Chapter 12 Sound
Intext Questions
Question 1: How does the sound produced by a vibrating object in a medium reach your ears?
Answer: When an object vibrates, it vibrates the neighbouring particle of the medium. These vibrating particles then vibrate to adjacent particles. Vibrations in an object create disturbance in the medium and consequently form compressions and rarefactions till it reaches our ears.
Region of high pressure and high density is called compression and the region of low pressure and low density is called rarefaction. As the object continues to vibrate, it produces a series of successive compression and rarefactions in the air, thus, propagating sound through the air and finally reaches our ears.
Question 2: Explain how sound is produced by your school bell.
Answer: The bell produces the sound when it is struck by air hammer. When the bell is struck by a hammer, it starts vibrating. Since, the vibrating objects produce sound, the bell produces sound.
Question 3: Why are sound waves called mechanical waves?
Answer: Sound waves are mechanical waves because they need a material medium for propagation like gases, liquids, solids.
Question 4: Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer: No, we will not be able to hear any sound produced by friend. This is because of the presence of vacuum. We know that, the sound needs a material medium to propagate. But in vacuum, there is no particle, hence the sound produced by friend will not reach us.
Question 5: Which wave property determines
a) loudness
b) pitch
Answer:
a) Amplitude; the amplitude of the sound wave determines the loudness or softness of a sound. If the amplitude is more, the sound will be loud. If the amplitude is less, the sound will be soft.
b) Frequency; the frequency of a sound wave determines pitch. The faster the vibration, the higher is frequency and higher is the pitch.
Question 6: Guess which sound has a higher pitch; guitar or car horn.
Answer: The faster the vibrations, higher is the frequency and higher is the pitch. Hence, guitar has a higher pitch.
Question 7: What are wavelength, frequency, time period and amplitude of a sound wave?
Answer:
• Wavelength: The distance between two consecutive compressions (C) or two consecutive rarefactions (R) is called the wavelength (λ). SI unit is metre.
• Frequency: The number of waves per unit time is called frequency (ν). SI unit is Hertz.
• Time period: The time taken to complete one wave is called time period (T). SI unit is second.
• Amplitude: The maximum displacement of the particle from its mean position is called amplitude (A). SI unit is metre.
Question 8: How are the wavelength and frequency of a sound wave related to its speed?
Answer: Speed = wavelength × frequency
Question 9: Calculate the wavelength of a sound wave whose frequency is 220 Hz and the speed is 440 m/s in a given medium.
Answer:
Given
Frequency = 220 Hz
Speed = 440 m/s
Wavelength = ?
We know that, speed = wavelength × frequency
Substituting the values, we get
440 = 220 × λ
λ = 440/220
λ = 2 m
Question 10: 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 the successive compression from the source?
Answer:
Given
Frequency = 500 Hz
Wavelength = 450 m
Time interval = ?
We know that, Frequency = 1/Time
Substituting the values, we get
500 = 1/T
T = 1/500
T = 0.002 s
Question 11: In which of three media, air, water or iron does the sound travel fastest at a particular temperature?
Answer: In iron, the sound travels the fastest at a particular temperature. This is due to the arrangement of particles. In solids, the particles are closely packed unlike liquids and gases.
Question 12: An echo is heard in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m/s?
Answer:
Given
Speed of sound = 342 m/s
Time taken for hearing echo = 3 s
Distance of the reflecting surface from the source = ?
We know that, 2d = v × t
Substituting the values, we get
2 × d = 342 × 3
d = 342 × 3/2
d = 513 m
Question 13: Why are the ceilings of concert halls curved?
Answer: The ceilings of concert halls are curved so that sound after reflection reaches all corners of the hall.
Question 14: What is the audible range of the average human ear?
Answer: The audible range of the average human ear is 20 - 20000 Hz.
Question 15: What is the range of frequencies associated with
i) Infrasound?
ii) Ultrasound?
Answer:
i) less than 20 Hz
ii) more than 20000 Hz
Question 16: 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 m/s, how far away is the cliff?
Answer:
Given
Speed of sound in salt water = 1531 m/s
Time taken to reach sonar pulse = 1.02 s
Distance between cliff and submarine = ?
