Question

Two tuning forks with natural frequencies $340\text{Hz}$ each move relative to stationary observer. One fork moves away from the observer while the other moves towards him at the same speed. The observer hears beats of frequency $3\text{Hz}$. Find the speed of the tuning fork (velocity of sound in air is $340\text{m/s}$).

• A. $6\text{m/s}$
• B. $4.5\text{m/s}$
• C. $1.5\text{m/s}$
• D. $3\text{m/s}$

Answer 1

The correct option is C $1.5\text{m/s}$

Given,
Velocity of sound, $v=340\text{m/s}$
Natural frequency of tuning fork, $f_0=340\text{Hz}$

Let,
$f_1=$ frequency observed by observer due to fork $1$,
$f_2=$ frequency observed by observer due to fork $2$,
$v_s=$ velocity of tuning fork,


As we know that,
$f_1=f_0\left(\frac{v}{v-v_s}\right)$ and

$f_2=f_0\left(\frac{v}{v+v_s}\right)$

According to the problem, the beat frequency,
$f_1-f_2=3$

Substituting the values we get,
$f_0\left(\frac{v}{v-v_s}\right)-f_0\left(\frac{v}{v+v_s}\right)=3$

$\Rightarrow [\frac{1}{(1-\frac{v_s}{v})}-\frac{1}{1+\frac{v_s}{v}}]f_0=3$

$\Rightarrow [{(1-\frac{v_s}{v})}^{-1}-{(1+\frac{v_s}{v})}^{-1}]f_0=3$

By binomial expansion with neglecting higher terms,

$\Rightarrow [(1+\frac{v_s}{v})-(1-\frac{v_s}{v})]f_0=3$

$\Rightarrow \frac{2v_s{f}_0}{v}=3$

$\Rightarrow v_s=\frac{3v}{2f_0}$

Substituting the values, we get

$v_s=\frac{\left(3\right)\left(340\right)}{\left(2\right)\left(340\right)}=1.5\text{m/s}$

Hence, correct option is (a).