Chapter 12
Sound
Intext question and answer
Page no 162
1. How does the sound produced by a vibrating object in a medium-reach your ear?
Ans: Sound is a kind of wave motion produced by any vibrating body. There is no actual moment of air molecules from the sound-producing bodies to our ears. But, sound energy travels through the vibration of air molecules in the form of sound waves. Vibrations in an object create disturbance in the medium and consequently produce waves consisting of compressions and rarefactions. Because of these compressions and refractions sound reaches our ear.
Ans: Sound is a kind of wave motion produced by any vibrating body. There is no actual moment of air molecules from the sound-producing bodies to our ears. But, sound energy travels through the vibration of air molecules in the form of sound waves. Vibrations in an object create disturbance in the medium and consequently produce waves consisting of compressions and rarefactions. Because of these compressions and refractions sound reaches our ear.
Page no 163
1. Explain how sound is produced by your school Bell.
Ans: When the school Bell is struck with an object like a hammer, it starts vibrating through a small distance. This creates compression and rarefactions in air and sound is produced.
2. Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Ans: Since sound waves require medium to propagate and there is no medium present on the moon. So, I’ll not able to hear the sound of my friend on the moon.
3. Why are sound waves called mechanical waves?
Ans: Since sound waves do some mechanical work while making a disturbance in medium, hence sound waves are called mechanical waves.
Page no 166
1. Which wave property determines (a) loudness, (b) pitch?
(a) The amplitude of sound waves determines loudness. Louder sound has greater amplitude and vice versa.
(b) The frequency of the sound waves determines pitch of the sound.
2. Guess which sound has a higher pitch – guitar or car horn?
Ans: Guitar has a higher pitch.
Page no 166
1. What is wavelength, frequency, time period, and amplitude of a sound wave?
Ans: Wavelength: Wavelength is the distance between two consecutive compressions or rarefaction of the wave. Its SI unit is m.
Frequency: The number of sound waves, produced in one second is called frequency. It is measured in hertz.
Time period: Time period is a time taken to produce one wave of sound.
Amplitude: Amplitude is the maximum displacement of the particles of the medium from their mean position.
2. How are the wavelength and frequency of a sound wave related to its speed?
Ans: The relation between frequency and wavelength of a sound wave is given as follows:
Velocity(V)= 2 wavelength x frequency
V is = Lambda xv
This means the speed is equal to the product of wavelength and frequency of the sound wave. This equation is also called “the wave equation” and applicable to all types of waves.
3. Calculate the wavelength of a sound-wave whose frequency is 220 HZ and speed is 440 M/S given medium.
Given :
Frequency, f= 220 Hz
Velocity, V= to 440 m/s
Wavelength, Lambda?
We know,
Speed= wavelength x frequency
v = Lamda x V
Lamda= 440 / 220
Thus, the wavelength of the sound wave is 2 m.
4. A person is listening to a tone of 500-Hertz sitting at a distance of 450m from the source of the sound. What is the time interval between successive compressions from the source?
Ans: Since the time interval between successive compressions is called time period or time interval. Given,
Frequency, (v) = 500 Hertz
We know that frequency, V = 1/t
Therefore,
Thus, t is = 1/ 500 = 0 .002 s
Thus, the time interval between two consecutive compressions of the given wave is 0.002 S.
1. Which wave property determines (a) loudness, (b) pitch?
(a) The amplitude of sound waves determines loudness. Louder sound has greater amplitude and vice versa.
(b) The frequency of the sound waves determines pitch of the sound.
2. Guess which sound has a higher pitch – guitar or car horn?
Ans: Guitar has a higher pitch.
Page no 166
1. What is wavelength, frequency, time period, and amplitude of a sound wave?
Ans: Wavelength: Wavelength is the distance between two consecutive compressions or rarefaction of the wave. Its SI unit is m.
Frequency: The number of sound waves, produced in one second is called frequency. It is measured in hertz.
Time period: Time period is a time taken to produce one wave of sound.
Amplitude: Amplitude is the maximum displacement of the particles of the medium from their mean position.
2. How are the wavelength and frequency of a sound wave related to its speed?
Ans: The relation between frequency and wavelength of a sound wave is given as follows:
Velocity(V)= 2 wavelength x frequency
V is = Lambda xv
This means the speed is equal to the product of wavelength and frequency of the sound wave. This equation is also called “the wave equation” and applicable to all types of waves.
