 # A stationary observer receives sound from two identical tuning forks, `
Question:

A stationary observer receives 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/sec. The oscillation frequency of each tuning fork is $v_{0}=1400 \mathrm{~Hz}$ and the velocity of sound in air is $350 \mathrm{~m} / \mathrm{s}$. The speed of each tuning fork is close to:

1. $\frac{1}{2} \mathrm{~m} / \mathrm{s}$

2. $1 \mathrm{~m} / \mathrm{s}$

3. $\frac{1}{4} \mathrm{~m} / \mathrm{s}$

4. $\frac{1}{8} \mathrm{~m} / \mathrm{s}$

Correct Option: , 3

Solution:

(3) From Doppler's effect, frequency of sound heard $\left(f_{1}\right)$ when source is approaching

$f_{1}=f_{0} \frac{c}{c-v}$

Here, $c=$ velocity of sound

$v=$ velocity of source

Frequency of sound heard $\left(f_{2}\right)$ when source is receding

$f_{2}=f_{0} \frac{c}{c+v}$

Beat frequency $=f_{1}-f_{2}$

$\Rightarrow 2=f_{1}-f_{2}=f_{0} c\left[\frac{1}{c-v}-\frac{1}{c+v}\right]$

$=f_{0} c \frac{2 v}{c^{2}\left[1-\frac{v^{2}}{c^{2}}\right]}$

For $c \gg>v$

$\Rightarrow v=\frac{2 c}{2 f_{0}}=\frac{c}{f_{0}}=\frac{350}{1400}=\frac{1}{4} \mathrm{~m} / \mathrm{s}$