength of the wave,

Let \(V_{\text {gas }}\)= Velocity of the wave in the gas.

⇒ \(V_{\text {air }}\) = Velocity of the wave in the air.

⇒ \(V_{\text {gas }}\) = 480 \(\lambda_2\) and \(V_{\text {air }}\) = 480  \(\lambda_1\)

Since \(V_{\text {gas }}\) =4x\(V_{\text {air }}\),\(\frac{480 \lambda_2}{4}\) = 480 \(\lambda_1\)

or,\(\frac{\lambda_2}{\lambda_1}\)\(=\frac{480 \times 4}{480}\)=\(\frac{4}{1}\)

⇒ \(\frac{\lambda_2}{\lambda_1}\)=\(\frac{4}{1}\)

(It is seen that when velocity of a wave increases, the wavelength also increases

proportionally, because from the relation, V = nλ, \(\frac{\mathrm{V}}{\lambda}\)  =n, which is a constant for constant

frequency. From this argument, it can be said that, \(\lambda_2\), is 4 times \(\lambda_1\)).

Question 57. A sound is made at one end of a hollow iron pipe 950 m long when two sounds are heard at the other end at interval of 2 seconds. If the velocity of sound through air at that time be 336 m/s, find the velocity of sound through iron.
Answer:

Given

A sound is made at one end of a hollow iron pipe 950 m long when two sounds are heard at the other end at interval of 2 seconds. If the velocity of sound through air at that time be 336 m/s,

If the velocity of sound through iron be V m/s, time required to travel

950 m\(=\frac{950}{V}\) and that required to travel through the same length of air=\(\frac{950}{336}\)s.

∴ 2=\(\frac{950}{336}\)−\(\frac{950}{V}\)

or, \(\frac{950}{V}\) =\(\frac{950}{336}\)−2

∴ V = 1148.3 m/s. So, velocity of sound through iron is 1148.3 m/s.

Question 58. Two sounds are heard at an interval of 2.5 s at one and of a hollow metallic pipe 1000 m long when the other end of it is struck gently. If the temperature of air at the time be 50°C and velocity of sound through air at 0°C be 332 m/s, find the velocity of sound through the material of the tube.
Answer:

Given

Two sounds are heard at an interval of 2.5 s at one and of a hollow metallic pipe 1000 m long when the other end of it is struck gently. If the temperature of air at the time be 50°C and velocity of sound through air at 0°C be 332 m/s

Increase in the velocity of sound through air at 50°C over that at 0°C is 61 cm x
50 = 3050 cm = 30.5 m.

∴ Velocity of sound through air at 500C = (320 + 30.5) m/s = 362.5 m/s.

If the velocity of sound through the material of the pipe be V,

⇒  \(\frac{1000}{362.5}\)−\(\frac{1000}{V}\) =2.5

or, \(\frac{1000}{362.5}\)−2.5=\(\frac{1000}{V}\)

or,0.26=\(\frac{1000}{V}\)

∴ V = 3846 m/s (approx.)

∴ The velocity of sound through the material of the pipe is 3846 m/s (approx.)

WBBSE Solutions For Class 9 Physical Science And Environment

• Chapter 1 Measurement
• Chapter 2 Force And Motion
• Chapter 3 Matter: Structure And Properties
• Chapter 4 Matter: Atomic Structure; Physical & Chemical Properties of Matter
  • 4.1 Atomic Structure
  • 4.2 Concept Of Mole
  • 4.3 Solution
  • 4.4 Acids, Bases & Salts
  • 4.5 Separation Of Components Of Mixtures
  • 4.6 Water
• Chapter 5. Energy In Action: Work, Power & Energy
• Chapter 6 Heat
• Chapter 7 Sound