Question

Consider two tuning forks with natural frequency $250\, Hz$. One is moving away and another is moving towards a stationary observer at same speed. If the observer hears beats of frequency $5 Hz$, then the speed of the tuning fork is: (Given, speed of sound wave is $350 \,m / s )$

• A. 2.5 m / s
• B. 3.5 m / s
• C. 5.0 m / s
• D. 2.0 m / s

Answer 1

Answer: (B)

Given, speed of sound, $v=350 \,m / s$, actual frequency, $n_{0}=250 \,Hz$ and number of beats heared, $x=5$
As source is moving towards the observer therefore, the apparent frequency,
$n_{1}=n_{0} \cdot \frac{v}{v-v_{s}}\,\,\,\,\,\,\dots(i)$
Similarly, as source is moving away from the observer therefore the apparent frcquency,
$n_{2}=\frac{n_{0} v}{v+v_{s}}\,\,\,\,\,\,\dots(ii)$
Beats heared by the observer, $x=n_{1}-n_{2}$
Hence, from the Eq. (i) and (ii), we get,
$x= n_{0} v\left[\frac{1}{v-v_{s}}-\frac{1}{v+v_{s}}\right]$
$x =n_{0} v\left[\frac{v+v_{s}-v+v_{s}}{v^{2}-v_{s}^{2}}\right] $
$\Rightarrow \, x=n_{0} v\left[\frac{2 v_{s}}{v^{2}-v_{s}^{2}}\right]$
$x\left(v^{2}-v_{s}^{2}\right)=n_{0} v\left(2 v_{s}\right)$
where, $v_{s}=$ speed of the tuning fork
$\Rightarrow \, v_{s}^{2}+\frac{2 n_{0} v}{5} \cdot v_{s}-v^{2}=0$
Now, putting the values, in above Eqs. we get
$v_{s}^{2}+35000 v_{s}-122500=0$
It is a quadratic equation, so the positive value of $v_{s}$ is
$v_{s}=\frac{7}{2}=3.5 \,m / s$