Question 1

What is sound and how is it produced?

Accepted Answer

Sound is defined as vibrations that travel through the air or another medium as an audible mechanical wave. It is produced from a vibrating body. The vibrating body causes the medium (water, air, etc.) around it to vibrate thus producing sound.

Question 2

Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound?

Accepted Answer

When a vibrating body swings forward, it pushes and compresses the air in front of it creating a region of high pressure. This region is called a compression. This compression starts to move away from the vibrating object.When the vibrating body swings backwards, it creates a region of low pressure called rarefaction. As the object swings / oscillates back and forth rapidly, a series of compressions and rarefactions is created in the air. These make the sound wave that propagates through air.

Question 3

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

Accepted Answer

Apparatus required: A strong jar, Electronic Bell, Cork, Vacuum pump. Procedure: Take an electronic bell and connect it to electrical supply. You should hear the bell ringing. This confirms that the bell is working. Now, take a jar and make a small hole in the bottom and insert a small pipe through it. Connect the other end of the pipe to a vacuum cleaner. Make a very small hole in the cork (just enough to insert wires) and insert wires connected to the electronic bell. Fit the cork as shown in the figure. Turn on the bell and the vacuum cleaner. You will observe no sound is heard.Conclusion:Since the bell was working but its sound could not be heard outside the jar when air was pumped out, it can be concluded that sound needs a medium for propagation.

Question 4

Why is sound wave called a longitudinal wave?

Accepted Answer

The vibration of medium that travel along or parallel to the direction of wave is called a longitudinal wave. In a sound wave, the particles of the medium vibrate in the direction parallel to the direction of propagation of disturbances.Hence, a sound wave is called a longitudinal wave.

Question 5

Which characteristics of the sound helps you to identify your friend by his voice while sitting with others in a dark room?

Accepted Answer

Timbre and pitch are the characteristics of sound which help to identify the sound of different voices. Thus, because of difference in timbre and pitch of the sound wave, a person can identify the voice of his friend sitting with others in a dark room.

Question 6

Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen. Why?

Accepted Answer

Speed of sound in air  (330 ms−1) is very less than speed of light in air  (3×108 m/s).Hence although flash and thunder are produced simultaneously, light (flash) reaches the earth surface a little earlier than sound (i.e thunder) .So thunder is heard a few seconds after the flash is seen.

Question 7

A person has a hearing range from 20Hz to 20KHz. What are the typical wavelengths of sound waves in air corresponding ton these two frequencies? Take the speed of sound in air as 344 ms−2.

Accepted Answer

Given :  Velocity of sound in air    v=344  ms−1Wavelength of the sound waves        λ=νv​For  ν1​=20  Hz,          λ1​=20344​ =17.2 mFor  ν2​=20  kHz,          λ2​=20×1000344​ =1.72 ×10−2 m

Question 8

Two children are at opposite ends of an aluminum 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 aluminum to reach the second child.

Accepted Answer

Let the distance between the two children be  dAlso let the time taken by sound to reach other child via air be  t1​ and  via  aluium rod be  t2​.Velocity of sound in air     va​=340  m/s∴     t1​=va​d​ =340d​         ..........(1)Velocity of sound in aluium rod     vR​=6420  m/s∴     t2​=vR​d​ =6420d​         ..........(2)Dividing equation (1) and (2),  we get        t2​t1​​=3406420​=18.88

Question 9

The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?

Accepted Answer

Frequency of source of sound         ν =100 HzThis means the source vibrates 100 times in 1 sec.∴ Number of times the source vibrates in 60 s is N = 60×100⟹ N=6000

Question 10

Does sound follow the same laws of reflection as light does? Explain.

Accepted Answer

Yes, sound also follows the law of reflection in the same manner as the light does.Law of reflection for sound-   The incident sound, the point of normal to the incident sound and the reflected sound all lie in same plane and also the angle of incidence with the normal &  the reflected angle with the normal  are equal.         ∠i=∠r

Question 11

When a sound is reflected from a distinct 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?

