Question
Two identical tuning forks vibrating at the same frequency 256Hz are kept fixed at some distance apart. A listener runs between the forks at a speed of 3.0m/s so that he approaches one tuning fork and recedes from the other figure. Find the beat frequency observed by the listener. Speed of sound in air = 332ms/s.

Answer


Here given velocity of the sources vs = 0

Velocity of the observer v0 = 3m/s

So, the apparent frequency heard by the man $=\Big(\frac{332+3}{332}\Big)\times256=258.3\text{Hz}.$

From the approaching tuning form = f'

$\text{f}'=\Big[\frac{(332-3)}{332}\Big]\times256=253.7\text{Hz}.$

So, beat produced by them $=258.3-253.7=4.6\text{Hz}.$

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