Question

A stationary observer receive sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears $2$ beats per second. The oscillation frequency of each tuning fork is $f_0=1400\text{Hz}$ and the velocity of sound in air is $350{\text{ms}}^{-1}$. The speed of each tuning fork is close to:

• A. $\frac{1}{2}{\text{ms}}^{-1}$
• B. $1{\text{ms}}^{-1}$
• C. $\frac{1}{4}{\text{ms}}^{-1}$
• D. $\frac{1}{8}{\text{ms}}^{-1}$

Answer 1

The correct option is C $\frac{1}{4}{\text{ms}}^{-1}$

As the observer is stationary,

From Doppler's effect, frequency of sound heard $\left(f_1\right)$ by observer when source is approaching will be

$f_1=f_0\frac{v}{v-v_s}$

Here, $v=$ velocity of sound
$v_s=$ velocity of source

Frequency of sound heard $\left(f_2\right)$ by observer when source is receding

$f_1=f_0\frac{v}{v+v_s}$

Beat frequency $f=|f_1-f_2|$

$\Rightarrow 2=|f_1-f_2|=f_{0}v[\frac{1}{v-v_s}-\frac{1}{v+v_s}]$
$\Rightarrow 2=f_{0}v\frac{2v_s}{v^2[1-\frac{v_s^2}{v^2}]}$
$\text{For}v>>v_s,\left(\frac{v_s^2}{v^2}\right)$ can be neglected

$\Rightarrow v_s=\frac{v}{f_0}=\frac{350}{1400}=\frac{1}{4}{\text{ms}}^{-1}$

Hence, option $\left(C\right)$ is correct.