Class 9 - Physics - Sound
What is sound and how is it produced?
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.
Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
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.
Cite an experiment to show that sound needs a material medium for its propagation.
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.
This proves sound needs a material medium to propagate, it can’t propagate through vacuum.
Why sound wave is called a longitudinal wave?
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.
Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
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.
Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
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.
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.
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
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.
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
The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
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
Does sound follow the same laws of reflection as light does? Explain.
Yes, sound waves obey same law of reflection which light obeys.
The laws of reflection of sound waves are as follows:-
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?
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:-
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.
Give two practical applications of reflection of sound waves.
The two practical applications of reflection of sound waves are following:-
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.
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
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.
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?
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.
What is reverberation? How can it be reduced?
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.
What is loudness of sound? What factors does it depend on?
Loudness is sound of high energy. Loudness of sound is measured in decibel (dB). Following are the factors on which Loudness depends:-
Explain how bats use ultrasound to catch a prey.
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.
How is ultrasound used for cleaning?
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.
Explain the working and application of sonar.
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:-
Applications of SONAR:-
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.
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.
Explain how defects in a metal block can be detected using ultrasound.
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.
Explain how the human ear works.
The auditory aspect of human ear is as follows:-