Question .1. The stem of a tuning fork is pressed against a table top. Answer the following questions :
(i) Would the above action produce any audible sound.
(ii) Does the above action cause the table to set into vibrations ?
(iii) If the answer above is yes, what type of vibrations are they ?
(iv) Under what conditions does the above action lead to resonance ?
Solution:
(i) Yes, the above action would produce audible sound.
(ii) Yes, the table top is set into forced vibrations.
(iii) The vibrations are forced vibration.
(iv) If the frequency of tuning fork is equal to the natural frequency of oscillation of the table top.
Question .2. State any two characteristics of a wave motion.
Solution:
The characteristics of wave motion:
(i) It is a periodic disturbance.
(ii) Energy transfer takes at a constant speed.
Question .3. If the amplitude of a wave is doubled, what will be the effect on its loudness?
Solution:
Loudness depends on the square of the amplitude of the wave. Thus when amplitude of the wave is doubled, the loudness becomes four times.
Question .4. Differentiate between forced vibrations and resonance.
Solution:
| Forced vibrations | Resonance |
1.The vibration that take place under the influence of external periodic forces are called forced vibration. | 1. The phenomena of setting a body into oscillations with its natural frequency by another vibrating body with the same frequency are called resonance or acoustic resonance. |
2. The external frequency may or may not be equal to the natural frequency of the body. | 2. The external frequency has the same frequency as the natural frequency of oscillation of the given oscillatory system. |
Question .5. How does the frequency and amplitudes affect a musical sound?
Solution:
| Effect of frequency to musical sound | Effect of amplitude to musical sound |
The frequency of a musical sound affects its ‘pitch’. The more the frequency of a musical sound, the sharper and shriller the music becomes. | The amplitude of a musical sound affects its loudness or intensity. The more the amplitude of the sound, the louder or more intense the music becomes. |
Question .6. Under what condition does resonance occur?
Solution:
Condition for resonance to take place is that the frequency of the external periodic force must be equal to the natural frequency of the vibrating system.
Question .7. Why is a loud sound heard at acoustic resonance?
Solution:
At acoustic resonance, the amplitude of vibration of the body becomes very large producing loud sound, as loudness is proportional to square of the amplitude.
Question .8. The rear view mirror of motor bike starts vibrating violently at some particular speed of motor bike.
(a) Why does this happen?
(b) What is the name of the phenomenon taking place?
(c) What could be done to stop the violent vibration?
Solution:
(a) This happens only when the frequency of vibration of the engine of bike becomes equal to the natural frequency of the rear view mirror.
(b) The name of this phenomenon is resonance.
(c) The speed of the bike should be change and this will change the frequency of vibration of the engine.
Question .9. Name the subjective property of sound related to its frequency and of light related to its wavelength.
Solution:
The subjective property of sound waves related to its frequency is pitch. The similar subjective property of light related to its wavelength is the colour of light.
Question .10. Two friends were playing on identical guitars whose strings were adjusted to give notes of the same pitch. Will the quality of the two notes be the same? Give a reason for your answer.
Solution:
The quality of the two notes need not be the same. The pitch of the note is determined by its frequency. The quality depends upon the number, distribution and relative intensity of the different harmonics and overtones.
Question .11. Give one example each of natural vibration, forced vibration and resonance.
Solution:
(i) Natural vibration : The vibration of a simple pendulum about its mean position.
(ii) Forced vibration : A sonometer wire under tension , vibrating under the influence of a vibrating tuning fork.
(iii) Resonance : A correctly adjusted length of a sonometer wire under proper tension, vibrating under the influence of a vibrating tuning fork.
Question .12. State three characteristics of a musical sound.
Solution:
The characteristics of musical sound are:
(i) loudness (intensity)
(ii) pitch (frequency)
(iii) quality (overall effect of all the harmonics present in the sound).
Question .13. How does musical sound differ from noise?
Solution:
| Musical sound | Noise |
1. It is a periodic disturbance which repeats itself at regular intervals. | 1. It is a non – periodic disturbance. |
2. There is no sudden change in amplitude or loudness. | 2. There are, very often, irregular sudden changes in amplitude or loudness or both. |
3. It is a sufficiently rapid so that it can be ensured as a sense of continuity. | 3. There is no any sense of continuity. |
Question .14. How does a stretched string on being set into vibration, produce a audible sound?
