Class 9 Science


Exercise Question Answer

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 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 in wave

Question 3: Cite an experiment to show that sound needs a material medium for its propagation.

Answer: Activity:

propagation of sound experiment bell jar

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

Or, Wavelength = Velocity/Frequency

When frequency = 20 Hz
Wavelength = 344/20 = 17.2 m

When frequency = 20 kHz = 20000 Hz
Wavelength = 344/20000 = 0.00172 m

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

= (Velocity of sound in aluminium)/(Velocity of sound in air)
= (5100 m per s)/(344 m per s) = 150:1

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:

When sound waves are 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 m2 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, `s = ut + ½ a\t^2`

Or, `s = ut + ½ g\t^2`

Here, s = 500 m and g = 10 m/s2
Or, `500 m = 0 xx t + ½ xx 10 ms^-2 xx t^2`

Or, `500 m = 5 ms^-2 xx t^2`
Or, `t = 10 s`

Now, we know that; Time = Distance/Speed
Or, Time `= (500 m)/(340 ms^-2) = 1.47 s`

So, total time to hear the sound of splash `= 10 s + 1.47 s = 11.47 s`

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 xx text(frequency)`

Frequency `= (339 m//s)/(0.015 m) = 22600 Hz`

Thus, frequency = 22600 Hz