Sound 1. (a) (b) Ans. (a) (b) (i) (ii) (iii) 2. Ans. 3. (a) (b) Ans. (a) (b)

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1 Sound 1. (a) What do you understand by the term sound energy? (b) State three conditions necessary for hearing sound. Ans. (a) Sound is a mechanical energy which produces sensation of hearing. (b) (i) There must be a vibrating body. (ii) There must be a material medium to carry sound energy. (iii) There must be a receiver i.e., ear so as to capture the sound vibrations. 2. Describe briefly an experiment to prove that vibrating bodies produce sound. Ans. Take a tuning fork and strike it against a rubber cork. Bring the tuning fork close to ear. A humming sound is heard. Again strike the tuning fork against the rubber cork and touch its prong to the stationary pith ball of pith ball electroscope. It is observed that pith ball repeatedly flies outward. Thus, experiment proves that sound is produced by a vibrating body. 3. (a) What do you understand by the term infra-sonic vibrations? (b) What do you understand by the term sonic vibrations? State the range of sonic vibrations for human ear. Ans. (a) The vibrations which are not perceived by human ear and have a range from 0 20 Hz are called infra-sonic vibrations. (b) The vibrations which produce sensation of hearing in human beings are called sonic vibrations. Their range is between 20 20,000 Hz. 1

2 4. (a) What do you understand by the term ultra-sonic vibrations? (b) Name three animals which can hear ultra-sonic vibrations. Ans. (a) The vibrations which are not perceived by human ear and their frequency range is above 20,000 Hz are called ultra-sonic vibrations. (b) (i) Bats (ii) Dogs (iii) Dolphins. 5. How do bats locate their prey during flight? Ans. The bats emit ultrasonic vibrations during flight. These vibrations on striking the prey are echoed back to the bats. On receiving the echo the bats home on the target. 6. What is Galton s whistle? To what use it can be put? Ans. Galton s whistle is a special kind of whistle which emits ultra-sonic vibrations between 20,000 Hz to 40,000 Hz. The vibrations of this range cannot be perceived by human ears, but can be easily perceived by dogs. Thus, the dogs can be trained to perform special tasks on hearing the sound from Galton s whistle. It is very useful, if intruders enter in someone s house. On blowing the whistle the dog will attack intruder and the intruder will not be able to hear the sound. 7. State four practical uses of ultra-sonic vibrations. Ans. (i) They are used in ultrasound scanning of various organs of human body. (ii) They are used for welding metals having high melting point. (iii) They are used for homogenising milk. (iv) They are used for scaring insects and rats from godowns. 2

3 8. Describe an experiment to prove that material medium is necessary for the propagation of sound. Ans. An electric circuit consisting of a cell, a switch, an electric bell is arranged inside a bell jar, which is placed on the platform of an evacuated pump as shown in diagram alongside. The switch of the electric circuit is pressed in, when a clear sound of bell is heard. Air is now removed from the bell jar by evacuating the pump. It is noticed that intensity of sound gradually decreases. When the bell jar is completely evacuated, it is noticed that no sound is heard when the hammer of bell is striking the gong. Thus, experiment clearly proves that material medium (in the present case air) is necessary for the propagation of sound energy. 9. Why do the astronauts talk to each other through radio telephone in space or on the surface of moon? Ans. It is because the space or moon has no air. Thus, if they speak without any external aid, sound will not be heard as material medium is necessary for the propagation of sound. However, when they talk through radio telephone, the electromagnetic waves are produced. These waves do not require any material medium for the propagation and hence, can be picked by the radio-receiver. 10. What are elastic waves? Name two kinds of elastic waves. Ans. The waves produced in a material medium are called elastic waves. Transverse waves and longitudinal waves are called elastic waves. 3

