Question

Figure shows a person standing somewhere in between two identical tuning forks. each vibrating at 512 Hz. If both the tuning forks move towards right a speed of 5.5 m s⁻¹, find the number of beats heard by the listener. Speed of sound in air = 330 m s⁻¹.

Answer 1

Given:
Frequency of tuning forks $f_0$ = 512 Hz
Speed of sound in air v = 330 ms⁻¹
Velocity of tuning forks $v_s$ = 5.5 ms⁻¹
The apparent frequency $\left(f_1\right)$ heard by the person from the tuning fork on the left is given by:

$f_1=\left(\frac{v}{v-v_s}\right)\times f_0$
On substituting the values in the above equation, we get:

$$\begin{gathered} f_1=\left(\frac{330}{330-55}\right)\times 512 \\ =520.68\mathrm{Hz} \end{gathered}$$

Similarly, apparent frequency ^{$\left(f_2\right)$} heard by the person from the tuning fork on the right is given by:
$f_2=\left(\frac{v}{v-v_s}\right)\times f_0$

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

$$\begin{gathered} f_2=\left(\frac{330}{330-5.5}\right)\times 512 \\ =503.60\mathrm{Hz} \end{gathered}$$

∴ beats produced
=$f_1-f_2$
= 520.68 − 503.60 = 17.5 Hz

As the difference is greater than 10 ( persistence of sound for the human ear is 1/10 of a second), the sound gets overlapped and the observer is not able to distinguish between the sounds and the beats.