Two identical tuning forks vibrating at the same frequency 256 Hz are kept fixed at some distance apart. A listener runs between the forks at a speed of `3.0 m s^-1` so that he approaches one tuning-fork and recedes from the other. find the beat frequency bserved by the listener. Speed of sound in air `= 332 m s^-1`.
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Answer

D

Explanation

Here given veolocity of he sources `v_s=0.
velocity of the observer `v_0=3m/s`
So, the apparent frequency heard by the man
`=((332+3)/332)xx256=258.3 Hz`
From the apparoching tuning fork `=f^1`
Similarly `f^11=((332-3)/332)xx256`
`=253.7 Hz`
So, beat produced by them
`=258.3-253.7=4.6Hz`