3. Calculate the wavelength of a sound-wave whose frequency is 220 HZ and speed is 440 M/S given medium.
Given :
Frequency, f= 220 Hz
Velocity, V= to 440 m/s
Wavelength, Lambda?
We know,
Speed= wavelength x frequency
v = Lamda x V
Lamda= 440 / 220
Thus, the wavelength of the sound wave is 2 m.
4. A person is listening to a tone of 500-Hertz sitting at a distance of 450m from the source of the sound. What is the time interval between successive compressions from the source?
Ans: Since the time interval between successive compressions is called time period or time interval. Given,
Frequency, (v) = 500 Hertz
We know that frequency, V = 1/t
Therefore,
Thus, t is = 1/ 500 = 0 .002 s
Thus, the time interval between two consecutive compressions of the given wave is 0.002 S.
Page no 166
1. Distinguish between loudness and intensity of sound
Ans: Loudness of sound: The loudness of sound is determined by the amplitude and the intensity of this sound wave is determined by the frequency of sound waves.
*Loudness of sound is determined by the amplitude.
*Loudness cannot be measured as a physical quantity because it is just a sensation that can be felt only.
* It is a subjective quantity. It depends upon the sensitivity of the human ear.
Intensity of this sound:
*Intensity of sound waves is determined by frequency of sound waves.
*It can be measured as a physical quantity.
*It is an object of physical quantity. It does not depend on the sensitivity of a human ear.
Page no 167
1. In which of the three media: air, water, iron, does the sound travel the fastest at a particular temperature?
Ans: At a particular temperature sound travels fastest in iron.
Page no 168
1. An echo returned in 3S. What is this distance of the reflecting surface from the source, given that the speed of sound is 342 m/s?
Ans: Given,
speed of sound= 342 m/s
Time = 3s
we know that,
The time taken by sound echo to travel in 3 seconds.
so, the time taken by sound to travel to or fro will be 3/2 = 1.5 seconds.
Distance= speed x time
Distance = 342 x 1.5 = 513 m.
Page no 169
1. Why are the ceilings of concert halls curved?
Ans: Since concrete halls are big, so the audience at the back rows of the Hall may not hear the clear sound of the speaker. To overcome this problem, the ceiling of the conquered halls is made concave. Concave ceiling helps the sound wave to reflect and sent to further distance which makes the concrete Hall enable to send clear sound to the audience even sitting in back rows of hall.
Page no 170
1. What is the audible range of the average human ear?
Ans: For an average human ear-command the audible range of frequency extends from 20 Hertz to 20,000 Hertz.
2. What is the range of frequencies associated with?
(a) infrasound
(b) ultrasound
Ans: (a) Infrasound less than 20 Hertz.
(b)Ultrasound more than 20 Hertz.
Page no 172
1. A submarine emits a sonar pulse which returns from the underwater Cliff in 1.02 seconds. If the speed of sound in saltwater is 1531 m/s, how far away is the Cliff?
Ans: To return the sonar pulse back, its wave has to travel two ways. Thus, the time taken by the sound to travel 1.02 seconds. So that the time taken by the sound to travel to or fro will be 1.02/2 = 0.51 seconds
Given,
Speed of sound wave =1531 m/s
Time, t= 0.51s
We know that,
Distance=speed x time
Distance=1531x 0.51 =780.81 m
So, the distance between the source and reflecting surface is 780.81
Important Points
- Sound is produced whenever an object is set into vibration, hence sound is a kind of mechanical energy.
- Sound travels as a longitudinal wave through a material medium.
- Sound travels as successive compressions and refractions in the medium.
- In sound propagation, it is the energy of sound that travels and not particles of the medium.
- Sound cannot travel in vacuum.
- The change in density from one maximum value to minimum value and against the maximum value makes one complete oscillation.
- The distance between two consecutive compressions or two consecutive refractions is called a wavelength.
- The time taken by the wave for one complete oscillation for the density or pressure of a medium is called time-period The unit of the time-periodis second.
- The number of complete oscillationsper unit-time is called the frequency. The speed, frequency, and wavelength of sound are related by the equation V is equal to frequency X wavelength.
- The speed of sound depends primarily on nature under temperature of the transmission medium.
- The law of reflection of a sound states that the direction in which the sound is incident and reflected makes equal angles with the normal to the reflective surface at the point of incidence and they’re on the same plane.
