First slide

Beats

Question

Two sources P and Q produce notes of frequency 660Hz each. A listener moves from P to Q with a speed of $1 m s^{- 1}$ . If the speed of sound is 330 $m s^{- 1}$ , then the number of beats heard by the listener per second will be

Easy

A

4
B

8
C

2
D

zero

Solution

The situation as shown in the figure.
$H e r e, Speed of listener, v_{L} = 1 {ms}^{- 1} Speed of sound, v = 330 {ms}^{- 1} Frequency of each source, ν = 660 Hz Apparent frequency due to P, v^{'} = \frac{v (v - v_{L})}{v} Apparent frequency due to Q, v^{''} = \frac{v (v + v_{L})}{v} Number of beats heard by the listener per second is v^{''} - v^{'} = \frac{v (v + v_{L})}{v} - v \frac{(v - v_{L})}{v} = \frac{2 {vv}_{L}}{v} = \frac{2 \times 660 \times 1}{330} = 4$

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