Class 9 Science Chapter 12
Sound NCERT Solutions
Page 162
1. How does the sound produced by a vibrating object in a medium reach your ear?
Answer:
When an object vibrates it forces the neighbouring particles of the medium to vibrate. These vibrating particles then forces the particles adjacent to them to vibrate.In this way, vibrations produced by an vibrating objects are transferred from one particle to another till it reaches the ear.
Page 163
1. Explain how sound is produced by your school bell.
Answer:
When the bell vibrates , it forces the adjacent particles in air to vibrate. When the bell continues to move, forward and backward , it creates a series of compression and rarefaction. The region of high pressure and high density is called compression. The region of low pressure and low density is called rarefaction. This makes the sound of a bell to propagate through air.
2. Why are sound waves called mechanical waves?
Answer:
Sound waves are called mechanical waves because they need a material medium to travel.They propagate through a medium because of interaction of the particles present in that medium.
3. Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer:
Sound waves need a material medium to travel. Since there is no atmosphere on moon therefore we will not be able to hear any sound produced by our friend.
Page 166
1. Which wave property determines (a) loudness, (b) pitch?
Answer:
a) Loudness is determined by amplitude of sound.
b) Pitch is determined by frequency.
2. Guess which sound has a higher pitch: guitar or car horn?
Answer:
Guitar will have higher pitch than car horn.Pitch of a sound is directly proportional to frequency.The frequency of vibration of sound produced by guitar is greater than that produced by a car horn.
1. What are wavelength, frequency, time period and amplitude of a sound wave?
Answer:
1)Wavelength: The distance between two consecutive compression or rarefaction is called Wavelength.
or
The minimum distance in which a sound wave repeats itself is called wavelength.
2)Frequency:The number of vibrations per second is called frequency.
or
Number of complete waves produced in one second.
3)Time Period(T):The time period required to produce one complete wave.
4)Amplitude:The magnitude of maximum disturbance in the medium on either side of mean value is called amplitude.
or
The maximum displacement of the particle of medium from their original undisturbed position, when wave passes through a medium is called amplitude.
2. How are the wavelength and frequency of a sound wave related to its speed?
Answer:
Speed = wavelength × frequency
v= λ × ν
3. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Answer:
Frequency of sound waves = 220 Hz
Speed of sound wave= 440 m/s
Speed = wavelength × frequency
v= λ × ν
λ = v/ ν
λ = 440/220 = 2 m
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?
Answer:
Frequency = 500 Hz
Frequency = 1/ Time period
Frequency = 1 /500 = 0.002 s
1. Distinguish between loudness and intensity of sound.
Answer:
| Loudness | Intensity of sound |
| It is the measure of the response of ear to sound. | It is the amount of sound energy passing through a unit area per sound. |
| It depends on amplitude | It depends on sound energy |
| It is not a physical quantity | It is a physical quantity. |
Page 167
1. In which of the three media, air, water or iron, does sound travel the fastest at a particular Q temperature?
Answer:
Speed of sound
Solid>Liquid>Gases
Sound travels fastest in iron and slowest in air.
Page 168
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 ms–1?
Answer:
Time for eco = 3s
Speed of sound= 342 m/s
Distance travelled by sound= time × speed
= 3 × 342 = 1026 m
In the given time interval, sound has 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 = 1026 / 2= 513 m
Page 169
1. Why are the ceilings of concert halls curved?
Answer:
Ceiling of concert halls are curved so that after reflection from the surface, sound spread uniformly in all direction.
Page 170
1. What is the audible range of the average human ear?
Answer: The audible range of the average human ear is 20 Hz -20,000 Hz.
2. What is the range of frequencies associated with
(a) Infrasound?
(b) Ultrasound?
Answer:
Infrasound has frequency less than 20 Hz.
Ultrasound has frequency more than 20,000 Hz
Page 172
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:
Time taken by the sonar ,t = 1.02 s
Speed of sound in slat water = 1531 m/s
Distance of the cliff from the submarine= Speed of sound × Time taken
= 1.02 × 1531
= 1561.62 m
Distance travelled by the sonar pulse during its transmission and reception in water = 2 × actual distance
Actual distance= 1561.62 / 2 = 780.31 m
Page 174,175
1. What is sound and how is it produced?
Answer:
Sound is produced by vibration.When a body vibrates, it forces the neighbouring particles of the medium to vibrate.This creates a disturbance in the medium which travels in the form of waves.This disturbance, when reaches the ear, produces sounds.
2. Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Answer:
When the vibrating body moves forward, it creates a region of high pressure in its vicinity.This region of high pressure is called as compression.When it moves backward , it creates a region of low pressure in its vicinity.This region is known as rarefaction.As the body continues to move forward and backward , it produces a series of compression and rarefaction.
3. Cite an experiment to show that sound needs a material medium for its propagation.
Answer:
Take an electric bell and an air tight glass bell connected to a vacuum pump.Suspend the bell inside the jar, and switch on the bell.We will be able to hear the bell ring.Now pump out the air from the glass jar.The sound of the bell will become low
and after some time we will not be able to hear any sound.This is so because almost all air has been pumped out of the glass jar.This shows that sound need a medium to travel.
4. Why is sound wave called a longitudinal wave?
Answer:
A longitudinal wave is a wave in which the particle movement is parallel to the direction of the wave propagation.Sound waves are longitudinal waves because particles of the medium through which the sound is transported vibrate parallel to the direction of sound wave moves.
5. Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer:
Quality of sound is the characteristic of the sound which helps us to identify our friend by his voice while sitting with others in a dark room.
6. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Answer:
The speed of sound is less than speed of light.Sound of thunder takes more time to reach Earth as compared to light.Hence a flash is seen before a thunder is heard.
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 ms–1.
Answer:
Speed = wavelength × frequency
v= λ × ν
a) Speed of sound in air = 344 m/s
frequency = 20 Hz
Wavelength = 344 /20 = 17.2 m
b) Speed of sound in air = 344 m/s
frequency = 20000 Hz
Wavelength = 344 /20000 = 0.0172 m
For humans, wavelength range is 0.0172 m to 17.2 m
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:
Speed of sound wave in aluminium= 6420 m/s
Speed of sound wave in air = 346 m/s
Let the length of aluminium rod = d
Time taken by sound wave to reach the other end
tAl= d/v
tAl=d/6420
Time taken by sound wave to reach other end
tair= d/v
tair=d/346
The ratio of time taken by the sound wave in air and aluminium
tair/tAl = (d/6420)/d/346
tair/tAl = 6420/346
tair/tAl = 18.55 : 1
9. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer:
Frequency = 100 Hz
Frequency = Number of oscillation/ Time taken
Number of oscillation= 100 × 60 s = 6000
The source vibrate 6000 times in 1 min.
10. Does sound follow the same laws of reflection as light does? Explain.
Answer:
Laws of reflection for sound are:
1) The incident sound wave, the normal to the reflecting surface and the reflected sound wave at the point of incidence lie in the same plane.
2)The angle of incidence is equal to the angle of reflection.
Thus we can say that sound follows the same laws of reflection as light does.
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:
The repetition of sound caused by reflection of sound waves is called echo.An echo is heard when the time interval between the original sound and reflected sound is at least 0.1 s. As the temperature increases the speed of sound in a medium also increases.On a hotter day, the velocity of sound is more.If the time interval between the original sound and reflected sound is greater than 0.1 s, then an eco is heard.
12. Give two practical applications of reflection of sound waves.
Answer:
The 2 practical applications of reflection of sound waves are:
1) It is used to measure the distance and speed of under water object.This technique is called as SONAR.
2) In a stethoscope, the sound of the patient’s heartbeat reaches the doctor’s ear by multiple reflections of sound.
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 ms–2 and speed of sound = 340 ms–1 .
Answer:
Height of the tower, S = 500 m
Velocity of sound= 340 m/s
Acceleration due to gravity = 10 m/s2
Initial velocity of the stone, u= 0 m/s
Time taken by the stone to fall to the base of tower = t1
S= ut1 + ½ at12
500 = 0 × t1 + ½ × 10 × t12
t1 = 10 s
Time taken by the sound to reach the top from a base of the tower , t2 = 500/ 340 = 1.47 s
Splash is heard at the top after time t = 10 + 1.47 = 11.47 s
14. A sound wave travels at a speed of 339 ms–1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Answer:
Speed of sound = 339 m/s
Wavelength of sound= λ= 0.015 m
Speed of sound= wavelength × frequency
339 = 0.015 × ν
ν = 22600 Hz
The frequency range of audible sound for human lies between 20 Hz to 20000 Hz.Since the frequency of the given sound is more than 20000 Hz, it is not audible.