Solution:
When stretched string is set into vibration, it forces surrounding air to vibrate. This vibrating air produces an audible sound.
Question .15. Will the sound be audible if the string is set into vibration on the surface of the moon? Give reason for your answer.
Solution:
No, we will not hear any audible sound on the surface of the moon. There is no atmosphere on moon and no medium for the propagation of sound.
Question .16. A vibrating tuning fork is placed over the mouth of a burette filled with water. The tap is opened and the water level gradually falls. It is observed that the sound becomes the loudest for particular length of air column :
(i) What is the name of the phenomenon taking place when this happens?
(ii) Why does the sound become the loudest ?
(iii) What is the name of the phenomenon taking place when sound is produced for another length of air column and is not the loudest?
(iv) How does the frequency of the loud sound compare with that of the tuning fork?
(v) State the unit for measuring loudness.
Solution:
(i) The phenomenon is called resonance.
(ii) The sound becomes loudest because, the natural frequency of the length of air column becomes equal to the frequency of the vibrating tuning fork.
(iii) The first harmonic mode of vibration of the air column.
(iv)Frequency of the loud sound is either equal to or odd integral multiple of the frequency of tuning fork.
(v) Decibel.
Question .17. What is Sonar? State the principle on which it is based.
Solution:
SONAR means sound navigation and ranging. Sonar is an instrument that use ultrasonic wave for sound ranging. It measures even short time intervals quite accurately.
Sonar works on the principle of echo. A strong and short (ultrasonic) sound signal is sent towards the bottom of the ocean. Echo of the signal is then detected and depth of ocean is calculated.
Question .18. State two ways by which the frequency of transverse vibrations of a stretched string can be decreased.
Solution:
The frequency of transverse vibration is given by
n = 1/2L √(T/m)
Where L = length of the vibrating string, T = tension in the string, m = mass per unit length of the string, Therefore, the frequency of transverse vibration of a stretched can be decreased by:
(i) increasing the length of the string (keeping T and m constant)
(ii) decreasing the tension in the string (keeping L and m constant).
Question .19. Explain why musical instruments like the guitar are provided with a hollow box.
Solution:
When the string of the musical instrument is set into vibration, forced vibration are produced in the air inside the box. As area of the hollow box is large and hence it sets a large volume of air into vibration of the same frequency as that of string. So, a loud sound is produced due to resonance. The sound produced by vibrating string is too weak to be heard.
Question .20. What is the relation between frequency, wavelength and speed of a wave?
Solution:
V = ν λ where, V = speed, ν = frequency and λ = wavelength.
Numerical – 1. A person standing between two vertical cliffs and 640 m away from the nearest cliff shouted. He heard the first echo after 4 seconds and the second echo 3 seconds later. Calculate
(i) the velocity of sound in air and
(ii) the distance between the cliffs.
Solution:
(i) For the first echo time to reach the nearest cliff = t = 2 s,
distance = S = 640 m.
V = S / t = 640/2 = 320 m s-1.
(ii) Let x be the distance of the second cliff, time = t = (4 + 3)/2 = 3.5 s.
Therefore, x = V × t = 320 × 3.5 = 1120 m.
Hence, distance between two cliff = 640 + 1120 = 1760m.
Numerical – 2. A longitudinal wave of wavelength 1 cm travels in air with a speed of 330 m s-1. Calculate the frequency of the wave. Can this wave be heard by a normal human being?
Solution:
Here, λ = 1 cm = 0.01 m, V = 330 m s-1.
Using, V = ν λ => ν = V/λ = 330/0.01 = 33000 Hz.
No, this wave can not be heard by a normal human being.
Numerical – 3. A certain sound has a frequency of 256 hertz and wavelength of 1.3 m. Calculate the speed with which this sound travels. What difference would be felt by a listener between this sound and another sound traveling at the same speed but of wavelength 2.6 m.
Solution:
(i) Here, ν = 256 Hz, λ = 1.3 m, V = ?
V = ν λ = 256 × 1.3 = 332.8 m s-1.
(ii) In second case, λ = 2.6 and V = 332.8 m s-1,
Hence, ν = V/λ = 332.8/2.6 = 128 Hz.
As this frequency is half the original one the sound will be less shrill than the original one.
