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 while receding from the other (figure). Find the beat frequency observed by the listener. Speed of sound in air = 332 m/s.

• A. 5.2
• B. 4.2
• C. 4.6
• D. 5.6

Answer 1

The correct option is C

4.6

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 from the tuning fork in front= $f^{\prime }=\left(\frac{332+3}{332}\right)\times 256=258.3$

the apparent frequency heard by the man from the tuning fork behind him=$f”=\left[\frac{(332-3)}{332}\right]\times 256=253.7Hz$.

So, beat produced by them = 258.3 - 253.7 = 4.6 Hz