4 11. The sound of an explosion on the surface of lake is heard by a boatman 100 m away and a diver 100 m below the point of explosion. (i) Of the two persons mentioned (boatman and diver), who would hear the sound first? (ii) Give reason for your answer in (i). (iii) If the sound takes t seconds to reach the boatman, approximately how much time it will take to reach the diver? Ans. (i) The diver hears the sound first. (ii) It is because sound travels at 1450 ms 1 in water and 330 ms 1 in air. (iii) The time in which sound reaches diver is s = seconds. 12. (a) State two important differences between light and sound waves. (b) What is approximate value of speed of sound in iron as compared to that in air? Illustrate your answer with simple experiment. Ans. (a) (i) Light waves are electromagnetic waves and hence, do not require material medium for propagation. However, sound waves are mechanical waves and, hence, require material medium for propagation. (ii) Light waves travel in air with a speed of ms 1, whereas sound waves travel in air at a speed of 330 ms 1. (b) The speed of sound in air is 330 ms 1 and in iron is 5100 ms 1. Ask a person to hit the rail at some predetermined time with a hammer. Put your ear on the rail. (The person must be 2 km or more from observer). You will notice that sound energy reaches very quickly to the ear through rails, however, it takes quite a lot of time before sound is heard through air. 4

5 13. How does a bat avoid obstacles in their way when in flight? Ans. The bat emits ultrasonic vibration in the course of its flight. These vibrations on striking an obstacle are reflected back to form echo. On hearing the echo the bat can locate the obstacle during flight and steers clear from it. 14. What do you understand by the following terms : (a) Wave, (b) Elastic wave, (c) Wave motion? Ans. (a) The disturbance produced in a medium by the to and fro motion of its particles about their mean position is called wave. (b) The waves produced in a material medium are called elastic waves. (c) The transfer of energy when the particles of a medium move about their mean positions is called wave motion. 15. State any two characteristics of wave motion. Ans. (a) A wave is caused due to periodic disturbance of particles of medium, and the wave by itself is periodic in nature. (b) It is the disturbance which travels outward and not the particles of medium. The particles of medium, simply vibrate either to and fro or up and down about their mean positions. 16. State three characteristics of medium necessary for the propagation of sound waves. Ans. (a) Medium must be elastic in nature. (b) Medium must have least frictional resistance between its molecules. (c) Medium must possess inertia of rest. 17. Explain, why lightning flash is seen before the crack of thunder. Ans. Light travels at a speed of ms 1 and hence, reaches instantly to an observer. However, sound travels at a very slow speed of 330 ms 1 and hence, reaches observer after sometime. Thus, flash of lightning is seen at once, but crack of thunder is heard later on. 5

6 18. Derive a relation between frequency and time period. Ans. Consider a vibrating body, having a time period T and frequency f. f vibrations are produced in = 1 s. 1 vibration is produced in 1 f s. But, time required to produce one vibration is called time period T T = 1 f. 19. Derive a relation between wave velocity; frequency and wavelength. Ans. Consider a wave, moving with a velocity v, such that f is its frequency, T the time period and λ the wavelength. In time T, the distance covered by wave = λ. In time one second the distance covered by wave = T λ But, f = 1 f v = f λ 20. The diagram shows a snap shot of a wave form of frequency 50 Hz in a string. The numbers in diagram represent disance in centimeters. For this wave motion find, (i) Wavelength (ii) Amplitude (iii) Wave velocity. Ans. (i) Wavelength = 20 cm. (ii) Amplitude = 4 cm. (iii) Wave velocity = f λ = = 1000 cms 1 = 10 ms 1. 6

7 21. Draw a diagram representing a wave of (a) amplitude 4 cm (b) wavelength 2 m. If the frequency of wave is 150 Hz, calculate its velocity. Ans. f = 150 Hz; λ= 2 m v = fλ = 150 s 1 2 m = 300 ms The diagram alongside shows a vibrating metal blade clamped at one end. P and R are the extreme positions occupied by the blade during its course of vibration, Q, being its position of rest. The vibrating blade produces a note of 480 Hz. (i) Mark on the diagram amplitude of vibration. (ii) If the velocity of sound in air is 320 ms 1, what is the wavelength of sound produced? Ans. (i) (ii) f = 480 Hz. v 320 λ= = = 0.66 m. f 480 7