- For hearing or distant sound, the time intervals between the original sound and the reflected one must be at least 0.1 S.
- The precipitation of sound in an auditorium is the result of the reproductive reflection of sound and collected reverberation.
- Sound properties such as pitch, loudness, and quality are data that remind by corresponding very properties please stop.
- Loudness is a physical-psychological response of the ear to an instant-sound intensity of sound. Next, the amount of sound energy passing each sound through the unit area is called the intensity of sound.
- Sound waves which frequency bill audible range are termed as inference and the audible range is called Arthur Sonic.
- Ultrasonic has many medical and industrial applications.
- Humans are able to hear with the help of an extremely sensitive device, the ear which allows the convert pressure variation into the audible frequency into electrical signals that travels to the brain via the auditor in knows.
Exercise question and Answer
1. What is sound and how it is produced?
Ans: Sound is a kind of energy produced in the form of phase when anything is set to vibrate it produces sound.
2. Describe with the help of a diagram how compression and rarefaction are produced in the air near a source of sound.
Ans: Compression and rarefaction in the air: Compression under reflection are produced because of the distribution in media caused by Soundwave. Sound-waves propagate because of compression and rarefaction of the particles of the medium.
When an object starts vibrating, it creates a disturbance in the medium. Because of the disturbance produced, the particles of medium come closer to each other compared to the normal position on the other hand-adjacent particles go further to each other will stop both happen simultaneously.
The region where particles come closer to each other is called compression and the region where particles go further to each other is called rarefaction. The next line in the given figure-comma straight line is showing the normal position of air particles. Denser lines are showing the region of compression and lesser than sir lines are showing the region of rarefaction of air particles.
1. What is sound and how it is produced?
Ans: Sound is a kind of energy produced in the form of phase when anything is set to vibrate it produces sound.
2. Describe with the help of a diagram how compression and rarefaction are produced in the air near a source of sound.
Ans: Compression and rarefaction in the air: Compression under reflection are produced because of the distribution in media caused by Soundwave. Sound-waves propagate because of compression and rarefaction of the particles of the medium.
When an object starts vibrating, it creates a disturbance in the medium. Because of the disturbance produced, the particles of medium come closer to each other compared to the normal position on the other hand-adjacent particles go further to each other will stop both happen simultaneously.
The region where particles come closer to each other is called compression and the region where particles go further to each other is called rarefaction. The next line in the given figure-comma straight line is showing the normal position of air particles. Denser lines are showing the region of compression and lesser than sir lines are showing the region of rarefaction of air particles.
3. Explain an experiment to show that sound needs a material medium for its propagation.
Ans. Take a glass Bell jar, connect it with a vacuum pump, and suspend an electric bill in it.
This happens because after creating a vacuum in the Bell jar there was no air present through which sound waves propagate.
This experiment shows that without medium sound cannot propagate and hence for the propagation of sound medium must be present.
Ans. Take a glass Bell jar, connect it with a vacuum pump, and suspend an electric bill in it.
- Connect the electric bell with the battery.
- Switch on the electrical Bell and hear the sound of Bell.
- Now remove the air completely from the Bell jar using a vacuum pump and observe the sound of the electric bell.
This happens because after creating a vacuum in the Bell jar there was no air present through which sound waves propagate.
This experiment shows that without medium sound cannot propagate and hence for the propagation of sound medium must be present.
4. Why is a sound wave called a longitudinal wave?
Ans: Since soundwave creates an oscillation in the particles of medium parallel to the distribution in the direction of propagation, the Soundwave is called Langley to dinner way.
5. Which characteristics of the sound help you to identify your friend by his voice while sitting with others in a dark room?
Ans: Timber and pitch are the characteristics of sound which helps to identify the sound of a different voice. Thus, because of sound in timber and pitch of the sound wave, I or any other can identify the voice of his friend sitting with others even in a dark room.
6. Flash and thunder are produced simultaneously. But under-thunder is heard a few seconds after the flash is seen, why?
Ans: This happens because of the difference in the velocity of light and sound waves. Light travels with much faster velocity than sound. That’s why Thunder is heard a few seconds after the flash of Thunder is seen instead of both are produced simultaneously.
7. A person has a hearing range from 20 Hertz to 20 kilohertz. Please note what are the typical wavelengths of a sound-wave in air corresponding to their two frequencies? Take the speed of sound in air as 344 m/s.
