A tuning fork of frequency `256 h_(Z)` is moving towards a well with a velocity of `5 m//s` . If the speed of sound is `330 m//s` , then the number of beats heard per second by a stationary observer lying between tuning fork and the well is

Answer

8

Explanation

The observer will hear a sound of the source moving away from him and another sound after reflection from the wall. the apparent frequencies of 5 these sound are
`V_(1)=(v V)/(V+u)=(256xx330)/((330+5))=252Hz`
`V_(2)=(v V)/(V+u)=(256xx330)/((330-5))=260Hz`
Therefore, the number of beats per second (beat frequency)=260-252=8.