Class 9 - Physics - Sound

Question1.

What is sound and how is it produced?

Answer:

Sound is a form of energy which is produced by vibration of objects. Vibrations are rapid to and fro motion of an object.

When a body vibrates, a disturbance is created in the medium by the particles.

This disturbance travels in the form of waves. This produces sound.

Physics Class 9 Sound Sound Wave 1

Question2.

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

Answer:

Compressions and rarefactions are produced because of the disturbance in medium caused by sound wave.

Sound waves propagate because of compressions and rarefactions of the particles of the medium.

Compressions are the regions of high pressure (high density of the particles).

Rarefactions are the regions of low pressure (low density of the particles).

When an object starts vibrating it creates disturbance in the medium.

Because of the disturbance of the particles of medium, they come closer to each other as compare to their normal position.

On the other hand adjacent particles go farther to each other. Both happen simultaneously.

The region where particles come closer to each other is called compressions and region where particles go farther from each other are called rarefactions.

Physics Class 9 Sound Compressions & Rarefactions

Physics Class 9 Sound Compressions & RareFaction 1

Question 3.

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

Answer:

By the following experiment it shows sound cannot travel through vacuum. It needs a medium to propagate.

Apparatus: - Electric bell, Airtight glass bell jar, Switch, vacuum pump.

Procedure: -

  1. Take an electric bell and an airtight glass bell jar. The electric bell is suspended inside the airtight bell jar.
  2. If we press the switch ’ON’ we will be able to hear the bell.
  3. Now when we connect the bell to the vacuum pump. All the air from the jar is pumped out gradually; the sound becomes fainter, although the same current is passing through the bell.
  4. After some time when more air is pumped out from the jar, we will hear a very feeble sound.
  5. And finally the sound will not be heard when all the air is completely removed from the jar.
  6. This happens as almost all air has been pumped out.

This proves sound needs a material medium to propagate, it can’t propagate through vacuum.

Physics Class 9 Sound Bell Jar Experiment

Question 4.

Why sound wave is called a longitudinal wave?

Answer:

Sound waves are longitudinal because particles of the medium travel parallel to the direction of propagation of waves.

In a longitudinal wave particles of the medium are oscillating in a direction parallel to the direction of propagation of wave.

 Physics Class 9 Sound Longitudinal Wave

Physics Class 9 Sound Spring

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:

Quality of sound is that characteristic which helps to identify a particular person.

Each person has its own quality of sound which will help us to identify our friend voice while 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?

Answer:

Flash and thunder both are produced simultaneously but thunder is heard a few seconds after the flash is seen because sound and light travel at different speeds.

 The speed of light is 3 x 108m/s while that of air is only 330m/s.

As time = (distance)/ (speed)

Sound takes more time than light to reach on earth.

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–1.

Answer:

Given:-

Speed of air v= 344 m s–1, Initial frequency ν1 = 20Hz and

Final velocity ν2 = 20 kHz = 20 x 103 Hz

Using λ = (v / ν)    where λ = wavelength, v = speed and ν= frequency

λ1 = (344/20) = 17.2m

λ2 = (344/20 x 103) =0.0172m

                                     

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:

Given:-

Speed of sound in air, v1 = 344m/s

Speed of sound in aluminium, v2 = 6420m/s

Let the length of the aluminium rod = d m.

Using Speed = (Distance)/ (Time)

Or, Time = (Distance)/ (Speed)

Time taken in air tair = (d/344) sec.

Time taken in aluminium ta = (d/6420) sec.

Required ratio (tair / ta) = (d/344)/ (d/6420) = 18.55:1

Question 9.

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

Answer:

Since the frequency of the source of sound is 100Hz.

Number of vibrations of the source in 1sec= 100

Number of vibrations of the source in 1 minute (i.e.60sec) = 100x60=6000

 Question 10.

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

Answer:

Yes, sound waves obey same law of reflection which light obeys.

The laws of reflection of sound waves are as follows:-

  1. The direction of incident sound wave, the direction of reflected sound wave, and normal at the point of incidence all lies on the same plane.
  2. The angle of incident sound wave is equal to the angle of reflected sound wave.

Physics Class 9 Sound 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:

Echo is the repetition of the same sound caused by the reflection of sound waves from a surface back to the listener.

To hear a distinct echo:-

  • Time interval between the original sound and the reflected sound must be at least 0.1s.
  • Minimum distance of the obstacle from the source of sound must be 17.2m.

In any medium if the temperature increases the speed of sound also increases.

Therefore on a hotter day, the time interval between the original sound and the reflected sound will decrease because velocity of sound is more.

 (Time taken = (Total distance/velocity)).

This means time taken by echo is less than 0.1sec as a result it won’t be heard.

Physics Class 9 Sound Echo

Question 12.

Give two practical applications of reflection of sound waves.

Answer:

The two practical applications of reflection of sound waves are following:-

  1. Working of a stethoscope is based on reflection of sound. The sound of patient’s heart reaches doctor’s ear by multiple reflections of sound.
  2. SONAR: - Reflection of sound is used to measure the distance and speed of underwater objects. This principle is used in SONAR’s.

Physics Class 9 Sound Sonar

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 s–2 and speed of sound = 340 m s–1.