Accepted Answer

Velocity of sound increases with temperature. Thus velocity of sound in air increases in the hotter day. Hence the time difference between the incident wave and the reflected wave decreases.Now a human ear requires a imum time interval of  101​th of a second to hear an echo. Thus, if the time interval is greater than 0.1 second on the hotter day, then an echo can be heard.But, on a hotter day, the time gap between the incidence of original and reflected sound is less than 0.1s due to higher velocity of sound. So, chances of hearing echo is less.

Question 12

Give two practical applications of reflection of sound waves.

Accepted Answer

(i) Reflection of sound is used to measure the speed and distance of underwater objects. This method is called SONAR.(ii) Working of a stethoscope - the sound of patient's heartbeat reaches the doctor's ear through multiple reflections of sound.

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 ms2 and speed of sound 340 ms−1 .

Accepted Answer

Let the time to hear the splash be   t secondsGiven :      Height of the tower   H=500  m  Thus time taken by the stone to reach the bottom         t1​=g2H​​   t1​=102×500​​ =10  sSpeed of sound     v=340  ms−1Thus time taken by sound to reach the top     t2​=vH​=340500​=1.47 s∴  Total time taken to hear the sound of splash      t=t1​+t2​ =10+1.47=11.47 s

Question 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?

Accepted Answer

Velocity of sound wave      v=339  ms−1Wavelength of the waves      λ=1.5 cm = 0.015 m∴   Frequency of the waves      ν=λv​=0.015339​=22.6 kHzAudible frequency range for human ear lies between  20  Hz  to  20  kHzThus this waves is not audible.

Question 15

What is reverberation? How can it be reduced?

Accepted Answer

Persistence of sound (after the source stops producing sound) due to the repeated reflection is called reverberation. Reverberation can be reduced by absorbing the sound using some materials as it reaches the wall and ceiling of the room and thus prevent the sound from getting reflected. Some materials which are used to reduce reverberation are fibre board, heavy curtains, plastics etc.

Question 16

What is loudness of sound? What factors does it depend on?

Accepted Answer

Loudness is a characteristic of a sound. It refers to how much energy a sound wave posses.It depends on amplitude of the sound wave. More the amplitude, more is the loudness of the sound.

Question 17

Explain how bats use ultrasound to catch a prey.

Accepted Answer

Bats have the ability to produce high-pitched ultrasonic squeaks. These squeaks get reflected by objects like preys and return to their ears. This helps a bat to know how far his prey is.

Question 18

How is ultrasound used for cleaning?

Accepted Answer

Objects that need to be cleansed are put in a cleaning solution and ultrasonic sound waves are passed through the solution. The high frequency of ultrasound waves helps in detaching the dirt from the objects. In this way ultrasound is used for cleaning purposes.

Question 19

Explain the working and application of a sonar.

Accepted Answer

Working: Sonar consists of transmitter and a detector and is installed in a boat or a ship as shown in Fig. The transmitter produces and transmits ultrasonic waves. These waves travel through water and after striking the object on the seabed, get reflected back and are sensed by the detector.The detector converts the ultrasonic waves into electrical signals which are appropriately interpreted. The distance of the object that reflected the soundwave can be calculate by knowing the speed of sound in water and the time interval between transmission and reception of the ultrasound. Let the time interval between transmission and reception of ultrasound signal be tt and the speed of sound through sea water by vv. The total distance, 2d2d travelled by the ultrasound is then, 2d=vxt2d=vxt.Applications : The sonar technique is used to detere the depth of the sea and to locate underwater hills, valleys, submarine, icebergs, sunkenship etc.

Question 20

A sonar device on a submarine sends out a signal and receives an echo 5s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.

Accepted Answer

We know that, Speed=timeDistance​Time =5sDistance =3625 mTotal distance to be travelled by sound =2×3625, as it is the distance taken by sonar wave to transmit and receive =7250 mSo, Speed =57250​=1450m/s

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Solution for Class 9, Science, Sound

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