Question

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

• 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, $v=350m/s$, actual frequency, $n_0=250Hz$ 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,
$x=n_{0}v\left[\frac{1}{v-v_s}-\frac{1}{v+v_s}\right]$
$x=n_{0}v\left[\frac{v+v_s-v+v_s}{v^2-v_s^2}\right]$
$\Rightarrow x=n_{0}v\left[\frac{2v_s}{v^2-v_s^2}\right]$
$x\left(v^2-v_s^2\right)=n_{0}v\left(2v_s\right)$
where, $v_s=$ speed of the tuning fork
$\Rightarrow v_s^2+\frac{2n_{0}v}{5}\cdot v_s-v^2=0$
Now, putting the values, in above Eqs. we get
$v_s^2+35000v_s-122500=0$
It is a quadratic equation, so the positive value of $v_s$ is
$v_s=\frac{7}{2}=3.5m/s$