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Two stationary sources A and B are sounding notes of frequency 340 Hz. A listener moves from A to B with a constant velocity u. If the speed of sound is 340 m/s. What must be the value of u. So that he hears 10 beats?

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10 m/s

5 m/s

2.5 m/s

20 m/s

answer is B.

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Detailed Solution

$f_{beat} = f_{1} - f_{2} \Rightarrow 10 = (\frac{340 + u}{340} - \frac{340 - u}{340}) \times 340 \Rightarrow 10 = 2 u \Rightarrow u = 5 m / s$

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