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Two sources A and B are sounding notes of frequency 680 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 per second?

A

3.5 m/s

B

3 m/s

C

2 m/s

D

2.5 m/s

Text Solution

Verified by Experts

The correct Answer is:
D
`v_(0) = u, v_(s) = 340, n = 680 Hz`

When the listener moves from A to B, he approaches a stationary source. The apparent frequency heard by the listener is
`n_(1) = ((v + v_(0))/(v)) xx n`
`= ((340 + u)/(340)) xx 680 - 680 + 2u`....(1)
As he is going away from A, the apparent frequency is less than n. It is given by
`n_(2) = (340 - u)/(340) xx 680`
`= (340 - u) xx 2 = 680 - 2u`...(2)
`n_(1) - n_(2) = 680 + 2u - 680 (-2u) = 4u`
`:. 10 = 4u " " :. u = 2.5 m//s`
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