Important Questions for CBSE Class 9 Science Chapter 12 – Sound

CBSE Important Questions of Sound Class 9 Science Chapter 12 

Science is a subject that requires deep study and understanding of the concepts associated. Hence, one must study properly to get a clear understanding of it and have strong command over all the topics. In this way, one can score well in the examinations.

The chapter ‘Sound’ deals with studying the properties and applications of sound in different mediums and sounds. You will also learn how the sound reaches our ears. The vital topics included in Chapter 12 Class 9 Science important questions are:

• Sound
• Production of Sound
• Propagation of Sound
• Reflection of Sound
• Range of Hearing
• Applications of Ultrasound
• Structure of Human Ear

The important questions Class 9 Science Chapter 12 covers all the topics listed above in a detailed manner. That gives students a complete idea of the chapter and its associated topics. Thus, ensuring they are a step ahead in their preparation.

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CBSE Class 9 Science Important Questions with Solutions

Also, get access to CBSE Class 9 Science Important Questions for other chapters too:

Sound Class 9 Important Questions with Answers

Science Class 9 Chapter 12 important questions include the following important questions and their solutions:

Question 1. The note is a sound

(a) mixing of several specified frequencies

(b) mixing of only two frequencies 

(c) only one frequency

(d) always unpleasant to hear

Answer:

The answer is (a) Mixing of several specified frequencies

Explanation:

Sound is the mixture of several frequencies that are generated by vibrating objects.

Question 2. A mechanical piano key struck softly, then again, but much louder this time. In the second case 

(a) the sound will be louder, but the pitch will be no different 

(b) the sound will be more audible, and the pitch will also be higher 

(c) the sound would be louder, but the pitch will be lower  

(d) both volume and tone are not affected

Answer: 

The answer is (a) the sound will be louder, but the pitch will be no different. 

Explanation:

Pitch depends on the frequency of the particular key, and loudness depends on the force with which the key is pressed.

Question 3. In SONAR, we tend to use

(a) ultrasonic waves

(b) infrasonic waves

(c) radio waves

(d) audible sound waves

Answer:

The answer is (a) ultrasonic waves

Question 4. Sound travels in the air when  

(a) the particles of the medium migrate from one place to another  

(b) there is not any moisture in the atmosphere  

(c) the disturbance moves  

(d) both particles and the disruptions you travel from one place to another.

Answer: 

The answer is (c) the disturbance moves 

Explanation:

Sound waves propagate by vibrating in their position, but the disturbance created by the vibration of the particles moves from one place to another.

Question 5. When we transform a weak sound into a strong sound, we increase its 

(a) frequency 

(b) amplitude 

(c) velocity

(d) wavelength

Answer: 

The answer is (b) amplitude

Explanation:

The intensity of the sound is proportional to its amplitude. As the amplitude increases, the weak sound changes to a strong sound.

Question 6. In the curve (Fig.), half the wavelength is

(a) A E

(b) B D

(c) D E

(d) A B

Answer:

The answer is (b) B D

Explanation:

Wavelength is the distance between the two given consecutive troughs. In the given graph, half the wavelength is BD.

Question 7. An earthquake produces which kind of sound before the primary shock wave begins

(a) audible sound

(b) infrasound

(c) ultrasound 

(d) none of the above

Answer:

The answer is (b) infrasound

Explanation:

Because of the infrared rays, few animals sense the earthquake tremors, and hence they behave abnormally before the earthquake begins. .

Question 8. Infrasound can be heard by

(a) human beings

(b) bat

(c) rhinoceros

(d) dog

Answer:

The answer is (c) rhinoceros.

Explanation:

Infrasound has a frequency of less than 20 Hz, so Rhinoceroses communicate using  5 Hz infrasound waves. Hence, rhinoceros is the correct answer.

Question 9. Before an orchestra plays in a music concert, a sitar player tries to adjust the tension and strum the string properly. By doing this, you adapt

(a) the intensity of sound only

(b) loudness of sound

(c) frequency of the sitar string with the frequency of other musical instruments

(d) the amplitude of sound only

Answer:

The answer is (c) the frequency of the sitar string with the frequency of other musical instruments.

Explanation:

Artists adjust the required frequencies before beginning to play instruments because the musical instruments will be tuned in by other available musical instruments to produce pleasant music.

