Ncert Solution Class 9 Science Chapter 11 Sound - Param Himalaya
Page 129
1. How does the sound produced by a vibrating object in a medium reach your ear?
Solution: When a body vibrates the air in its neighborhood is alternately compressed and rarefied. The compressed air has higher pressure than surrounding air. It therefore pushes the air particles near it causing compression to move forward. A rarefaction or low pressure is created at the original place.
These compressions and rarefaction causes particles in the air to vibrate about their mean position. The energy is carried forward in these vibration. This is how sound travels.
2. Explain how sound is produced by your school bell.
Solution: When the school bell vibrates, it pushes the adjacent particles of air to vibrate. This causes the disturbance in the wave when the bell moves forward it pushes the air particles in front of it and creates a region of high pressures known as compression and when the bell moves backwards, it creates a region of low pressure known as rarefaction.
The continuous forward and backward movement of the bell produces a series of compression and rarefaction. This makes the sound of a bell to propagate through the air.
3. Why are sound waves called mechanical waves?
Solution: Sound waves are mechanical waves because they need a material medium for propagation, like air or liquids like water, or metals like silver.
4. Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Solution: No, I would not be able to hear any sound produced by my friend on the moon. Sound requires a medium, such as air, to travel through, and the moon lacks a significant atmosphere. Since there is no air or medium to transmit sound waves, any sounds my friend makes would not be heard by me on the moon. In the absence of a medium for sound propagation, communication through sound waves is not possible in the vacuum of space.
Page 132
1. Which wave property determines
(a) loudness, (b) pitch?
Solution:
a) Loudness is determined by the amplitude of the sound. Greater the amplitude more will be the loudness.
b) Pitch is determined by frequency. Higher is the frequency, greater will be the pitch.
2. Guess which sound has a higher pitch: guitar or car horn?
Solution: Guitar
Example 11.1 A sound wave has a frequency of 2 kHz and wave length 35 cm. How long will it take to travel 1.5 km?
Solution:
Given
Frequency, f = 2 KHz = 2000 Hz
Wavelength, $\lambda$ = 35 cm = 0.35 m
We know that speed , v of the wave = wavelength × frequency
$t = \frac{d}{v} = \frac{1.5 \times 1000 m}{700 m/s}$
$t= \frac{15}{7}s = 2.1s$
Thus sound will take 2.1 s to travel a distance of 1.5 km.
1. What are wavelength, frequency,time period and amplitude of a sound wave?
Solution:
Wavelength: The distances between two consecutive compressions or rarefaction of a wave. Its S.I unit is meter.
Frequency: One compression and one rarefaction constitutes one vibration. The number of vibration in a second is called frequency. Its unit is Hertz.
Amplitude: When waves are produced, the particles vibrate about their mean position. The maximum displacement from its mean position of a particle is called its amplitude. It is measured in meters.
Time period: The time taken by the wave to complete one oscillation i.e., the time between two consecutive compressions or rarefactions is called time period.
2. How are the wavelength and frequency of a sound wave related to its speed?
Solution:
Speed = Wavelength × frequency
$$v = \lambda \times f$$
3. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Solution:
Given , f = 220 Hz , v = 440 m/s
Speed = wavelength × frequency
$v= \lambda \times f$
$\lambda = \frac{v}{f}$
$\lambda = \frac{440}{220} = 2 Hz$
4. 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?
Solution :
The time interval between two successive compressions is equal to the time period of the wave. The time period is reciprocal of the frequency of the wave and is given by the relation:
$T = \frac{1}{f} = \frac{1}{500}$
T= 0.02 sec
Page 133
1. Distinguish between loudness and intensity of sound.
Solution :
Loudness :
1. It is subjective quantity. A sound may be loud for one person but may be feeble for another person.
2. It cannot be measured as it is just a sensation which can be felt only
3. Loudness of sound is measured by the unit decibel (dB)
4. It depends on the sensitivity of ears.
Intensity :
1. It is an objective physical quantity which does not change for person to person.
2. It can be measured as a physical quantity.
3. Intensity of sound is measured by the unit watt per square metre $W/m^{2}$.
4. It does not depends on the sensitivity of ears.
2. In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Solution:
From the given media, sound travels fastest in iron at a certain temperature.
Example 11.2 A person clapped his hands near a cliff and heard the echo after 2 s.What is the distance of the cliff from the person if the speed of the sound, v is taken as 346 m/s ?
Solution:
Given,
Speed of sound, v = 346 m/s
Time taken for hearing the echo,t = 2 s
Distance travelled by the sound = v × t = 346 m/s × 2 s = 692 m
In 2 s sound has to travel twice the distance between the cliff and the
person. Hence, the distance between the cliff and the person
= 692 m/2 = 346 m.
Page 134
1. An echo is heard in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m/s?
Solution :
Speed of sound, V = 342 m s − 1
Echo returns in time, t = 3s
Distance = speed × time
D=V ×t = 342 × 3 = 1026 m
In the given time interval, sound has to travel a distance that is twice the distance of the reflecting surface and the source.
Hence, the distance of the reflecting surface from the source
d= D/2 = 1026/2 m = 513 m .
Page 135
1. Why are the ceilings of concert halls curved?
Solution :
Reflection of Sound:
1. The act of sound waves bouncing back from a surface is referred to as sound reflection.
2. The ceilings of concert hall are curved to ensure that sound reaches all corners of the space uniformly following reflection from the curved surface.
Page 136
1. What is the audible range of the average human ear?
Solution:
The audible range of an average human ear lies between 20 Hz to 20,000 Hz. Humans cannot hear sounds having frequency less than 20 Hz and greater than 20,000 Hz
2. What is the range of frequencies associated with
(a) Infrasound? (b) Ultrasound?