8 23. A sound wave of frequency 640 Hz travels 800 m in 2.5 s. Calculate : (a) speed of sound (b) wavelength of sound wave. Dis tan ce 800 m Ans. (a) Speed of sound = = Time 2.5 s = 320 ms 1. (b) Speed of sound (v) = fλ. 320 = 640 λ λ = 320 = 0.5 m A television station broadcasts at a frequency of 4500 MHz. If the speed of television waves is ms 1. Calculate the wavelength of television waves. Ans. v = ms 1, f = 4500 MHz = Hz. Wavelength, λ = 8 v 3 10 = f = m = 0.067m. 25. A longitudinal wave of wavelength 0.03 cm travels in air with a speed of 330 ms 1.Calculate the frequency of the wave. Can this wave be heard by normal human ear. Give a reason for your answer. Ans. v = 330 ms 1 ; λ = 0.03 cm = m. 4 v Frequency, f = = = λ = 1,100,000 Hz. The above wave cannot be heard. It is because, the maximum range to which human ear can hear is 20,000 Hz. As the frequency of 1,100,000 Hz is far in excess therefore it will not produce any sound effect in ear. 8

9 26. A sound wave has a frequency of 2000 Hz and wavelength 17 cm. If the wavelength increases to 51 cm, what is the frequency, the nature of material through which sound is propagating remains same. Ans. Initial wavelength (λ 1 ) = 17 cm. Initial frequency (f 1 ) = 2000 Hz. Final wavelength (λ 2 ) = 51 cm. Final frequency (f 2 ) =? For a given material velocity of sound is a constant quantity. f 2 λ 2 = f 1 λ 1 f 2 51 cm = 2000 Hz 17 cm. f 2 = = Hz A disturbance in air has wavelength of 22 m and speed 330 ms 1. Calculate the frequency of disturbance. State whether above disturbance is audible to normal human ear. Give one reason for your answer. Ans. v = 330 ms 1 λ = 22 m. 1 v 330 ms Frequency, f = = λ 22 m = 15 Hz. The above disturbance is not audible. It is because human ear is sensitive to frequencies between 20 Hz to 20,000 Hz. 28. An ultraviolet radiation has wavelength 150 Å. If the speed of ultraviolet radiation is ms 1, calculate (i) frequency of radiation in MHz (ii) Time period. {1 Å = m]. Ans. (i) λ = 150 Å = m = m; v = ms v Frequency, f = 3 10 ms λ = Hz Frequency in MHz = 6 10 = MHz. (ii) Time period, T = 1 1 = f = second. 9

10 29. The wavelength of waves produced on the surface of water is 20 cm. If the wave velocity is 24 ms 1, calculate : (a) number of waves produced in one second (b) time required to produce one wave. Ans. (a) Wavelength (λ) = 20 cm = 0.2 m Wave velocity (v) = 24 ms 1 Number of waves produced in one second = Frequency 1 v 24 ms = = = 120 Hz. λ 0.2 m (b) Time required to produce one wave = Time period (T) 1 1 = = s = second. f A thin metal plate is placed against the teeth of a cog wheel. If the cog wheel is rotated at a constant speed of 360 rotations per minute and has 80 teeth, calculate : (a) frequency of note produced (b) speed of sound, if the wavelength is 0.7 m (c) what will be the effect, if the speed of cog wheel is halved? Ans. (a) Number of rotations of cog wheel in 1 minute (60 s) = Number of rotations of cog wheel in 1second = 60 s = 6 s 1. Frequency of note produced = No. of rotations per second Teeth in cog wheel = 6 s 1 80 = 480 s 1 = 480 Hz. (b) Speed of sound = Frequency Wavelength = 480 s m = ms 1. (c) The frequency of note produced is halved. Thus, in turn will lower the pitch of sound, i.e., a bass note is produced. 10

11 31. A vibrating tuning fork can produce note of wavelength 0.83 m and has a time period of seconds. Calculate the wave velocity note and the distance covered by sound wave in 0.08 s. Ans. Time period = s 1 1 Frequency (f) = = = 400 Hz. 3 T Wave velocity = f λ = = 332 ms 1. Distance covered by sound in 0.08 seconds = 332 ms s = m. 32. The distance between one crest and one trough produced on the surface of water is 0.04 m. If the waves are produced at a rate of 180 per minute. Calculate : (a) time period (b) wave velocity. Ans. (a) Number of waves produced in 1 minute (60 s) = 180 Number of waves produced in 1 second = s = 3 s 1 Frequency of waves = 3 s Now, Time periond (T) = = = s = 0.33 s. 1 f 3s 3 λ (b) Distance between one crest and one trough = 2 λ = 0.04 m 2 λ = = 0.08 m Wave velocity = λ = T 1/3s 0.08 m = 0.24 ms 1. 11