We know that, 2d = v × t
Substituting the values, we get
2 × d = 1531 × 1.02
d = 1531 × 1.02/2
d = 780.81 m
Question 17: Distinguish between loudness and intensity of sound.
Answer:
• The amount of sound energy passing each second through unit area is called the intensity of sound.
• The loudness of a sound is determined by its amplitude.
Exercise Questions
Question 1: What is sound and how is it produced?
Answer: Sound is a form of energy that enables us to hear. Sound is produced by vibrating bodies.
Question 2: Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Answer:
When a vibrating object moves forward, it pushes and compresses the air in front of it creating a region of high pressure. This region is called a compression (C). This compression starts to move away from the vibrating object. When the vibrating object moves backwards, it creates a region of low pressure called rarefaction (R). As the object moves back and forth rapidly, a series of compressions and rarefactions is created in the air. These make the sound wave that propagates through the medium. Compression is the region of high pressure and rarefaction is the region of low pressure.
Question 3: Cite an experiment to show that sound needs a material medium for its propagation.
Answer: Take an electric bell and an airtight glass bell jar. The electric bell is suspended inside the airtight bell jar. The bell jar is connected to a vacuum pump. If you press the switch you will be able to hear the bell. Now start the vacuum pump. When the air in the jar is pumped out gradually, the sound becomes fainter, although the same current is passing through the bell. After some time when less air is left inside the bell jar you will hear a very feeble sound. If the air is removed completely, we won’t be able to hear any sound. This shows that sound needs a material medium for its propagation.
Question 4: Why is sound wave called a longitudinal wave?
Answer: The particles of sound wave do not move from one place to another but they simply oscillate back to forth about their position of rest. This is exactly how sound wave propagates, hence sound waves are longitudinal waves.
Question 5: Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer: Quality of sound is a characteristic that helps us identify the voice of a particular person. Two people may have the same pitch and loudness but their qualities will be different.
Question 6: Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Answer: The speed of sound in air is 344 m/s while the speed of light is 3 × 10⁸ m/s. The speed of sound is less than the speed of light. Due to this, thunder takes more time to reach the Earth as compared to flash. Hence, the flash and thunder are produced simultaneously but light is seen first and later the thunder.
Question 7: A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 m/s.
Answer:
Given
Hearing range of sound = 20 Hz - 20000 Hz
Speed of sound in air = 344 m/s
Wavelength = ?
We know that, speed = wavelength × frequency
Substituting the values, we get
344 = λ × 20
λ = 344/20
λ = 17.2 m
344 = λ × 20000
λ = 344/20000
λ = 0.0172 m
Question 8: Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.
Answer:
Let the length of aluminium rod by d.
Speed of sound wave at 25° in aluminium = 6420 m/s
Time taken to reach other end = d/6420 → 1
Speed of sound wave at 25° in air = 346 m/s
Time taken to reach other end = d/346 → 2
Dividing eq 1 and 2, we get
6420/346 = 18.55 s
Question 9: The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer:
Given
Frequency of sound = 100 Hz
We know that, Frequency = No. of oscillations/Time
Substituting the values, we get
100 = x/60
x = 100 × 60
x = 6000 oscillations
Question 10: Does sound follow the same laws of reflection as light does? Explain.
Answer: Yes, sound follows the same laws of reflection as light does. The reflected sound wave is equal to the incident sound wave. The reflected sound wave, the normal, and incident sound wave lie at the same plane.
Question 11: When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?
Answer: An echo is heard when the time interval between the reflected sound and the original sound is at least 0.1 s. As the temperature increases, the speed of sound in a medium also increases. On hotter day, the time interval between the reflected and original sound will decrease and an echo is audible only if the time interval between the reflected sound and original sound is greater than 0.1 s.
Question 12: Give two practical applications of reflection of sound waves.
Answer:
• Stethoscope is a medical instrument used for listening to sounds produced within the body, mainly in the heart or lungs. In stethoscopes the sound of the patient’s heartbeat reaches the doctor’s ears by multiple reflection of sound.
• The ceilings of concert halls, conference halls and cinema halls are curved so that sound after reflection reaches all corners of the hall.
Question 13: A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 m/s² and speed of sound = 340 m/s.
Answer:
Given
Height of tower = 500 m
Velocity of sound = 340 m/s
Acceleration due to gravity = 10 m/s²
Initial velocity of the stone = 0 m/s
Time taken (t₁) by stone to fall into pond = ?
