Question

A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of $2Hz$. Find the speed of each tuning fork, if their oscillation frequency is $680Hz$ and the velocity of sound in air is $340m/s$ [Use $g=10m/s^2$]

• A. $1m/s$
• B. $1.5m/s$
• C. $2m/s$
• D. $0.5m/s$

Answer 1

The correct option is D $0.5m/s$

We know that: $n^{\prime }=n\left(\frac{v\pm v_0}{v\mp v_S}\right)$

Given,
Velocity of sound in air, $v=340m/s$
Actual frequency, $n=680Hz$
Beat frequency, $n_1-n_2=2Hz$
Find the speed of tuning fork, $v_S=?$

Frequency from approaching tuning fork,

$n_1=n\left(\frac{v}{v-v_S}\right)\dots \left(i\right)$
Frequency from receding tuning fork,
$$\begin{array}{cc}{n}_2& =n\left(\frac{v}{v-v_S}\right)\dots \left(ii\right)\\ n_1-n_2& =n\left(\frac{v}{v-v_S}\right)-n\left(\frac{v}{v+v_S}\right)\\ 2& =680\left(\frac{340}{340-v_S}\right)-680\left(\frac{340}{340+v_S}\right)\\ 2& =680{(1-\frac{v_S}{340})}^{-1}-680{(1+\frac{v_S}{340})}^{-1}\end{array}$$
Using Binomial expansion, we have

$2=680[1+\frac{v_S}{340}-1+\frac{v_S}{340}]$

$2=680\left[\frac{2v_S}{340}\right]$


$v_S=\frac{2}{4}$

$v_S=0.5m/s$

Final answer: (c)