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Q.

A man is watching two trains, one leaving and the other coming in, with equal speed of 4 m/s. If they sound their whistles, each of natural frequency of240 Hz the number of beats heard by the man (velocity of sound in air = 320 m/s) will be equal to

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a

6

b

3

c

zero

d

12

answer is A.

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Detailed Solution

When the train is coming to the observern′=VV−Vsn=320320−4×240 n′′=VV+Vsn=320320+4×240 ≅237Hz When the train is moving away from observer  Beat frequency =n′−n′′=243−237=6
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