We know that, s = ut + 1/2at²
Substituting the values, we get
500 = 1/2 × 10 × t²
t² = 500/5
t = √100
t = 10 s
Time taken (t₂) by the sound to reach the top from tower base = ?
= 500/340
= 1.47 s
t = t₁ + t₂ = 10 + 1.47 = 11.47 s
Question 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? Will it be audible?
Answer:
Given
Speed of sound wave = 339 m/s
Wavelength = 1.5 cm (0.015 m)
We know that, speed = wavelength × frequency
Substituting the values, we get
339 = 0.015 × ν
ν = 339/0.015
ν = 22600 Hz
Since, the audible range of hearing for humans is 20 - 20000 Hz, we cannot hear the sound.
Question 15: What is reverberation? How can it be reduced?
Answer: The persistence of sound due to repeated reflection and its gradual fading away is called reverberation of sound. In an auditorium or big hall excessive reverberation is highly undesirable. To reduce reverberation, the roofs and walls of the auditorium are generally covered with sound-absorbent materials like compressed fibreboard, rough plaster or draperies.
Question 16: What is loudness of sound? What factors does it depend on?
Answer: Loudness is a measure of amplitude. The loudness of a sound is determined basically by its amplitude.
Question 17: Explain how bats use ultrasound to catch a prey.
Answer: Bats search out prey and fly in dark night by emitting and detecting reflection of ultrasonic waves. The high pitched ultrasonic squeaks of the bats are reflected from the obstacles or prey and returned to the bat’s ear. The nature of reflections tells the bat where the obstacle or prey and what is it like.
Question 18: How is ultrasound used for cleaning?
Answer: Ultrasound is generally used to clean parts located in hard-to-reach places, for example, spiral tube, odd shaped parts, electronic components etc. Objects to be cleaned are placed in a cleaning solution and ultrasonic waves are sent into the solution. Due to the high frequency, the particles of dust, grease and dirt get detached and drop out. The objects thus get thoroughly cleaned.
Question 19: Explain the working and application of a sonar.
Answer: The SONAR stands for Sound Navigation And Ranging. It is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects. Sonar consists of a transmitter and a detector and is installed in a boat or a ship. The transmitter produces and transmits ultrasonic waves. These waves travel through water and after striking the object on the seabed, get reflected back and are sensed by the detector. The detector converts the ultrasonic waves into electrical signals which are appropriately interpreted. The distance of the object that reflected the sound wave can be calculated by knowing the speed of sound in water and the time interval between transmission and reception of the ultrasound. The sonar technique is used to determine the depth of the sea and to locate underwater hills, valleys, submarine, icebergs, sunken ship etc.
Question 20: A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.
Answer:
Given
Time taken to reach echo = 5 s
Distance of object from submarine = 3625 m
Speed of sound in water = ?
We know that, 2d = v × t
2 × 3625 = v × 5
v = 2 × 3625/5
v = 1450 m/s
Question 21: Explain how defects in a metal block can be detected using ultrasound.
Answer: Ultrasounds can be used to detect cracks and flaws in metal blocks. Metallic components are generally used in construction of big structures like buildings, bridges, machines and also scientific equipment. The cracks or holes inside the metal blocks, which are invisible from outside reduces the strength of the structure. Ultrasonic waves are allowed to pass through the metal block and detectors are used to detect the transmitted waves. If there is even a small defect, the ultrasound gets reflected back indicating the presence of the flaw or defect
Question 22: Explain how the human ear works.
Answer: The outer ear is called ‘pinna’. It collects the sound from the surroundings. The collected sound passes through the auditory canal. At the end of the auditory canal there is a thin membrane called the ear drum or tympanic membrane. When a compression of the medium reaches the eardrum the pressure on the outside of the membrane increases and forces the eardrum inward. Similarly, the eardrum moves outward when a rarefaction reaches it. In this way the eardrum vibrates. The vibrations are amplified several times by three bones (the hammer, anvil and stirrup) in the middle ear. The middle ear transmits the amplified pressure variations received from the sound wave to the inner ear. In the inner ear, the pressure variations are turned into electrical signals by the cochlea. These electrical signals are sent to the brain via the auditory nerve, and the brain interprets them as sound.
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