Page No. 162
Ques.1. How does the sound produced by a vibrating object in a medium reach your ear?
Ans. When the object vibrates, it sets the neighboring particles to vibrate. These particles exert force on other particles and pass on the energy to other parts of medium. The particles do not get transported, but only the disturbance or energy is transferred. In this way, sound reaches our ears.
Page No. 163
Ques.1. Explain how sound is produced by your school bell.
Ans. When the hammer hits the gong of the bell, it starts vibrating. These vibrations set the particles of surrounding air vibrating. The disturbance travels in all directions and the sound propagates.
Ques.2. Why are sound waves called mechanical waves?
Ans. Sound waves are produced by oscillations of particles of the medium. So they require a material medium for their propagation. Thus they are called mechanical waves.
Ques.3. Suppose you and your friends are on the moon. Will you be able to hear any sound produced by your friend?
Ans. No, it is not possible to hear any sound on moon. There is no medium such as air on moon to carry sound waves. Sound cannot travel through vacuum as it is a mechanical wave.
Page No. 166
Ques.1. Which wave property determines
(a) loudness, (b) pitch?
Ans. (a) Loudness is determined by intensity or amplitude of sound waves .
(b) Pitch is determined by frequency of sound wave.
Ques.2. Guess which sound has a higher pitch: guitar or car horn?
Ans. Guitar, because it has higher frequency.
Ques.3. What are wavelength, frequency, time period and amplitude of a sound wave?
Ans. Wavelength: The distance between two consecutive compressions or two consecutive rarefactions is called the wavelength. It is devoted to A.
Frequency: The number of complete waves produced per second is called frequency of the wave.
Time Period: Time taken to complete one complete vibration is called time period.
Amplitude: The maximum displacement of the particles of the medium from their mean positions during the propagation of a wave is called amplitude of the wave.
Ques.4. How are the wavelength and frequency of a sound wave related to its speed?
Ans. Speed of sound = Frequency × Wavelength
Ques.5. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 ms1 in a given medium.
Ans. Given: frequency, v = 220 Hz; speed,
v = 440 ms-1
Ques.6. 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? [Speed o f sound in air = 330 ms-1]
Ans. Given: f = 500 Hz, v = 330 ms-1, d = 450 m
Time between two successive compressions is called time period.
Ques.7. Distinguish between loudness and intensity of sound.
|1. Loudness is a measure of response of our ear to the sound.||1. It is the energy passing per second per unit are a normal to the direction of energy flow.|
|2. It depends on intensity as well as sensitivity of ear. Therefore , it is not absolute but relative.||2. It does not depend upon the sensitivity of ear.|
|3. The unit of loudness is decibel.||3. The unit of intensity is watt/m2.|
Page No. 167
Ques.1. In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Ans. Sound travels fastest in solids due to their high elasticity, so out of given media sound travels fastest in iron.
Page No. 168
Ques.1. An echo is returned in 3s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 ms1 ?
Ans. If d is the distance of reflecting surface from the source and t is the time interval of echo-return, then
2d = vt ⇒ d = vt/2
Here, v - 342 ms-1, t = 3 s
Page No. 169
Ques.1. Why are the ceiling of concert halls curved?
Ans. Ceilings of concert halls are curved so that sound after reflection reaches all corners of the hall.
Page No. 170
Ques.1. What is the audible range of the average human ear?
Ans. Audible range for human ear = 20 Hz - 20 kHz
Ques.2. What is the range of frequencies associated with:
Ans. (a) Infrasound: Sound waves between the frequencies 1 to 20-Hz
(b) Ultrasound: Sound waves of the frequencies above 20,000 Hz.
Page No. 172
Ques.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 ms1, how far away is the cliff?
Ans. Given: t = 1.02 s, v = 1531 ms-1
Page No. 174
Ques.1. What is sound and how is it produced?
Ans. A sound is a form of energy that produces the sensation of hearing. It is produced by oscillation/ vibration of particles of a material medium.
Ques.2. Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Ques.3. Cite an experiment to show that sound needs a material medium for its propagation.
Ans. Fix an electric bell inside a jar connected to a vacuum pump. Close the mouth of jar and ring the bell. Slowly remove the air using the vacuum pump, so that vacuum is created in the jar. The sound will end when all the air is pumped out. This demonstrates the requirement for a medium to carry sound.
Ques.4. Why is sound wave called a longitudinal wave?
Ans. In sound wave, the vibration of the particles of the medium are along the direction of wave propagation, so sound waves in air are longitudinal.
Ques.5. Which characteristic of the sound helps you to identify your friend by his voice while sitting with Miens in a dark room?
Ans. It is the quality of sound (or waveform) which helps us to identify the voice of our friend without seeing him.
Ques.6. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Ans. The speed of light is 3 x 108 m/s while that of sound in air is only 330 m/s.
