A man is watching two trains one leaving and other approaching with equal velocities of 4 ms^(-1). If they sound their whistles each of natural frequency 240 hz, the number of beats heard per seceond by the mas will be (velocity of sound in air 320 ms^(-1) ) :

A man is watching two trains one leaving and other approaching with eq

1.	A man is watching two trains one leaving and other approaching with equal velocities of 4 ms^(-1). If they sound their whistles each of natural frequency 240 hz, the number of beats heard per seceond by the mas will be (velocity of sound in air 320 ms^(-1) ) :
Answer» 6 12 4 None of these. SOLUTION :here `V_(1) = ((320)/(320 - 4)) 240 = 243` Hz. , and `v_(2) = ((320)/(320 + 4)) 240 = 237 Hz. `, `THEREFORE` beat frequency , b = `v_(1) - v_(2)` = 243 - 237 = 6 Hz. Hence the correct choice is (a).

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A man is watching two trains one leaving and other approaching with equal velocities of 4 ms^(-1). If they sound their whistles each of natural frequency 240 hz, the number of beats heard per seceond by the mas will be (velocity of sound in air 320 ms^(-1) ) :

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