Question

Two turning forks with natural frequencies $340Hz$ each move relative to a stationary observer. One fork moves away from the observer, while the other moves towards the observer at the same speed. The observer hears beats of frequency $3Hz$. Find the speed of the turning forks.

• A. $1.5m/s$
• B. $2m/s$
• C. $1m/s$
• D. $2.5m/s$

Answer 1

The correct option is A $1.5m/s$

Let $v=$ speed of sound and $v_S=$ speed of tuning forks.
Apparent frequency of fork moving towards the observer is
$n_1=\left(\frac{v}{v-v_s}\right)n$
Apparent frequency of the fork moving away from the observer is
$n_2=\left(\frac{v}{v+v_s}\right)n$
If $f$ is the number of beats heard per second.
then $f=n_1-n_2$
$\Rightarrow f=\left(\frac{v}{v-v_s}\right)n-\left(\frac{v}{v+v_s}\right)n$
$\Rightarrow f=\frac{v(v+v_s)-v(v-v_s)}{v^2-v_s^2}\left(n\right)$
$\Rightarrow \frac{2v{v}_{s}n}{v^2-v_s^2}=f\Rightarrow 2\left(\frac{v_s}{v}\right)=n=f\{if{v}_s<<v\}$
$\Rightarrow v_s=\frac{fv}{2n}$
putting $v=340m/s,f=3,n=340Hz$ we get,
$v_s=\frac{340\times 3}{3\times 340}=1.5m/s$.