Ans: Given, the velocity of sound=344m/s
We know that,
Velocity= wavelength/ frequency
In case of frequency of 20 Hz,
Wavelength=344/20=17.2m
In case of frequency of20000Hz
Wavelength=344/20000Hz =0.0172m
8. Two children are at opposite ends of an aluminum-rod. One strikes the end of the road with a stone. Find the ratio of times taken by sound waves in aluminum and in the air to reach the second child.
Ans: We know that,
The speed of sound in air = 344 m/s.
The speed of sound in aluminum = 5100 m/s.
Hence, the ratio of time taken by the sound to travel through the air and through aluminum will be = velocity of sound in aluminum/velocity of sound in air = 5100 / 344 = 14.82
9. The frequency of a source of sound is 100 Hertz. How many times does it vibrate in a minute?
Ans: Given, frequency = 100 Hz.
This means the source of sound vibrates 100 times in one second.
Therefore, the number of vibrations in one minute, i.e., In 60 seconds =100X60= 6000 times.
10. Does sound follow the same laws of reflection as light does? Explain.
Ans: Yes, the sound wave follows the same laws of reflection as the light does. The laws of reflection of sound is as follows the incident Soundwave, the reflected sound wave and the normal at the point of the incident, all lie in the same plane.
The angle of incidence of sound wave and angle of reflection of sound wave to the normal are equal.
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: The sound of echo depends upon the distance from the source of sound and reflecting-surface. The distance between both must be equal 2 or more than 17.2 meters. If the given distance is more than 17.2 meters—then-one can hear the echo sound on a hotter day also.
On the hotter day the velocity of sound increases thus it is necessary to hear the sound of the echo, the distance should be more than 17.2 meters. If the given distance is equal to 17.2 meters, then to hear the sound on a hotter day would not be possible.
12. Give two practical applications of reflection of sound waves.
Ans: Bulk Horan and stethoscope are examples of practical applications of reflection of sound waves.
In bulk Horan, sound is amplified and sent to the desired direction because of reflection. In a stethoscope also, sound is sent in the desired direction because of its reflection characteristic.
13. As a stone is dropped from the top of a tower 500-meters high into a pond of water at the base of the tower. When is the splash here at the top? Given, g=10m/s2 and speed of sound = 340m/s.
Ans: Given,
Height of tower = 500 meter
Acceleration due to gravity, g =10 m/s
Velocity of sound =340 m/s
Thus, to calculate the time of splash-sound, first of all time taken to reach the stone in the water is to be calculated.
We know that,
S=ut+1/2gt2
Substituting the values, we get
500 =oXt+1/2X10Xt2
500=5Xt2
500/ 5 = t2
T=10s
Now, we know that
Distance= speed x time
Time =500/ 340=1.47 s
Thus, the total time taken to hear the sound of splash = 10 + 1.47 s = 11.47 s
14. A sound wave travels at a speed of 339 m/s. If its wavelength is 1.5 cm, what is the frequency of the wave?
Ans: Given,
The velocity of sound, v = 339 ms
Wavelength = 1.5 cm =0. 015 m
Frequency = f =?
We know that, speed = wavelength x frequency
339 = 0.015 m x frequency
Frequency = 339/0.015 = 22600 Hz.
Thus, frequency = 22600 Hz.
15. What is reverberation? How can it be reduced?
Ans: The persistence of sound waves for a long time due to multiple reflections from the walls, ceiling, and floor of the hall is known as reverberation.
The reverberation can be decreased by using sound observing materials in the auditorium.
The windows of the auditorium are covered with heavy curtains.
The special tiles known as acoustic tiles are used for the flooring of the auditorium.
Padded seats are arranged in the Hall.
Some plants employ some plants in pots that are also arranged in the hall to reduce the reverberation time.
16. What is loudness of sound? What factors does it depend on?
Ans: Loudness is a measure of the sound energy reaching the ear per second. It depends on the amplitude of the vibrating body producing the sound. Greater the amplitude of sound waves, louder the sound will be. A sound produced from the body vibrating with a large amplitude creates a loud sound. On the other hand, a sound from the vibrating body with a small amplitude produces a feeble or soft sound. Loudness is a subjective quantity as it depends on the sensitivity or the response of our ears.
