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 3.0m s⁻¹ 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 m s⁻¹.

Answer 1

Given:
Speed of sound in air v = 332 ms⁻¹
Velocity of the observer $v_0$ = 3 ms^$-$1
Velocity of the source $v_s$ = 0
Frequency of the tuning forks $f_0$ = 256 Hz
The apparent frequency $\left(f_1\right)$ heard by the man when he is running towards the tuning forks is

$f_1=\left(\frac{v+v_0}{v}\right)\times f_0$

On substituting the values in the above equation, we get:

$f_1=\left(\frac{332+3}{332}\right)\times 256=258.3\mathrm{Hz}$

The apparent frequency $\left(f_2\right)$ heard by the man when he is running away from the tuning forks is

$f_2=\left(\frac{v-v_0}{v}\right)\times f_0$

On substituting the values in the above equation, we get:

$$\begin{gathered} f_2=\left(\frac{332-3}{332}\right)\times 256 \\ =253.7\mathrm{Hz}. \end{gathered}$$

∴ beats produced by them
= $f_2-f_1$
=258.3 − 253.7 = 4.6 Hz