Two tuning forks A and B each of natural frequency $85 H z$ move with velocity $10 m/s$ relative to stationary observer 'O'. Fork A moves away from the observer while the fork B moves towards him as shown in the figure. A wind with a speed $10 m/s$ is blowing in the direction of motion of fork A. Find the beat frequency measured by the observer in Hz. [Take speed of sound in air as $340 m/s]$

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Solution

$f_{b e a t} = f_{A} - f_{B}$
From doppler's effect
Apparent frequency for A
$f_{A} = \frac{340 - 10}{340 - 10 + 10} \times 85 = \frac{330}{4} Hz$
$= \frac{330}{4} Hz$
Similarly, apparent frequency for B
$f_{B} = \frac{340 + 10}{340 + 10 - 10} \times 85 = \frac{350}{4} Hz$
$f_{b e a t} = f_{A} - f_{B} = 5 Hz$

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fbeat=fA−fB From doppler's effect Apparent frequency for A fA=340−10340−10+10×85=3304Hz =3304Hz Similarly, apparent frequency for B fB=340+10340+10−10×85=3504 Hz fbeat=fA−fB=5 Hz