Question 10. The wavelength of the sound wave has units: 7

 (a) metres

(b) metres/sound 

(c) (meters)²

(d) meters/second²

Answer: (a) metres

Question 11. Light is a

(a) longitudinal wave

(b) transverse wave

(c) both

(d) none

Answer: (b) Transverse wave

Question 12. In compression, pressure density is

(a) high

(b) less

(c) remains the same

(d) maybe a) or b) depending upon disturbance 

Answer: (a) High

Question 13. The frequency of ultrasonic sound waves is

(a) greater than 20 HZ

(b) greater than 20,000 HZ

(c) greater than 2 HZ

(d) greater than 2 MHZ

Answer:  (b) Greater than 20,000 Hz

Question 14. Stethoscopes work on the principle of: 

(a) multiple reflections of the sound 

(b) ultrasounds 

(c) both a and b 

(d) none of the above

Answer: (a) multiple reflections of the sound

Question 15. The Audible Range of the human ear is:

(a) 20 HZ – 20,000 HZ

(b) 20 HZ – 20 MHZ

(c) 20 HZ – 20 kHZ

(d) Both a) and c)

Answer: (d) Both a) and c)

Question 16. The order of bones is the human ear from outside to inside:

(a) Hammer, Stirrup Anvil

(b) Stirrup, Hammer, and Anvil 

(c) Anvil, Stirrup, and Hammer

(d) Hammer, Anvil, and Stirrup

Answer: (a) Hammer, stirrup Anvil

Question 17. Which of the following is used in echocardiography?

(a) ultrasound waves

(b) X-Ray waves

(c)  infrasound waves

(d) both a) and c)

Answer: (a) Ultrasound waves

Question 18. Infrasound is produced by:

(a) bats

(b) dogs

(c) rhinoceros

(d) rats

Answer: (c) Rhinoceros

Question 19. Speed of sound is maximum in:

(a) solids

(b) liquids

(c) gases

(d) plasma 

Answer: (a) Solids

Question 20. The innermost part of the ear is called as

(a) cochlea

(b) pinna

(c) hammer

(d) anvil

Answer: Cochlea

Question 21. The graph (Fig.) shows the displacement versus time relation for a disturbance travelling with a velocity of 1500 m s–1. Calculate the wavelength of the disturbance.

Answer:

T= 2×10⁶s

Frequency is given by

v= 1/T

= 105 Hz

Wavelength is given by

λ = v/V = 5×10⁵m

Question 22. Which of the two graphs above (a) and (b) (Fig.) representing the human voice is likely to be the male voice? Justify your answer.

Answer:

The pitch of the male voice is lighter than the pitch of the female; hence the given graph represents the male voice.

Question 23. S. I. units of frequency is:

Answer: The S.I.-derived unit for the frequency is Hertz (Hz). One hertz means that a particular event repeats once per second. The SI unit for the given period is the second.

Question 24. A girl sits in the middle of a  12 m × 12 m park. On the left is a building that connects to the park, and on the right of the park is a path that adjoins the park. A sound is created on the road by a cracker. Is it possible for the girl to listen to the echo of this sound? Justify your answer.

Answer:

An echo will be detected if the gap between the original sound and the reflected sound received at the listener’s end is around 0.1 seconds.

Sound velocity × time interval

= 344×0.1

= 34.4 m

where sound reflects from the building and reaches the girl that is way smaller than the desired distance. Thence can not be heard.

Question 25. Why do we hear the sound produced by the humming bees while the sound of vibrations of a pendulum is not heard?

Answer:

Humming bees produce pleasant sounds by beating their wings, and the frequency of the sound they make is in the range of 20Hz to 20000 Hz, which is audible. On the other hand, the pendulum makes the sound less than 20 Hz, below the audible range; hence, we don’t hear the sound of pendulum vibrations.

Question 26. If any explosion occurs at the bottom of a lake, what type of shock waves in the water will take place?

Answer:

The answer is longitudinal waves.

Question 27. The sound produced by a thunderstorm is heard 10 s after the lightning is seen. Calculate the approximate distance of the thundercloud. (Given the speed of sound = 340 ms^–1)

Answer:

Distance = speed ×time

Here speed = 340ms^-1

Time = 10s

= 340× 10 = 3400m.

Question 28. For hearing the loudest ticking sound heard by the ear, find the angle x in Fig.

Answer:

The angle of incidence is consistently found to be equal to the angle of reflection.

The angle of incidence = 900 – 500=400

The angle of reflection = angle of incidence = 400

So, the angle x is 400

Question 29. Why are the ceiling and wall behind the stage of good conference halls or concert halls made curved?

