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Page 162

Question: 1 - How does the sound produced by a vibrating object in a medium reach your ear?
Answer:- Vibrations in an object create disturbance in the medium and consequently compressions and rarefactions. Because of these compressions and rarefactions sound reaches to our ear.

Page 163

Question: 1 - Explain how sound is produced by your school bell.
Answer:- School bell starts vibrating when heated which creates compression and rarefaction in air and sound is produced.

Question: 2 - Why are sound waves called mechanical waves?
Answer: Since sound waves do some mechanical work while making disturbance in medium, hence sound waves are called mechanical wave.

Question: 3 -Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer: Since sound waves require medium to propagate and there is no medium present on the moon. So, I will not able to hear the sound of my friend on the moon.

Page 165

Question: 1 - Question: 1 - Which wave property determines (a) loudness, (b) pitch?
Answer: (a) Amplitude of sound waves determines loudness. Louder sound has greater amplitude and vice versa.
(b) Frequency of the sound waves determined pitch of the sound.

Question: 2 - Guess which sound has a higher pitch: guitar or car horn?
Answer: Sound of the car horn has higher pitch.

Question: 3 - What are wavelength, frequency, time period and amplitude of a sound wave?
Answer: Wavelength: Wavelength is the distance between two consecutive compressions or rarefaction of wave.
Frequency: The number of sound wave produced in one second is called frequency.
Time period: Time period is the time taken to produce one wave of sound.
Amplitude: Amplitude is the maximum displacement along the mean position of the particles of medium.

Question: 4 - How are the wavelength and frequency of a sound wave related to its speed?
Answer: The relation between frequency and wavelength of sound wave is given as follows:
Velocity (v)=Wavelength(λ)X Frequency (ν), v = λX ν
This means the speed is equal to the product of wavelength and frequency of the sound wave.
This equation is also called the ‘wave equation’ and applicable to all types of wave.
Question: 5 - Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Answer:
Given,
Frequency (ν) = 220 Hz
Velocity (v) = 440m/s
Wavelength (λ) =?
We know;

NCERT CBSE In text Questions and Answer 5

Thus, wavelenght = 2m

Question: 6 - A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
Answer: Since, the time interval between successive compressions is called time period or time interval.
Here given,
Frequency (ν) = 500Hz
T (Time period) =?
We know that;

NCERT CBSE In text Questions and Answer 6

Thus, time interval between two consecutive compression of the given wave = 0.2 s

Question: 7 - Distinguish between loudness and intensity of sound.
Answer: Loudness of sound is determined of amplitude and intensity of the sound wave is determined by frequency of sound waves.

Page 167

Question: 1 - In which of the three media; air, water or iron does the sound travel the fastest at a particular temperature?
Answer:At particular temperature sound travels fastest in iron.

Page 168

Question: 1 - An echo returned in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m /s?
Answer:To return an echo sound has to cover distance of two way.
Here, given,
Speed of sound = 342 m/s
Time = 3s
Thus,
Distance = speed X time
 Distance = 342m/s×3 s=1026 m
Thus,the distance between the source and reflecting surface=1026÷2=513m

Page 169

Question: 1 -Why are the ceilings of concert halls curved?
Answer: Since, concert halls are big, so audience at the back rows of the hall may not hear clear sound of speaker. To overcome this problem, the ceiling of the concert halls is made concave. Concave ceiling helps the sound wave to reflect and send to farther distance which makes the concert hall enable to send clear sound to the audience even sitting in back rows of hall.

Page 170

Question: 1 - What is the audible range of the average human ear?
Answer: 20 Hz to 20000 Hz
Question: 2 - What is the range of frequencies associated with
(a) Infrasound
(b) Ultrasound
Answer:
(a) Infrasound: Less than 20 Hz
(b) Ultrasound: More than 20000 Hz

Page 172

Question: 1 - A submarine emits a SONAR pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff ?
Answer:To return the SONAR pulse back, its wave has to travel two way.
Here, given,
Velocity (v) of sound wave = 1531m/s
Time (T) = 1.02 s
Thus, Distance = speed X time
Distance=1531 ms^(-1)×1.02 s=1561.62 m
So,the distance between the source and reflecting surface=1561.62÷2=780.81m

Exercise Questions (NCERT Book) – Sound Class nine

Question: 1 - What is sound and how is it produced?
Answer: Sound is a kind of energy produced in the form of waves. When anything is set to vibration, it produces sound.

