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 16-E10). Find the beat frequency observed by the listener. Speed of sound in air = 332 m/s.

Answer 1

Here given velocity of the sources $v_s=0.$

Velocity of the observer, $v_0=3m/s$

So, the apparent frequency heard by the man

$=\left(\frac{332+3}{332}\right)\times 256=258.3Hz$

From the approaching tuning fork $=f^1$

Similarly, $f^{11}=\left(\frac{332-3}{332}\right)\times 256$

=253.7 Hz.

So, beat produced by them

=258.3-253.7=4.6 Hz