Answer:

The ceiling and the wall behind the stage of the conference or concert halls are curved to ensure the audience can equally hear the reflected sound.

Question 30. Graph each with two separate diagrams   

(i) Two sound waves with equal amplitude but different frequencies?  

(ii) Two sound waves with equal frequency but different amplitudes.  

(iii) Two sound waves have different amplitudes and also different wavelengths.

Answer:

Question 31. Establish the relationship between the speed of sound, its wavelength and frequency. If the velocity of sound in air is 340 m/s. calculate

(i) wavelength when the frequency is 256 Hz.

(ii) frequency when the wavelength is 0.85 m

Answer: 

The relationship between Sound speed, wavelength and frequency is as follows:

Speed = distance/time 

V = Wavelength/Time  

V = Wavelength × 1/Time 

Frequency = 1/Time  

V = Wavelength × Frequency  

1) Wavelength = Velocity/Frequency

  =340 /256

  = 1.32m  

2) Frequency = Speed / Wavelength

  =340/0.85

  = 400 HZ

Question 32. Draw a curve showing the changes in density or pressure with distance for a sound-induced disturbance. Mark the position of compression and rarefaction on this curve. Also, define wavelengths and time periods with this curve.

Answer:

Question 33: What is SONAR?

Answer: 

The SONAR technique is used 

for determining the depth of the sea and locating underwater hills, valleys, submarines, icebergs, sunken ships, etc.

Question 34: Differentiate between infrasonic & ultrasonic sound waves.

Answer: 

Sound waves that have frequencies below the audible range are termed “infrasonic”, and those above the audible range are known as “ultrasonic”.

Question 35. Guess which sound has a higher pitch: guitar or car horn?

Answer:

When the ear responds to the wavelength of the sound, it is called pitch. The higher the frequency of the sound, the higher the pitch. In comparison, the guitar operates at a higher frequency, so the guitar’s sound would be higher pitched.

Question 36. In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature? 

Answer: The fastest speed of sound is in solids that decrease in liquids. But sound travels much more slowly in the air. Amongst the given options above, the fastest medium for sound is iron. 

Question 37. What is the audible range of the average human ear?

Answer: For the average human ear to function, the range of particular frequencies varies from 20Hz to 20,000 Hz, which is considered an audible range. Frequency lower than 20 Hz and higher than 20K Hz is not audible to humans. 

Question 38. Which sound characteristic helps you to identify your friend by his voice while sitting with others in a dark room?

Answer: Timbre and pitch are the characteristics of sound that help to identify the sound of different voices. Thus, due to timbre and pitch, a person can determine the voice of others when sitting in a dark room.

Question 39. A person’s hearing range is 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in the air that correspond to these two frequencies? Take the speed of sound in air as 344m/s.

Answer: 

For 20 Hz sound waves, the wavelength would be given as:

v=n×λ  

λ=v/n=344/20=17.2m  

For 20kHz sound waves, the wavelength would be given as:

v=n×λ  

λ=v/n=344 ms−1/20000 Hz=0.0172 m  

Question 40. Two children are at opposite ends of the aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times the sound wave takes in the air and in aluminium to reach the second child.

Answer:.

As the speed of sound in air =344 m/s  

And the speed of sound in aluminium = 6420 m/s  

we already know that v=  distance/time thus time =d/v  

= time taken by sound wave in air/time taken by sound wave in aluminium

= d/344:d/6420

= 6420/344

=18.66/1  

The sound would take 18.66 times more time to travel through the air than in aluminium in reaching other boys.

Question 41. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?

Answer: The frequency of the source of sound is 100 Hz, meaning the sound source vibrates 100 times in one second. Thus, the vibrations made by the sound source in 1 min (60 sec) would be 6000.

Question 42. Does sound follow the same laws of reflection as light does? Explain.

Answer 42. Yes, sound obeys the same laws of reflection as light since the directions in which sound is incident and reflected are at the same angle to the normal to the reflective surface at the given point of incidence. So all three are in the same plane.

Question 43. When a sound bounces off a distant object, it creates an echo. Keep the distance between the reflecting surface and the sound-generating source the same. Hear the echo on a hotter day? 

 Answer 43. The perception of sound lasts about 0.1 s in the human brain. To hear a clear echo, the time interval between the original sound and the reflected sound must be at least 0.1 s. 