17. Explain how bats use ultrasound to catch a break?
Ans: Bats can produce ultrasonic waves by flapping their wings and can also detect these waves. The ultrasonic waves produced by the bats after reflection from the obstacles like building guide them to remain away from the obstacles during their flights. Hence, they can fly at night without hitting the obstacles. This helps them to catch their prey at night with the help of ultrasonic waves. The ultrasonic waves produced by a bat spread out. These waves—after reflecting from a prey reach the bat. Thereby, the bat can easily locate its prey. The ears of the bat are so sensitive and drained that they not only get information of the distance of the obstacle but also the nature of reflecting surface.
18. How ultrasound is used for cleaning the objects?
Ans: The method used for cleaning the objects with the help of ultrasonic waves is known as ultrasonic cleaning. The instrument or device—as electronic components, spiral tubes to be cleaned—is dipped in liquid. The ultrasonic waves are passed through this liquid. Due to the high frequency of these waves, the dirt or any other impurity is removed from the parts of the instrument which cannot be approached directly.
19. Explain the working and application of sonar.
Ans: The word SONAR stands for “sound navigation and ranging “. It is a device that is used in the ships to locate rocks, icebergs, submarines, old ships sunk in the sea. It’s also used to measure the depth of the sea. It is based on the principle of the reflection of the sound wave.
A beam of ultrasonic waves from the transmitter of SONAR-fitted on the ship is sent towards the bottom of the sea at regular intervals. When these beams are intercepted by an object, they are reflected back from the bottom of the sea. The reflected-beam is received by the receiver or the detector of these—SONAR in the ship. The shape and position of objects under the sea and ocean is detected on the basis of speed and nature of reflected ultrasound waves. That time taken by the ultrasonic waves to go from the ship to the bottom of the sea and then back to the ship is noted.
Let it” t “seconds. Therefore, the time taken by the ultrasonic waves to go from the ship to the bottom of the sea is t/2 seconds.
Using the following formula,
S = t/2V, we can find the depth of the sea.
Ans: Since soundwave creates an oscillation in the particles of medium parallel to the distribution in the direction of propagation, the Soundwave is called Langley to dinner way.
5. Which characteristics of the sound help you to identify your friend by his voice while sitting with others in a dark room?
Ans: Timber and pitch are the characteristics of sound which helps to identify the sound of a different voice. Thus, because of sound in timber and pitch of the sound wave, I or any other can identify the voice of his friend sitting with others even in a dark room.
6. Flash and thunder are produced simultaneously. But under-thunder is heard a few seconds after the flash is seen, why?
Ans: This happens because of the difference in the velocity of light and sound waves. Light travels with much faster velocity than sound. That’s why Thunder is heard a few seconds after the flash of Thunder is seen instead of both are produced simultaneously.
7. A person has a hearing range from 20 Hertz to 20 kilohertz. Please note what are the typical wavelengths of a sound-wave in air corresponding to their two frequencies? Take the speed of sound in air as 344 m/s.
Ans: Given, the velocity of sound=344m/s
We know that,
Velocity= wavelength/ frequency
In case of frequency of 20 Hz,
Wavelength=344/20=17.2m
In case of frequency of20000Hz
Wavelength=344/20000Hz =0.0172m
8. Two children are at opposite ends of an aluminum-rod. One strikes the end of the road with a stone. Find the ratio of times taken by sound waves in aluminum and in the air to reach the second child.
Ans: We know that,
The speed of sound in air = 344 m/s.
The speed of sound in aluminum = 5100 m/s.
Hence, the ratio of time taken by the sound to travel through the air and through aluminum will be = velocity of sound in aluminum/velocity of sound in air = 5100 / 344 = 14.82
9. The frequency of a source of sound is 100 Hertz. How many times does it vibrate in a minute?
Ans: Given, frequency = 100 Hz.
This means the source of sound vibrates 100 times in one second.
Therefore, the number of vibrations in one minute, i.e., In 60 seconds =100X60= 6000 times.
10. Does sound follow the same laws of reflection as light does? Explain.
Ans: Yes, the sound wave follows the same laws of reflection as the light does. The laws of reflection of sound is as follows the incident Soundwave, the reflected sound wave and the normal at the point of the incident, all lie in the same plane.
The angle of incidence of sound wave and angle of reflection of sound wave to the normal are equal.
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: The sound of echo depends upon the distance from the source of sound and reflecting-surface. The distance between both must be equal 2 or more than 17.2 meters. If the given distance is more than 17.2 meters—then-one can hear the echo sound on a hotter day also.
