Two identical tuning forks vibrating at the same frequency 256 Hz… - Vidyadip

Question

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 $3.0\ m/s$ 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 $= 332\ ms/s.$

✓

Answer

Here given velocity of the sources $v_s = 0$
Velocity of the observer $v_0 = 3\ m/s$
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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