Numerical – 4. A sound wave of wavelength 0.332 m has a time period of 10-3 s. If the time period is decreased to 10-4 s, calculate the wavelength and frequency of the new wave.
Solution:
(i) Here, λ = 0.332 m, T = 10-3 s.
Using V = λ/T = 0.332/10-3 = 332 m s-1.
(ii) T = 10-4 s, λ = ? , ν = ?
Using V = λ/T => λ = VT = 332 × 10-4 = 0.0332 m.
And ν = 1/T = 1/10-4 = 104 Hz.
Numerical – 5. Radio waves of speed 3 × 108 m s-1 are reflected off the moon and received back on the earth, the time elapsed between the sending of the signal and receiving it back on the earth’s surface is 2.5 seconds. What is the distance of the moon from the earth?
Solution:
Here, c = 3 × 108 m s-1, t = 2.5 s, d = ?
This is a case of an echo, so, d = (c × t)/2
= (3 × 108 × 2.5)/2 = 3.75 × 108 m s-1.
Numerical – 6. An observer stands at a distance of 850 m from the cliff and fires a gun. After what time gap will he hear the echo, if sound travels at a speed of 350 m s-1 in air?
Solution:
Here, c = 350 ms-1, d = 850 m, t = ?
This is a case of an echo, so, 2d = c × t => t = 2d/c
= (2 × 850)/350 = 4.86 s.
Numerical – 7. A pendulum has a frequency of 5 vibration per second. An observer starts the pendulum and fires a gun simultaneously. He hears the echo from the cliff after 8 vibrations of the pendulum. If the velocity of sound in air is 340 m s-1, what is the ×distance between the cliff and the observer?
Solution:
Here, ν = 5, then T = 1/ν = 1/5 = 0.2 s.
Therefore, time for 8 vibrations = t = 8 × 0.2 = 1.6 s, c = 340 m s-1, d = ?
This is a case of an echo, so, d = (c × t)/2 = (340 × 1.6)/2 = 272 m.
Numerical – 8. The wavelength of waves produced on the surface of water is 20 cm. If the wave velocity is 24 m s-1, calculate :
(i) the number of waves produced in one second and
(ii) the time required to produce one wave.
Solution:
Here, λ = 20 cm = 0.20 m, V = 24 m s-1, ν = ?, T = ?
(i) Using formula, V = ν λ, we have
ν = V / λ = 24/0.20 = 120 Hz.
(ii) T = 1 / ν = 1 / 120 s.
Numerical – 9. A radar is able to detect the reflected waves from an enemy aeroplane, after a time interval of 0.20 milliseconds. If the velocity of the waves is 3 × 10^8 m s-1, calculate the distance of the plane from the radar.
Solution:
Here, t = 0.02 millisecond = 0.02 × 10-3 s, c = 3 × 108 m s-1, d = ?
Where d is the distance of the plane from the radar.
Using, 2d = c × t => d = (c × t)/2 = (3 × 108 × 0.02 × 10-3)/2
= 6000/2 = 3000 m = 3 km.
Numerical – 10. A man standing in front of a vertical cliff fires a gun. He hears the echo after 3 seconds. On moving closer to the cliff by 82.5 m, he fires again. This time, he hears the echo after 2.5 seconds. Calculate :
(i) the distance of the cliff from the initial position of the man.
(ii) the velocity of sound.
Solution:
Let distance of the cliff from the original position be d and V is the velocity of sound, then the time t after which the first echo was heard is given by
t = 2d/V —- —- —- —- —- (i)
In the second case, the time t’ is given by
t’ = 2(d – 82.5)/V —- —- —- —- (ii)
dividing (ii) by (i) we get t’/ t = (d – 82.5) / d
Or, 2.5 / 3 = 1 – 82.5/d => 82.5 / d = 1 – 2.5 / 3 = 0.5 / 3 = 1/6
Or, d = 82.5 × 6 = 495 m.
V = 2d / t = 2 × 495/3 = 990 / 3 = 330 m s-1.
Numerical – 11. A radar sends a signal to an aeroplane at a distance 45 km away with a speed of 3 × 10^8 m s-1. After how long is the signal received back from the aeroplane?
Solution:
Here, d = 45 km = 45 × 103 m, V = 3 × 108 m s-1, t = ?
Using V = 2d/t we get t = 2d / V = 2 × 45 × 103 / 3 × 108
= 3/104 = 0.0003 s.