 Therefore, the total distance that the sound travels from the point of origin to the reflecting surface and vice versa must be at least (344 m/s)0.1 s=34.4 m 

 In order to hear clear echoes, the minimum distance from the obstacle to the sound source must be half that, i.e. 17.2 m. The speed of sound increases with increasing temperature. 

 Therefore,  the speed of sound is faster on a warmer day, so echoes due to multiple sound reflections may be heard more than once. The result will be that people will not hear a clear echo.

Question 44. Give two practical applications of the reflection of sound waves. 

• Megaphones or loudhailers 

 Answer 44(i). Megaphones or loudhailers, horns, and musical instruments such as trumpets are designed to send sound in a specific direction without spreading it in all directions. 

• Stethoscope 

 Answer 44(ii). The stethoscope is a medical instrument used to listen to sounds made in the body, primarily in the heart or lungs. With stethoscopes, the sound of the patient’s heartbeat reaches the doctor’s ear through multiple sound reflections. 

Question  45. A stone is dropped from the top of a 500m high tower into a pool of water at the base of the tower. When can you hear the splashing above? Given the speed of sound. 

 Answer 45. v2=u2+2gh 

 v2=0+2×10×500 

 v2=10000 

 v2=10000=100 ms−1 

 we also know that v=u+gt=0+10t 

 100= 10t100 = 10t, or the time it takes for the rock to reach the surface of the pond (t)=100/10=10 sec(t) 

 So the time it takes for the sound to travel from the surface of the pond to the surface to reach = d/v=500/340 =1.47 sec 

, so the total time  taken for a tip splash = 10 + 1.47 = 11.47 s 

Question  46. A sound wave travels at 339 ms–1. If its wavelength is 1.5 cm, what is the wave’s frequency? Will it be audible? 

 Answer 46. 

 Since we know that v=λV 

 339=0.015 m×v 

 v=339/0.015=22600 Hz 

 Because the resulting frequency is outside the audible range of humans (20 Hz to 20 kHz), the sound is therefore inaudible to human ears. 

Question  47. What is reverberation? How can it be reduced? 

Answer 47. Reverberation is the repeated sound reflection that causes the sound to persist  in the environment for a long time. 

 To reduce reverberation, the ceiling and walls of the auditorium are usually covered with sound-absorbing materials such as fiberboard, rough plaster or curtains. Seating materials are also chosen based on their sound-absorbing properties.

Question 48. What is the loudness of sound? What factors does it depend on?

Answer 48. Loudness refers to however loud or soft a sound seems to a listener. The sound intensity is, in turn, determined by the intensity of the sound waves. Intensity could be a measure of the quantity of energy in sound waves. The unit of intensity is the decibel (dB). Therefore, the intensity of sound waves determines the loudness of the noise. Intensity comes from two factors: the amplitude of the sound waves and the distance they have travelled from the sound source.

The amplitude is a measure of the size of sound waves.

 Depends on the amount of energy that triggered the waves.

 Higher amplitude waves have more energy and intensity and therefore sound louder.

Question 49. Explain how bats use ultrasound to catch prey.

Answer 49. Bats use ultrasound to catch prey through the phenomenon of reflection of sound waves.

 The bats generate the ultrasonic wave, which is reflected by the prey and returns to the bats’ ears. That allows the bat to determine the distance and position of the prey.

Question 50. How is ultrasound used for cleaning?

Answer 50. Ultrasound cleans hard-to-clean parts of objects such as coiled tubing, irregularly shaped machinery, etc. The object to be cleaned is placed in the cleaning solution, and ultrasound waves pass through the solution. Due to their high frequency, the ultrasonic waves whirl up the cleaning solution. Therefore, the objects are thoroughly cleaned.

Question 51. Explain how your school bell produces sound.

Answer 51. The sound of the bell depends upon the bell’s vibration when it is rung. If the bell starts ringing, it forces the molecules around the air to vibrate. This produces waves. Thus, the compression is produced, but the rarefaction makes the sound echo through the air. 

Hence the vibrating molecules put pressure on one another. Air particles are disturbed and they start moving forward and backwards.  

Question 52. Why are sound waves called mechanical waves?

Answer 52. Mechanical waves generate at the wave speed through a particular medium (solid, liquid, or gas). A sound wave is an associated example of a mechanical wave. Sound waves cannot travel through a vacuum. It needs some medium to pass, like air, water, or metal, similar to mechanical waves. That’s why; a sound wave is known as a mechanical wave.

Question 53. Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?

