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Two sources A and B are sending notes of frequency 680 Hz. A listener moves from A to B with a constant velocity 'u'. If the speed of sound in air is 340 ms^–1, what must be the value 'u' so that he hears 10 beats per second ?

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2.0 m-s^–1

2.5 m-s^–1

3.0 m-s^–1

3.5 m-s^–1

answer is B.

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Detailed Solution

$n'_{A} = \frac{v - u}{v} n$

$n'_{B} = \frac{v + u}{v} n$

$∆ n = \frac{2 un}{V} \Rightarrow 10 = \frac{2 u (680)}{340}$

$∴ u = 2.5 {ms}^{- 1}$

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