As Time = Distance/speed
So sound takes more time than light to reach on earth. That is why thunder is heard later than flash is seen.
Ques.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 ms1.
Ans. Wavelength of sound of frequency 20 Hz
wavelength of sound of frequency 20,000 Hz
Ques.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. Given velocity of sound in air and aluminium are 346 ms1 and 6420 ms1 respectively.
Ans. ∵ Time taken = Distance/speed
Ques.9. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Ans. ∵ 1 minute = 60 s
Vibrations in 1 s = frequency = 100.
Vibrations in 60 s = 60 x 100 = 6000 times
Ques.10. Does sound follow the same laws of reflection as light does? Explain.
Ans. Yes, sound and light follow the same laws of reflection given below :
(a) Angle of incidence at the point of incidence = Angle of reflection.
(b) At the point of incidence the incident sound wave, the normal and the reflected sound wave lie in the same plane.
Ques.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 remain the same. Do you hear echo sound on a hotter day?
Ans. Speed of sound is increased as temperature rises. Thus, echo reaches faster on a hotter day and it becomes difficult to detect the echo. Thus, echo can't be heard distinctly.
Ques.12. Give two practical applications of reflection of sound waves.
Ans. Applications of reflection of sound waves are
(a) to locate underwater hidden objects such as rocks and icebergs through SONAR.
(b) to detect any undesired objects in the sky near airports and borders of the country.
Ques.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.
Ans. Time is taken by stone to reach the water
t = 10 s
Time for the splash sound to travel back
t1 = 1.47 s
Time for the splash to be heard = 10 + 1.47 = 11.47 s from the time the stone is dropped.
Ques.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?
Ans. Given: Velocity of sound, v = 339 ms-1
Wavelength, λ = 1.5 cm - 0.015 m
Frequency, v =
= 22.6 kHz
This sound is not audible to human beings as frequency is higher than audible range.
Ques.15. What is reverberation? How can it be reduced?
Ans. The time from generation of sound until its loudness reduces to zero is called reverberation time. The process due to which the persistence of sound is caused is called reverberation. This is reduced in an auditorium using sound absorbent materials and good absorbers of sound.
Ques.16. What is loudness of sound? What factors does it depend on?
Ans. Loudness of sound is a measure of response of sound to our ear. Loudness of sound is not simply the energy reaching the human ear, but it also tells about the sensitivity of human ear detecting this energy. Loudness of sound is measured in decibel (dB). As energy reaching the ear depends on square of amplitude, so loudness of sound depends on two factors:
(i) Amplitude of sound waves and (ii) Sensitivity of ear.
Ques.17. Explain how bats use ultrasound to catch their prey.
Ans. The high pitched ultrasonic waves sent by the bats to the prey are received by them as they get reflected from the prey. This makes bats to move in right direction to catch their prey.
Ques.18. How is ultrasound used for cleaning?
Ans. Ultrasound is used to clean the hard to reach places such as spiral tubes, electronic components etc. Object to be cleaned is placed in the cleaning solution and ultrasonic waves are sent into it. The high frequency of ultrasound detaches the dust, grease and dirt from the object and it gets thoroughly cleaned.
Ques.19. Explain the working and application of a SONAR.
Ans. SONAR expands into Sound Navigation and Ranging with ultrasonic waves. It is used to locate underwater obstacles or enemy submarines etc. It is used by defence services to maintain a track of any trouble due to the enemies from sea. Sonar works on the principle of finding time taken by ultrasonic sound to travel back to the ship after hitting any object.
It is given by, t = 2d/v where v is the velocity of wave in the media and d is the distance of the object from the ship. It has a transmitter and receiver to send and receive ultrasonic waves. It is also used to find depth of ocean at any point.
Ques.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.
Ans. Given: Time, t = 5s;
Distance between submarine and object
= 3625 m.
Distance travelled by sound
= 3625 m x 2 (sound + echo) = 7250 m
Speed of sound in water =
= 1450 m/s.
Ques.21. Explain how defects in a metal block can be detected using ultrasound.
Ans. Ultrasonic waves are allowed to pass through the metal. If the block is flawless, it will pass through it. If there is crack or deformity, the wave gets reflected. Time taken by wave to return back is measured and helps to locate the flaw. If wave is not reflected, it means metal has no deformity.
Ques.22. Explain how the human ear works.
(a) Outer ear: It has a funnel-shaped ear pinna that collects the sound and transfers it to the eardrum. Eardrum starts vibrating when sound waves reach it.
(b) Middle ear: It has small bones called hammer, anvil, and stirrup. These bones amplify the vibrations of the eardrum.
(c) Inner ear: The amplified vibrations are transferred to the cochlea in which cochlear fluid converts them into signals which are transferred via the auditory nerve to the brain.