Question

A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one receded with the same speed (much less than the speed of sound). The observer hears $2beats/sec$. The oscillation frequency of each tuning fork is $v_0=1400Hz$and the velocity of sound in air is $350m/s$. The speed of each tuning fork is close to

• A. $\frac{1}{4}m/s$
• B. $1m/s$
• C. $\frac{1}{2}m/s$
• D. $\frac{1}{8}m/s$

Answer 1

The correct option is A

$\frac{1}{4}m/s$

Step 1: Given data

No. of beats heard by the observer in one second $n=2$

The oscillation frequency of each tuning fork, $v_0=1400Hz$

Velocity of sound in air, $c=350m/s$

Step 2: Find the speed of each tuning fork, $\mathit{v}$

If, $f_1=$frequency of the tuning fork which is approaching to the observer,

Then, $f_1=\frac{v_{0}c}{c-v}$

If, $f_2=$frequency of the tuning fork which is receding

Then, $f_1=\frac{v_{0}c}{c+v}$

No. of beats heard by the observer in one second$n=\left|f_1-f_2\right|$

$$\begin{gathered} \Rightarrow n=\frac{v_{0}c}{(c-v)}-\frac{v_{0}c}{(c+v)} \\ \Rightarrow n=\frac{1400\times 350}{350-v}-\frac{1400\times 350}{350+v} \end{gathered}$$

As, beats$n=2$

Therefore, $\frac{1400\times 350}{350-v}-\frac{1400\times 350}{350+v}=2$

$v=\frac{1}{4}m/s$

Hence, option (A) is correct.