First slide

Doppler's effect

Question

A source of sound S of frequency 500 Hz situated between a stationary observer O and a wall W, moves towards the wall with a speed of 2 m/s. If the velocity of sound is 332 m/s, then the number of beats per second heard by the observer is (approximately)

Difficult

A

8
B

6
C

4
D

2

Solution

For direct sound source is moving away from the observes so frequency heard in this case
$n_{1} = n (\frac{v}{v + v_{s}}) = 500 (\frac{332}{332 + 2}) = 500 (\frac{332}{334}) Hz$
The other sound is echo, reaching the observer from the wall and can be regarded as coming from the image of source formed by reflection at the wall. This image is approaching the observer in the direction of sound.
Hence for reflected sound, frequency heard by the observer is
$n_{2} = n (\frac{v}{v - v_{S}}) = 500 (\frac{332}{332 - 2}) = 500 (\frac{332}{330}) Hz$
Beats frequency $= n_{2} - n_{1} = 500 \times 332 (\frac{1}{330} - \frac{1}{334}) = 6 .$

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