On the hotter day the velocity of sound increases thus it is necessary to hear the sound of the echo, the distance should be more than 17.2 meters. If the given distance is equal to 17.2 meters, then to hear the sound on a hotter day would not be possible.
12. Give two practical applications of reflection of sound waves.
Ans: Bulk Horan and stethoscope are examples of practical applications of reflection of sound waves.
In bulk Horan, sound is amplified and sent to the desired direction because of reflection. In a stethoscope also, sound is sent in the desired direction because of its reflection characteristic.
13. As a stone is dropped from the top of a tower 500-meters high into a pond of water at the base of the tower. When is the splash here at the top? Given, g=10m/s2 and speed of sound = 340m/s.
Ans: Given,
Height of tower = 500 meter
Acceleration due to gravity, g =10 m/s
Velocity of sound =340 m/s
Thus, to calculate the time of splash-sound, first of all time taken to reach the stone in the water is to be calculated.
We know that,
S=ut+1/2gt2
Substituting the values, we get
500 =oXt+1/2X10Xt2
500=5Xt2
500/ 5 = t2
T=10s
Now, we know that
Distance= speed x time
Time =500/ 340=1.47 s
Thus, the total time taken to hear the sound of splash = 10 + 1.47 s = 11.47 s
14. A sound wave travels at a speed of 339 m/s. If its wavelength is 1.5 cm, what is the frequency of the wave?
Ans: Given,
The velocity of sound, v = 339 ms
Wavelength = 1.5 cm =0. 015 m
Frequency = f =?
We know that, speed = wavelength x frequency
339 = 0.015 m x frequency
Frequency = 339/0.015 = 22600 Hz.
Thus, frequency = 22600 Hz.
15. What is reverberation? How can it be reduced?
Ans: The persistence of sound waves for a long time due to multiple reflections from the walls, ceiling, and floor of the hall is known as reverberation.
The reverberation can be decreased by using sound observing materials in the auditorium.
The windows of the auditorium are covered with heavy curtains.
The special tiles known as acoustic tiles are used for the flooring of the auditorium.
Padded seats are arranged in the Hall.
Some plants employ some plants in pots that are also arranged in the hall to reduce the reverberation time.
16. What is loudness of sound? What factors does it depend on?
Ans: Loudness is a measure of the sound energy reaching the ear per second. It depends on the amplitude of the vibrating body producing the sound. Greater the amplitude of sound waves, louder the sound will be. A sound produced from the body vibrating with a large amplitude creates a loud sound. On the other hand, a sound from the vibrating body with a small amplitude produces a feeble or soft sound. Loudness is a subjective quantity as it depends on the sensitivity or the response of our ears.
17. Explain how bats use ultrasound to catch a break?
Ans: Bats can produce ultrasonic waves by flapping their wings and can also detect these waves. The ultrasonic waves produced by the bats after reflection from the obstacles like building guide them to remain away from the obstacles during their flights. Hence, they can fly at night without hitting the obstacles. This helps them to catch their prey at night with the help of ultrasonic waves. The ultrasonic waves produced by a bat spread out. These waves—after reflecting from a prey reach the bat. Thereby, the bat can easily locate its prey. The ears of the bat are so sensitive and drained that they not only get information of the distance of the obstacle but also the nature of reflecting surface.
18. How ultrasound is used for cleaning the objects?
Ans: The method used for cleaning the objects with the help of ultrasonic waves is known as ultrasonic cleaning. The instrument or device—as electronic components, spiral tubes to be cleaned—is dipped in liquid. The ultrasonic waves are passed through this liquid. Due to the high frequency of these waves, the dirt or any other impurity is removed from the parts of the instrument which cannot be approached directly.
19. Explain the working and application of sonar.
Ans: The word SONAR stands for “sound navigation and ranging “. It is a device that is used in the ships to locate rocks, icebergs, submarines, old ships sunk in the sea. It’s also used to measure the depth of the sea. It is based on the principle of the reflection of the sound wave.
A beam of ultrasonic waves from the transmitter of SONAR-fitted on the ship is sent towards the bottom of the sea at regular intervals. When these beams are intercepted by an object, they are reflected back from the bottom of the sea. The reflected-beam is received by the receiver or the detector of these—SONAR in the ship. The shape and position of objects under the sea and ocean is detected on the basis of speed and nature of reflected ultrasound waves. That time taken by the ultrasonic waves to go from the ship to the bottom of the sea and then back to the ship is noted.