Answer 53. A sound wave needs the medium to propagate. Be it for solid, gas, or liquid medium. The moon resides in a vacuum; as it has no air, it has no medium to generate sound waves. Thus, it is not feasible to hear any sound on the moon.

Question 54. Which wave property determines

(a) Loudness,

Answer 54(a). The loudness of the sound is used for the decibel unit (dB). Loudness means the intensity of a particular wave or molecule. The determination of the loudness is dependent on the magnitude of the wave. In short, the amplitude of the wave determines the loudness of sound.

(b) Pitch

Answer 54(b). Pitch means the response of the sound by ear that goes hand in hand. The higher the  be the pitch, the higher will be the frequency. Hence, the pitch of the sound is entirely dependent on the frequency of the sound.

Question 55. How are the wavelength and frequency of a sound wave related to its speed? 

Answer 55. The longer the wavelength, the longer the frequency waves. In terms of speed, it depends on the medium through which the sound wave is propagating. The more inflexible an inelastic medium, the faster the sound travels. 

 The equation is formed as VW = fλ, where VW is the speed of sound, f is its frequency, and λ is its wavelength. 

Question  56. Calculate the wavelength of a sound wave with a frequency of 220 Hz and a speed of 440 m/s in a given medium. 

 Answer 56.The frequency of the sound wave, n= 220 Hz 

 The speed of the sound wave, v = 440 m/s 

 For a sound wave, 

 speed = wavelength x frequency 

 v = λ × n 

 440 = λ × 220 

 λ = 440/220 = 2 m 

 Therefore, the wavelength for the sound wave is 2 m. 

Question  57. A person hears a  500 Hz tone sitting  450 m from the sound source. What is the time interval in between successive compressions of the source? 

 Answer 57.The time interval between two successive compressions of the 

 source  

T = 1/v = 1/500 = 0.002 seconds. 

Question  58. Explain how sonar works and how it is used. 

 Answer 58. SONAR stands for  Sound Navigation and Ranging. It is the device that uses ultrasonic waves to estimate and measure underwater objects’ specific distance, direction and speed

 This is how the SONAR works: 

• It consists of a transmitter and detector and is installed on a ship or boat. 
• The transmitter in SONAR generates and transmits powerful ultrasonic waves. 
• Ultrasonic waves travel through water, and after hitting the target, the beam bounces off the seabed and is received by the underwater detector (mounted on a ship). The detector will then convert the waves into electrical signals, which are correctly interpreted.
• The time between sending as well as receiving the signal is also noted.   

Few Applications of SONAR Technology  

The core use of sonar technology is to estimate the range, depth and direction of arrival of objects. Below are the different applications of sonar technology:

• Speciality sonars are used on ships and submarines for underwater communications. 
• Medical imaging to detect cysts and cancer cells is done with sonar, and this method is known as a sonogram.

Question 59. How does the sound produced by an object vibrating in a medium reach your ear? 

Answer 59. 

When an object vibrates, it needs the surrounding particles of the medium to vibrate. The particles adjacent to the vibrating particles are forced to vibrate. Therefore, the sound produced by an object vibrating in a medium is transmitted from particle to particle until it reaches your ear.

Question 60. Explain how the sound of your school bell is made. 

Answer 60. 

 if the school bell is struck with a hammer, it reciprocates, producing compression and rarefaction due to vibration. That is how the school bell sounds.

Question 61. Why are sound waves called mechanical waves? 

Answer 61. 

Sound waves will always require a medium to propagate to interact with the particles. Therefore, sound waves are called mechanical waves.

Question 62. Suppose you and your friend are on the moon. Will you be able to hear any noise produced by your friend? 

Answer 62. 

 No. Sound waves need a medium to propagate. Due to the absence of the atmosphere on the moon and the inability of sound to travel in a vacuum, I will not be able to hear any sound generated by my friend. 

 Question  63. Which does wave property determine (a) loudness and (b) pitch? 

 Answer 63. 

(a). Amplitude: The loudness of the sound and its amplitude are directly related. The larger the amplitude, the louder sound. 

(b). Frequency: The pitch of the sound and its frequency are directly related. When the pitch is high, the sound frequency is also high. 

Question  64. Guess which sound is higher in pitch: guitar or car horn? 

 Answer 64. 

The pitch of the sound is directly proportional to its frequency. Hence, the guitar has a higher sound compared to a car horn. 

Question  65. What are a sound wave’s wavelength, frequency, time period, and amplitude? 

 Answer 65. 

(a) Wavelength can be defined as the distance between two consecutive rarefactions or two consecutive compressions. The SI unit for wavelength is a meter (m). 