15. What is reverberation? How can it be reduced?
Answer:
The persistence of sound in big hall due to repeated reflections from walls, ceiling, floor of the hall is called reverberation.
The excessive reverberation in big halls can be reduced by
(1) Carpets on floors
(2) Heavy curtains on doors and windows.
(3) Roof and wall are covered by sound absorbent material like compressed fibreboard, rough plasters.
16. What is loudness of sound? What factors does it depend on?
Answer:
Loudness is the measure of sound energy reaching the ear per second. Greater the sound energy reaching our ears per second, louder the sound will appear to be.
Loudness of sound depends upon amplitude, intensity of sound and sensitivity of the ear.
17. Explain how bats use ultrasound to catch a prey.
Bats fly in darkness of night without colliding with other objects by method of echolocation.
18. How is ultrasound used for cleaning?
Answer:
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.
19. Explain the working and application of a sonar.
Answer:
It is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects.The transmitter produces and transmit ultrasonic waves.These waves travel through water and after striking the objects on sea bed, get reflected back and are sensed by the detector.The detector converts the ultrasonic waves into electrical signal which are interpreted.The time taken for the echo to return to ship is measured by sonar apparatus.The distance of under water object is then calculated from time taken by echo to return.The time it takes for an echo to return is used to find out how far away something is.
20. A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.
Answer:
Time taken to hear the echo= 5 s
Distance of the object from submarine, d = 3625 m
Total distance travelled by the sonar waves during the transmission and reception in water = 2d
Velocity of sound in water , v = 2d /t = (2 × 3625) / 5 = 1450 m/s
21. Explain how defects in a metal block can be detected using ultrasound.
Answer:
Ultrasounds can be used to detect cracks and flaws in metal blocks. Metallic components are used in the construction of big structures like buildings, bridges, machines and scientific equipments. The ultrasound waves are allowed to pass through metal block to which are fitted detectors to detect the waves. If the metal block has a very small defect, such as an air bubble or a crack, then the ultrasound waves are reflected from such spots. The reflected ultrasonic waves indicates that metal block is defective.
22. Explain how the human ear works.
Answer:
Part of the ear which we see outside
(1) Pinna:Board part
(2) Ear canal:2-3 cm long passage
(3) Ear drum(Tympanum):At the end of ear canal is thin,elastic and circular membrane.
Middle Part
(1) It consist of three small and delicate bones called hammer,anvil and stirrup .They are linked to each other.
(2) Eustachian Tube:The lower part of middle ear has a narrow tube going to the throat.It connects middle ear to throat and ensures that air pressure inside middle ear is same as that on the outside.
Inner Part
(1) Cochlea:Coiled tube
One side is connected to middle ear through elastic membrane over window.The cochlea is filled with a liquid.The liquid contains nerve cells which are sensitive to sound.the other side of cochlea is connected to auditory nerve which goes into the brain.
Working
Sound waves—->Pinna—–>ear canal—–>Fall on ear drum
Sound waves consist of compression(high pressure regions)and rarefaction(low pressure region).
When compression of sound waves strikes ear drum the pressure on the outside of membrane increases and forces the eardrum to move inward.
When rarefaction of sound waves strikes ear drum the pressure on the outside of membrane decreases and forces the eardrum to move outward.
When sound waves fall on ear drum,it starts vibrating back and forth rapidly.Vibrating ear drum causes hammer to vibrate.
Hammer—–>Anvil—–>Stirrup(Vibration passes)—–>Oval window—-> Liquid of cochlea.
The vibrating liquid of cochlea sets up electrical impulse in nerve cell present in it.These impulses are carried by auditory nerve to the brain.The brain interprets these electrical impulses as sound and we get sensation of hearing.
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