Question

A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes 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. $\frac{1}{2}m/s$
• B. $\frac{1}{4}m/s$
• C. $1m/s$
• D. $\frac{1}{8}m/s$
  

Answer 1

The correct option is B $\frac{1}{4}m/s$

Formula used:
$f=\frac{(v+v_0)}{(v-v_s)}{f}_0$
Beat frequency $=f_1-f_2$
Apparent frequency of source (1),
$f_1=f\left(\frac{v}{v-v_s}\right)f_0$
Apparent frequency of source (2),
$f_2=f\left(\frac{v}{v+v_s}\right)f_0$
Beat frequency $=f_1-f_2$
$\Delta f=v(\frac{1}{v-v_s}-\frac{1}{v+v_s})f_0$
$\Delta f=v{f}_0\left(\frac{v+v_s-v+v_s}{v^2-v_s^2}\right)=\frac{2v{f}_0{v}_s}{v^2-v_s^2}$
Since $v_s\ll v$
$v^2-v_s^2\approx v^2$
$\Rightarrow \Delta f=\frac{\left(2v{f}_0{v}_s\right)}{v^2}=\frac{\left(2f_0{v}_s\right)}{v}$
$\Rightarrow 2=\frac{2\times 1400\times v_s}{350}$
$\Rightarrow v_s=\frac{1}{4}m/s$

Final answer: (d)