Answer:

Given:-

Height of the tower, s = 500 m

Velocity of sound, v = 340 m s−1

Acceleration due to gravity, g = 10 m s −2

Initial velocity of the stone, u = 0 (since the stone is initially at rest)

Time taken by the stone to fall to the base of the tower, t1

According to second law of equation of motion:-

s=ut1 + (1/2) gt12

 500 = 0 x t1 + (1/2) x 10 x t12

  t12 =100

  t1 = 10s

Now, time taken by the sound to reach the top from the base of the tower,

  t2= (500 / 340) =1.47 s

Therefore, the splash is heard at the top after time, t

Total time taken t= t 1 + t 2 = 10 + 1.47 = 11.47 s.

Question 14.

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?

Answer:

Given:-

Speed of sound, v= 339 m s- 1

Wavelength of sound, λ= 1.5cm = 0.015 m

Speed of sound = Wavelength x Frequency= λ x v

∴ v= (v / λ) = (339 / 0.015) = 22600 Hz

Since the frequency of the given sound is more than 20,000 Hz, it is not audible.

Question 15.

What is reverberation? How can it be reduced?

Answer:

Reverberation is a sound produced in a closed area due to repeated reflection of sound.

It happens generally in a closed area.

Time interval between original and reflected sound is less than 0.1s.

For example: - In auditorium.

To reduce reverberation, (1) the roof and walls of the auditorium are generally covered with sound-absorbent materials like compressed fibreboard, rough plaster or draperies.

(2) The seat materials are also selected on the basis of their sound absorbing properties.

Physics Class 9 Sound Conference Hall

Question 16.

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

Answer:

Loudness is sound of high energy. Loudness of sound is measured in decibel (dB). Following are the factors on which Loudness depends:-

  1. Amplitude of sound wave.
  2. Sensitivity of ear.

Physics Class 9 Sound Sound Wave

Question 17.

Explain how bats use ultrasound to catch a prey.

Answer:

Bats search out for their preys by emitting and detecting reflections of ultrasonic waves.

The high pitched ultrasonic squeaks of bat are reflected from the obstacle or prey and return to bat’s ear.

 

This allows bats to know the distance of the prey from them.

Physics Class 9 Sound Bat

Question 18.

How is ultrasound used for cleaning?

Answer:

To clean any objects, it is placed in a cleaning solution and ultrasonic waves are sent into the solution.

Due to high frequency, the particles of dust, grease and dirt get detached and drop out. The objects are thoroughly cleaned.

Ultrasound waves have this unique property that they can pass through the obstacle and can go inside.

Question 19.

Explain the working and application of sonar.

Answer:

SONAR is an acronym of Sound Navigation and Ranging. Sonar is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects.

It works on the principle of echo.

Sonar consists of a transmitter and a detector and is installed in a boat or a ship, as shown in the figure.

Working of sonar:-

  1. The transmitter produces and transmits ultrasonic waves which get reflected by the underwater objects after striking them.
  2. These waves are received by the detector which can convert these waves into appropriate electrical signals.
  3. The distance of the object that reflected the sound wave can be calculated by knowing the speed of sound in water and the time interval between transmission and reception of the ultrasound.
  4. Let t=time interval between transmission and reception of the ultrasound signal. v = speed of sound through seawater.
  5. Total distance travelled by the ultrasound = 2d.
  6. Therefore 2d = v x t.

Applications of SONAR:-

  1. To determine the depth of the sea.
  2. To locate underwater hills, valleys, submarine, icebergs, sunken ship etc.

Physics Class 9 Sound Sonar

Question 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:

Given:-

Time t = 5s

Distance = 3625m

Total distance travelled by the sonar waves during the transmission and reception in water= 2d

Velocity of sound in water, v= (2d / t) = (2 x 3625 / 5) = 1450ms-1.

Question 21.

Explain how defects in a metal block can be detected using ultrasound.

Answer:

Ultrasound waves are high frequency sound waves.

They have one unique property that they can travel along well defined path even with obstacles.

Ultrasound waves can be transmitted through a metal block if metal block does not have any defects in it and is detected by detector.

But, if a metal block has defects, it won’t allow ultrasound waves to pass through it;

as a result, ultrasound waves will not be reach detector. Therefore defects in the metal block can be detected by the ultrasound.

Physics Class 9 Sound Metal Block 1

Physics Class 9 Sound Metal Block 2

Question 22.

Explain how the human ear works.

Answer:

The auditory aspect of human ear is as follows:-

  1. The human ear consists of three parts – the outer ear, middle ear and inner ear.
  2. The outer ear is called ‘pinna’. It collects the sound from the surroundings and directs it towards auditory canal.
  3. At the end of the auditory canal there is a thin membrane called the ear drum or tympanic membrane.
  4. When a compression reaches the eardrum, the pressure on the outside of the membrane increases and forces eardrum inward.
  5. Similarly, the eardrum moves outward when a rarefaction reaches it. In this way the eardrum vibrates.
  6. The vibrations are amplified several times by three bones (the hammer, anvil and stirrup) in the middle ear.
  7. The middle ear transmits the amplified vibrations of the sound wave to the inner ear.
  8. In the inner ear, the pressure variations are turned into electrical signals by the cochlea.
  9. These electrical signals are sent to the brain via the auditory nerve, and the brain interprets them as sound.

Physics Class 9 Sound Ear

Share this with your friends  

Download PDF


You can check our 5-step learning process


.