Question: 2 - Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Answer:-
Compression and rarefaction in air – Compression and rarefaction are produced because of the disturbance in medium caused by sound wave. Sound wave propagates because of compression and rarefaction of the particles of the medium.
When an object starts vibrating, it creates disturbance in medium. Because of the disturbance particles of medium come closer to each other compare to their normal position on the other hand adjacent particles go farther to each other. Both happen simultaneously.
The region where particles are come closer to each other is called compression and region where particles go farther to each other is called rarefaction.
In the given figure straight lines are showing the normal position of air particles. Dense lines are showing the region of compression and less dense lines are showing region of rarefaction of air particles.

Compression and rarefaction - sound wave- class nine science cbse ncert11

Question: 3 - Cite an experiment to show that sound needs a material medium for its propagation.
Answer:- Activity:-
Take a glass bell jar, connect it with vacuum pump and suspend an electric bell in it.
Connect electric bell with a battery.
Switch on the electric bell and hear the sound of bell.
Now remove the air completely from the bell jar using vacuum pump and observe the sound of electric bell.

propagation of sound experiment class nine scinece cbse ncert12

It is observed that sound of electric bell does not come out after pumping out air from the bell jar.
This happens because after creating vacuum in the bell jar there were no air present through which sound wave can propagate.
This experiment shows that without medium sound cannot propagate and hence for the propagation of sound medium must be present.

Question: 4 - Why is sound wave called a longitudinal wave?
Answer:
Since sound wave creates oscillation in the particles of the medium parallel to the disturbance in the direction of propagation, thus sound waves are called longitudinal wave. This would be more clear by taking the definition of longitudinal wave into account.
Longitudinal wave: When oscillation is created parallel to the disturbance of the particles of medium in the direction of propagation.

Question: 5 - Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer: Timbre and pitch are the characteristics of sound which help to identify the sound of different voice. Thus, because of difference in timbre and pitch of the sound wave I or any other can identify the voice of his friend sitting with others even in dark room.

Question: 6 - Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Answer: This happens because of the difference in the velocity of light and sound waves. Light travels with much faster velocity than sound. That’s why thunder is heard a few seconds after the flash of thunder is seen instead of both are produced simultaneously.

Question: 7 - A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 m/s.
Answer:
Given, velocity of sound = 344 m/s
We know that
Velocity = wavelength X frequency

NCERT CBSE In text Questions and Answer 7

Question: 8 - Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.
Answer:
We know that,
The speed of sound in air = 344 m per second
The speed of sound in aluminium = 5100 m per second
Hence, the ratio of time taken by the sound to travel through air and through aluminium

NCERT CBSE In text Questions and Answer 8

Question: 9 - The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer:
Given, frequency = 100 Hz
This means the source of sound vibrates 100 times in one second.
Therefore, number of vibrations in 1 minute, i.e. in 60 seconds = 100 x 60 = 6000 times.
Question: 10 - Does sound follow the same laws of reflection as light does? Explain.
Answer: Yes, the sound wave follows the same laws of reflection as the light does. The laws of reflection of sound are as follows:
The incident sound wave, the reflected sound wave and the normal at the point of incident, all lie in the same plane.
The angle of incidence of sound wave and angle of reflection of sound wave to the normal are equal.
When sound waves reflected from a surface, the angle of incidence is equal to the angle of reflection to the normal and the incident wave, normal and reflected wave are in the same plane. This can be proved by experiment.
Thus, sound wave obeys the laws of reflection.

Question: 11 - When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?
Answer: To hear the sound of echo depends upon the distance from source of sound and reflecting surface. The distance between both must be equal to or more than 17.2 meter. If the given distance is more than 17.2 meter then one can hear the echo sound on a hotter day also.
Although, in hotter day the velocity of sound increases, thus it is necessary to hear the sound of echo the distance should be more than 17.2 meter. If the given distance is equal to 17.2, then to hear the sound in hotter day would not be possible.

Question: 12 - Give two practical applications of reflection of sound waves.
Answer: Bulb horn and Stethoscope are examples of practical applications of reflection of sound waves.
In bulb horn sound is amplified and sent to the desired direction because of reflection. In stethoscope also sound is sent to the desired direction because of its reflection characteristic.

Question: 13 - A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 m–2 and speed of sound = 340 m/s.
Answer:
Given,
Height of tower = 500 m
g = 10m/s
Velocity of sound = 340 m/s2
Thus, to calculate the time of splash sound, first of all time taken to reach the stone in the water is to be calculated.
We know that,

NCERT CBSE In text Questions and Answer 13

Question: 14 - A sound wave travels at a speed of 339 m/s. If its wavelength is 1.5 cm, what is the frequency of the wave?
Answer:
Given,
Velocity (v) of sound = 339 m/s
Wavelength (λ) = 1.5 cm = 0.015 m
Frequency (ν) =?
We know that, speed = wavelength X frequency
339 m/s = 0.015 m X frequency

NCERT CBSE In text Questions and Answer 14

Thus, frequency = 22600 Hz

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