(b) Frequency: Frequency is defined as the number of oscillations per second. The SI unit of frequency is the hertz (Hz). 

 (c) Amplitude: Amplitude can be defined as the maximum height reached by the trough or crest of a sound wave. 

(d) Time period: The time period is the time required to produce one complete sound wave cycle.

Question 66. How are the wavelength and frequency of a sound wave related to its speed?

 Answer 66. 

the wavelength, velocity, and frequency are related as follows: 

 velocity = wavelength x frequency 

 v = λ ν,  the higher the speed of a sound, the greater its wavelength for a given frequency.

Question  67. Find the wavelength of a sound wave with a frequency of 220 Hz and a velocity of 440 m/s in a specific medium. 

Answer 67. 

Assuming 

  sound wave frequency = 220 Hz. 

  sound wave velocity = 440 m/s. 

 calculate wavelength. 

 We know that 

 velocity = wavelength × frequency 

 v = λ ν 

 440 = wavelength × 220 

 wavelength = 440/220 

 wavelength = 2 

Here, the wavelength of the sound wave is 2 meters.

Question 68. A person hears a  500 Hz tone sitting at a distance of 450 m from the sound source. What is the time interval between successive compressions of the source? 

Answer 68. 

 The time interval between successive compressions of the source is equal to the time period, and the time period is reciprocal to the frequency. Therefore, it can be calculated as follows: 

 T= 1/F 

 T= 1/500 

 T = 0.002 s. 

Question 69. Distinguish between loudness and intensity of the sound. 

Answer 69. 

The amount of sound energy that travels through an area per second is called the intensity of a sound wave. Loudness is defined by its amplitude. 

Question  70. In which of the three media, air, water, or iron, does sound propagate faster at a given temperature? 

Answer 70. 

Sound travels faster in solids than in any other medium. Therefore, at a given temperature, sound propagates faster in iron and slower in gas. 

Question  71. An echo can be heard after 3 s. What is the distance from the reflecting surface to the source at a sound speed of  342 ms-1?

Answer 71. 

Velocity of sound (v) = 342 ms-1 

 Echo returns in time (t) = 3 s 

 Distance travelled by sound = v × t = 342 × 3 = 1026 m 

 In the interval of a given time, the sound travels a distance twice the distance between the reflecting surface and the source. 

 Here, the distance of the reflecting surface from the source =1026/2 = 513 m. 

Question  72. Why are the ceilings of concert halls curved? 

Answer 72. 

 The ceilings of the concert halls are curved to uniformly spread sound in all directions after bouncing off the walls. 

Question  73. What is the audible range of the average human ear? 

Answer 73. 

20 Hz to 20,000 Hz. Any sound below 20 Hz or above 20,000 Hz is inaudible to human ears. 

Question 74. What is the frequency range associated with (a) infrasound? (b) ultrasound?

Answer 74. 

(a). 20Hz 

(b). 20,000 Hz. 

Question 75. A submarine emits a sonar pulse that returns from an underwater cliff in 1.02 s. If the speed of sound in the salt water is 1531 m/s, how far away is the cliff? 

Answer 75. 

Time (t) for the return of the sonar pulse  = 1.02 s 

 Speed ​​of sound (v)  in salt water = 1531 m s-1 

 Distance travelled by the sonar pulse = Speed ​​of sound × time it takes 

 = 1531 x 1.02 = 1561.62 m 

The distance from cliff to submarine = (total distance traveled by sonar pulse) / 2

 = 1561.62 / 2 

 = 780.81 m.

Question 76. What is sound, and how is it generated?  

Answer 76. 

Sound is created by vibration. A body vibrates, causing neighbouring particles in the medium to vibrate. This creates a disturbance in the medium that travels in waves and reaches the ear. Hence the sound is generated.  

Question 77. Using a diagram, describe how compressions and rarefactions occur in the air near a sound source.  

Answer 77. 

When the school bell is struck with the hammer, it moves back and forth, creating compression and rarefaction due to vibration. As it moves forward, it creates high pressure in its surroundings. This area of ​​high pressure is called compression. When it moves backwards, it creates an area of ​​low pressure. This area is called dilution.

Question 78. Cite an experiment to show that sound requires a material medium to propagate.  

Answer 78. 

Take an electric bell and hang it in an empty bell equipped with a vacuum pump (as shown in the picture below).