Solution :
(a) The frequencies of infrared sound are lower than 20 hertz.
(b) The frequency of ultrasound exceeds 20,000 hertz.
Exercise Solution:
1. What is sound and how is it produced?
Solution :
When an object vibrates, it forces the neighbouring particles of the medium to vibrate. These vibrating particles then force the particles adjacent to them to vibrate. In this way, vibrations produced by an object are transferred from one particle to another till it reaches the ear.
2. Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Solution :
Take an electric bell and an airtight glass bell jar. The electric bell is suspended inside the airtight bell jar. The bell jar is connected to a vacuum pump If you press the switch you will be able to hear the bell. Now start the vacuum pump. When the air in the jar is pumped out gradually, the sound becomes fainter, although the same current is passing through the bell. After some time when less air is left inside the bell jar you will hear a very feeble sound. Now if we evacuate the bell jar no sound is heard.
Result: The above mentioned activity shows that sound needs a medium to propagate.
3. Why is sound wave called a longitudinal wave?
Solution:
Sound wave is called a longitudinal wave because sound waves travel in the air through compressions and rarefactions.
4. Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Solution:
Pitch of the sound wave.
5. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Solution :
Since speed of thunder (sound) is much less (332 m/s) as compared to speed of flash (light) which is about 
6. 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–1 .
Solution :
For 20 Hz sound waves the wavelength would be
$\lambda$= v/n = 344/20 = 17.2 m
For 20 kHz sound waves the wavelength would be
/20000 Hz = 0.0172 m
7. 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.
Solution:
Since speed of sound in air = 344 m/s
and speed of sound in aluminium = 6420 m/s
we know that v = distance/time therefore time = d/v
$T_{air} = \frac{d}{v_{air}}$
$T_{aluminium} = \frac{d}{v_{aluminium}}$
time taken by sound wave in air/time taken by sound wave in aluminium
$\frac{T_{air}}{T_{aluminium} } = \frac{\frac{d}{v_{air}}}{\frac{d}{v_{aluminium}}}$
$= \frac{T_{aluminium}}{T_{air}}$
$= \frac{6420}{344}$
$\frac{T_{air}}{T_{aluminium} }$ = 18.66
the sound will take 18.66 times more time through air than in aluminium in reaching other boy.
8. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Solution:
Frequency of source of sound being 100 Hz means the sound source vibrates 100 times in one second.
therefore vibrations made by sound source in 1 min (60 sec) = 100 x 60 = 6000
9. Does sound follow the same laws of reflection as light does? Explain.
Solution :
Ans.Yes, sound also follows the same laws of reflection as light does. Their laws are:
(i)The direction of incidence sound, The direction of reflected sound and normal placed at the point of incidence are all on the same plane.
(ii)The angle of incidence and the angle of reflection are equal.
10. 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?
Solution :
$\therefore Time = \frac{Distance}{Speed(velocity)}$
that is, time and velocity (speed) have an inverse ratio. Increasing the temperature of any medium increases the speed of sound. Therefore, on a hot day, due to the higher temperature, the speed of sound will increase and we will hear the echo sooner than the cold days.
11. Give two practical applications of reflection of sound waves.
Solution:
Two practical applications of reflection of sound waves.
(i) Echo. (ii) Reverberation
12. 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 s–2 and speed of sound = 340 m s–1
Solution:
Case 1 : u = 0 , s = 500 m , g = 10 $m/s^{2}$
$s = ut + \frac{1}{2} gt^{2}$
$500 = 0 + \frac{1}{2} \times 10 \times t_{1}^{2}$
$t_{1}^{2} = 100$
$t_{1} = 10 sec$
Time taken by stone to reach the pond surface $t_{1}$ = 10 sec
Case -2 Again , s = 500 m , v = 340 m/s
$s = vt_{2}$
$t_{2} = \frac{s}{v} = \frac{500}{300}$
$t_{2} = 1.47 sec$
so the total time taken for splash being heard at the top
$t=t_{1} + t_{2}$
t= 10 + 1.47 = 11.47 sec
13. A sound wave travels at a speed of 339 m s–1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Solution :
Since we know that
Speed = frequency × wavelength
Frequency= speed / wavelength
ν = 339/0.015 = 22600 Hz
14. What is reverberation? How can it be reduced?
Solution:
The repeated reflection of sound due to which sound persists for a long time is called reverberation. To reduce reverberation, the roof and walls of the auditorium are generally covered with sound-absorbent materials like compressed fibreboard, rough plaster or draperies. The seat materials are also selected on the basis of their sound absorbing properties.
15. What is loudness of sound? What factors does it depend on?
Solution :
Ans.Loudness of a sound is the intensity of sound. It depends on it’s amplitude. Such a sound which has more energy is called it’s loudness. The loudness depends on the following factors:
(i)Amplitude (ii)Energy (iii)Intensity
16. How is ultrasound used for cleaning?
Solution :
Ultrasound is generally used to clean parts located in hard-to-reach places, for example, spiral tube, odd shaped parts, electronic components etc. Objects to be cleaned are placed in a cleaning solution and ultrasonic waves are sent into the solution. Due to the high frequency, the particles of dust, grease and dirt get detached and drop out. The objects thus get thoroughly cleaned.
17. Explain how defects in a metal block can be detected using ultrasound.
Solution : Ultrasounds can be used to detect cracks and flaws in metal blocks. Metallic components are generally used in construction of big structures like buildings, bridges, machines and also scientific equipment. The cracks or holes inside the metal blocks, which are invisible from outside reduces the strength of the structure. Ultrasonic waves are allowed to pass through the metal block and detectors are used to detect the transmitted waves. If there is even a small defect, the ultrasound gets reflected back indicating the presence of the flaw or defect