Waves
- Two vibrating tuning forks produce progressive waves given by y1 = 4 sin 500 πt and y2 = 2 sin 506 πt. Number of beats produced per minute is
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Equation of progressive wave is given by
y = A sin2πf t
Given y1 = 4sin500 πt and y2 = 2sin506πt.
Comparing the given equations with equation of progressive wave, we get
2f1 = 500 f1 = 250
2f2 = 506 f2 = 253
Beats = f2– f1 = 253 – 250 = 3 beats/sec
= 3 × 60 = 180 beats/minute.Correct Option: B
Equation of progressive wave is given by
y = A sin2πf t
Given y1 = 4sin500 πt and y2 = 2sin506πt.
Comparing the given equations with equation of progressive wave, we get
2f1 = 500 f1 = 250
2f2 = 506 f2 = 253
Beats = f2– f1 = 253 – 250 = 3 beats/sec
= 3 × 60 = 180 beats/minute.
- A source of sound S emitting waves of frequency 100 Hz and an observor O are located at some distance from each other. The source is moving with a speed of 19.4 ms–1 at an angle of 60° with the source observer line as shown in the figure. The observor is at rest. The apparent frequency observed by the observer is (velocity of sound in air 330 ms–1)
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Here, original frequency of sound, f0 = 100 Hz
Speed of source Vs = 19.4 cos 60° = 9.7
From Doppler's formula
f1 = f0f1 = f0 
v - v0 
v - vs f1 = 100 v - 0 v - (19.7) f1 = 100 v = 100 V 1 - 9.7 1 - 9.7 V 330
= 103Hz
Apparent frequency f1 = 103 HzCorrect Option: A
Here, original frequency of sound, f0 = 100 Hz
Speed of source Vs = 19.4 cos 60° = 9.7
From Doppler's formula
f1 = f0f1 = f0 v - v0 v - vs f1 = 100 v - 0 v - (19.7) f1 = 100 v = 100 V 1 - 9.7 1 - 9.7 V 330
= 103Hz
Apparent frequency f1 = 103 Hz
- A speeding motorcyclist sees trafic jam ahead of him. He slows down to 36 km/hour. He finds that traffic has eased and a car moving ahead of him at 18 km/hour is honking at a frequency of 1392 Hz. If the speeds of sound is 343 m/s, the frequency of the honk as heard by him will be :
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According to Doppler's effect
Apparent frequencyn' = n v - v0 = 1392 343 - 10 = 1412 Hz v - vs 343 + 5 Correct Option: C
According to Doppler's effect
Apparent frequencyn' = n v - v0 = 1392 343 - 10 = 1412 Hz v - vs 343 + 5
- A train moving at a speed of 220 ms–1 towards a stationary object, emits a sound of frequency 1000 Hz. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected by the driver of the train is :
(speed of sound in air is 330 ms–1)
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Frequency of the echo detected by the driver of the train is
(According to Doppler effect in sound)ƒ' = v - u ƒ v - u
where ƒ = original frequency of source of sound
ƒ′ = Apparent frequency of source because of the relative motion between source and observer.ƒ' = 330 - 220 1000 = 5000 Hz 330 - 220 Correct Option: C
Frequency of the echo detected by the driver of the train is
(According to Doppler effect in sound)ƒ' = v - u ƒ v - u
where ƒ = original frequency of source of sound
ƒ′ = Apparent frequency of source because of the relative motion between source and observer.ƒ' = 330 - 220 1000 = 5000 Hz 330 - 220
- The driver of a car travelling with speed 30 m/sec towards a hill sounds a horn of frequency 600 Hz. If the velocity of sound in air is 330 m/s, the frequency of reflected sound as heard by driver is
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f ′ is the apparent frequency received by an observer at the hill. f ′′ is the frequency of the reflected sound as heard by driver.f' = v f, v - 30 f" = v + 30 f, v + 30 f = 360 × 600 v v - 30 300
= 720 HzCorrect Option: B
f ′ is the apparent frequency received by an observer at the hill. f ′′ is the frequency of the reflected sound as heard by driver.f' = v f, v - 30 f" = v + 30 f, v + 30 f = 360 × 600 v v - 30 300
= 720 Hz