 First, you hear the bell ringing. Now use the vacuum pump to pump some air out of the bell. You will notice that the ringtone decreases. If you keep pumping the air out of the hood, the glass pitcher will run out of air after a while. Now try to ring the bell. There is no sound, but you can see that the prong of the bell is still vibrating. If there is no air in the bell jar, a vacuum is created. Sound cannot travel through a vacuum. This experiment shows that sound requires a material medium to propagate.  

Question 79. Why is a sound wave called a longitudinal wave? 

Answer 79. 

The vibration of the medium that propagates parallel to the wave direction or along the wave direction is called a longitudinal wave. The direction of the particles in the medium oscillates parallel to the direction of propagation of the disturbance. Thus, a sound wave is called a longitudinal wave.  

Question 80. While sitting in a dark room with other people, what sound characteristics help you identify your friend from their voice?  

Answer 80. 

Sound quality is a feature that helps us identify a specific person’s voice. Two people may have the same tone and volume, but their qualities will differ

Question 81. Lightning and thunder occur simultaneously. But the thunder is heard a few seconds after the lightning. Why? Answer 81. 

The speed of the sound is 344 m/s, while the speed of light is 3 × 108 m/s. The rate of light is slower compared to the speed of light. Because of this, thunder takes a longer time to reach  Earth compared to the speed of light, which is faster. Therefore, lightning is seen before every thunder.

Question 82. A person’s hearing range is 20 Hz to 20 kHz. What are the typical wavelengths for the sound waves in the air that correspond to these two frequencies? Assume the speed of the sound in the air to be 344 m s−1.  

Answer 82.

For sound waves,

  velocity = wavelength × frequency  

v = λ × v  

the velocity of the sound wave in the air = 344 m/s  

(a) For v = 20 Hz  

λ1 = v/v1 = 344/ 20 = 17.2 m  

(b) For v2 = 20,000 Hz  

λ2 = v/v2 = 344/20,000 = 0.0172 m  

Hence, the human hearing wavelength ranges from 0.0172 m to 17.2 m.

Question 83. Two children are on opposite ends of an aluminium pole. Hit the end of the pole with a stone. Find the ratio of the times it takes for the sound wave in the air and in the aluminium to reach the second child.  

Answer 83. 

Consider the length of the aluminium pole = d  

speed of the sound wave at 25°C, V Al = 6420 ms-1  

the time it takes to reach another end  

T Al = d/ (V Al) = d / 6420  

The velocity of sound in air, V air = 346 ms-1

The time it takes for sound to reach each end,  

T air = d/ (V air) = d/346  

Hence the ratio of time, the sound in aluminium and air

 requires  T air / t Al = 6420 / 346 = 18.55  

Question 84. The frequency of a sound source is 100 Hz. How many times does it oscillate in one minute?  

Answer 84.

Frequency = (number of oscillations) / total time  

number of the oscillations = frequency × total time    

sound frequency = 100 Hz  

total time = 1 min (1 min = 60 s)  

number of the oscillations or vibrations = 100 × 60 = 6000  

The source vibrates 6000 times in one minute and generates a frequency of 100 Hz.  

Question 85. Does sound follow the same laws of reflection as light? To explain.  

Answer 85. 

Yes, sound follows the same law of reflection as light. The reflected sound wave and incident sound waveform at an equal angle with the surface normal at the point of impact. Also, the reflected sound wave, normal to the point of the incidence and incident sound wave, all lie in the same plane.  

Question 86. When a sound bounces off a distant object, it creates an echo. Keep the distance between the reflecting surface and the sound-generating source the same. Do you hear echoing noises on a hotter day?  

Answer 86. 

An echo is heard if the time interval between the reflected sound as well as the original sound is at least 0.1 seconds when the temperature increases, and so does the speed of sound in a medium. On a warmer day, the interval between the reflected and original sounds decreases. An echo only becomes audible if the time interval between the reflected sound and the original sound is greater than 0.1 s.

Question 87. Give two practical applications of the reflection of sound waves.

Answer 87. 

(i) The reflection of sound is used to measure the speed and distance of objects underwater. This method is called SONAR.

(ii) How a stethoscope works: A patient’s heartbeat reaches the doctor’s ear through multiple sound reflections.  

Question 88 A stone is dropped from the top of a  500 m high tower into a pool of water at the tower’s base. When can you hear the splashing above? Given g = 10 m s−2 and the speed of sound = 340 m s−1.

Answer 88. 

