Question

Two tuning forks with natural frequencies $340\,Hz$ 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 $3\, Hz$. Find the speed of the tuning forks.

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

Answer 1

The Correct Option is A

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 $f =n_{1}-n_{2}$
$\Rightarrow f=\left(\frac{v}{v-v_{s}}\right) n-\left(\frac{v}{v+v_{s}}\right) n$
$\Rightarrow f=\frac{v\left(v+v_{s}\right)-v\left(v-v_{s}\right)}{v^{2}-v_{s}^{2}}(n)$
$\Rightarrow \frac{f v}{2 n}$ putting $v=340 m/s, f=3,340\,Hz$ we get
$v_{s}=\frac{340 \times 3}{3 \times 340}=1.5\, m/s$