Let it” t “seconds. Therefore, the time taken by the ultrasonic waves to go from the ship to the bottom of the sea is t/2 seconds.
Using the following formula,
S = t/2V, we can find the depth of the sea.
Where V is the speed of an ultrasonic wave in water and S is the depth of the sea.
The method of finding the distance of an object using echo is called echo-ranging. While the method of finding the depth of the sea-using echo is called echo depth-ranging. In this case, SONAR is called a fathometer.
20. A sonar device on a submarine sends out a signal and receives an ecosystem later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.
Ans: Given, the time taken by the ultrasonic sound waves to travel from the ship to the seabed, and back to the ship in 5 seconds. So, the time taken by the ultrasonic sound to travel from the ship to sea bed will be half of this time, which is 5/2 is = 2.5 seconds. This means that the sound takes 2.5 seconds to travel from the ship to the bottom of the sea.
Distances = 3625 m
We know that,
Speed = distance / time
Substituting these values in the above formula, we get
Speed = 3625/2.5 = 1450 m/s,
Thus, the speed of sound in seawater is 1450 m/S.
21. Explain how defects in a metal block can be detected using ultrasound?
Ans: The beams of ultrasonic waves are thrown on metal under investigation to detect the crack or fault. The ultrasonic waves get reflected from all the parts. However, the intensity of the ultrasonic waves gets reflected from the fault or crack is different from the intensity of the waves reflected from the other part of the metal. Thus, the position of the fault or a crack in the metal can be easily located. In fact, the picture of the metal produced by the reflection of ultrasonic waves is taken on the screen or the monitor. By analyzing, the position of the crack or fault in the metal is detected.
The method of finding the distance of an object using echo is called echo-ranging. While the method of finding the depth of the sea-using echo is called echo depth-ranging. In this case, SONAR is called a fathometer.
20. A sonar device on a submarine sends out a signal and receives an ecosystem later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.
Ans: Given, the time taken by the ultrasonic sound waves to travel from the ship to the seabed, and back to the ship in 5 seconds. So, the time taken by the ultrasonic sound to travel from the ship to sea bed will be half of this time, which is 5/2 is = 2.5 seconds. This means that the sound takes 2.5 seconds to travel from the ship to the bottom of the sea.
Distances = 3625 m
We know that,
Speed = distance / time
Substituting these values in the above formula, we get
Speed = 3625/2.5 = 1450 m/s,
Thus, the speed of sound in seawater is 1450 m/S.
21. Explain how defects in a metal block can be detected using ultrasound?
Ans: The beams of ultrasonic waves are thrown on metal under investigation to detect the crack or fault. The ultrasonic waves get reflected from all the parts. However, the intensity of the ultrasonic waves gets reflected from the fault or crack is different from the intensity of the waves reflected from the other part of the metal. Thus, the position of the fault or a crack in the metal can be easily located. In fact, the picture of the metal produced by the reflection of ultrasonic waves is taken on the screen or the monitor. By analyzing, the position of the crack or fault in the metal is detected.
22. Explain how the human ear works.
Ans: The external ear catches sound waves and channelizes them to the eardrum, via the ear canal. During compressions (high-pressure region), the pressure increases outside the eardrum which forces the eardrum to move inwards. During rarefactions (low-pressure regions), the pressure decreases outside the eardrum which forces the eardrum to move outwards. Thus, a vibration is produced in the eardrum. Further, the three bones i.e., malleus, incus and stapes amplify the sound-wave, by vibrating in turns. In the inner ear, the vibrations are converted into electrical signals. These signals are transmitted by the auditory nerve to the brain. Finally, the ear interprets those signals as sound.
Ans: The external ear catches sound waves and channelizes them to the eardrum, via the ear canal. During compressions (high-pressure region), the pressure increases outside the eardrum which forces the eardrum to move inwards. During rarefactions (low-pressure regions), the pressure decreases outside the eardrum which forces the eardrum to move outwards. Thus, a vibration is produced in the eardrum. Further, the three bones i.e., malleus, incus and stapes amplify the sound-wave, by vibrating in turns. In the inner ear, the vibrations are converted into electrical signals. These signals are transmitted by the auditory nerve to the brain. Finally, the ear interprets those signals as sound.