Height (s) of the tower = 500 m  

Velocity ​​of sound (v)  = 340 m s−1  

Acceleration  due to gravity (g) = 10 m s−1  

Initial velocity (u) of the stone = 0  

Time (t1) of the stone needed to fall to the base of the tower 

According to the second equation for motion:

  s= ut1 + (½) g (t1)2  

500 = 0 x t1 + ( ½) 10 (t1) 2. 

(t1)2 = 100. 

t1 = 10 s  

time (t2) for the sound to travel from the base of the tower to the top  

= 500/340 = 1.47 s.  

t = t1 + t2  

t = 10 + 1.47  

t = 11 .47 sec  

Question 89. A sound wave travels at a speed of 339 m s-1. What is the frequency of the wave if its wavelength is 1.5 cm? Will it be audible?  

Answer 89. 

Speed of sound (v) = 339 m s−1  

Wavelength of sound (λ) = 1.5 cm or 0.015 m  

Formula of speed of sound = wavelength × frequency  

v = v = λ X v  

v = v / λ = 339 / 0015 = 22600 Hz.  

The frequency of the sound that humans can hear is in the range of 20 Hz to 20,000 Hz. The frequency of sound specified is greater than 20,000 Hz, i.e. not audible.

Question 90. What is reverberation? How can it be reduced?  

Answer 90. 

Multiple continuous sound reflections in a large enclosed space are reverberations. It can be reduced by covering the walls and ceilings of enclosed spaces with sound-absorbing materials such as loose wool and fiberboard.  

Question 91 What is the intensity of the sound? What factors does it depend on?

Answer 91. 

Loud noises have a lot of energy. The loudness is directly related to the amplitude of the vibrations. It is proportional to the square of the amplitude of sound vibrations.

Question 92. Explain how bats use ultrasound to catch prey.  

Answer 92. bats can produce high-pitched ultrasonic chirps. These chirps are reflected off objects such as prey and return to their ears. This helps a bat know how far away its prey is.  

Question 93. How is ultrasound used for cleaning?  

Answer 93.  The objects that are to be cleaned are placed in a cleaning solution, and then ultrasonic waves are passed through the solution. The high frequency of the ultrasound waves helps loosen dirt from objects. In these ways, ultrasound is used for cleaning purposes.  

Question 94. Explain the operation and application of a sonar.  

Answer 94. SONAR is the abbreviation for Sound Navigation as well as Ranging. It’s an acoustic device used to measure underwater objects’ direction, speed and depth, viz. Shipwrecks and submarines with ultrasound.

 It is also used to determine the depth of oceans and seas.

 An ultrasonic beam is generated that travels through the seawater, emitted by the transducer. The reflection creates an echo that is detected and recorded by the detector. It is then converted into electrical signals. The distance from the underwater object represented by “d” is calculated from the time (represented as “t”) it takes for the echo to return at speed (represented as “v”) and is expressed as

  2d = v × t.

 This method of distance measurement is also known as echo ranging.

Question 95. A sonar on a submarine emits a signal and receives an echo 5 s later. Calculate the speed of sound in water when the distance from the object to the submarine is 3625 m.  

Answer 95. Time (t) needed to hear the echo = 5 s  

Distance (d) from object to submarine = 3625 m  

Total distance  SONAR travels in the water during reception and transmission  = 2d  

Velocity (v)  sound in water = 2d/t = (2 × 3625) / 5

  = 1450 ms-1

Question 96. Explain how defects in a metal block can be detected using ultrasound.

Answer 96. 

Defective metal blocks do not allow the ultrasound to pass through and are not reflected back. This technique is used to detect defects in metal blocks. Create a setup as shown in the figure, with the ultrasound going through one end and the detectors placed at the other end of a metal block. Hence the defective part of the metal block does not transmit ultrasound; the sensor will not detect it. Defects in metal blocks could be detected with the help of ultrasound.

Question 97. Explain how the human ear works.

Answer 97. 

Various sounds generated by particles in our environment are picked up by the pinna, which transmits these sounds through the ear canal to the eardrum. The eardrum begins to vibrate back and forth when sound waves hit it. The vibrating eardrum initiates the vibration of the small bone hammer. These vibrations travel from the hammer to the third bony stirrup through the second bony anvil. The stirrup hits the membrane of the oval window to transmit its vibration to the cochlea. The liquid in the cochlea generates electrical impulses in nerve cells. These electrical impulses are transferred to the brain by the auditory nerve. The brain interprets them as sound, and hence we have the sensation of hearing.

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