Question

A police man on duty detects a drop of $10\%$ in the pitch of the motion of car as it crosses him. If the velocity of sound is $330m/s$. Calculate the speed of the car:

• A. $17.4m/s$
• B. $20.4m/s$
• C. $18.6m/s$
• D. $16.4m/s$

Answer 1

The correct option is D $16.4m/s$

We know by Doppler effect the

$f^{\prime }=f\frac{(v\pm v_o)}{(v\pm v_s)}$

While $f^{\prime }$ frequency is the observed $f$ is the original frequency $v$ is the speed of sound $v_o$ is the observers speed and $v_s$ is the sources speed.

i) When the car is approaching the man

Then, $f_1=f\frac{(v+o)}{(v-v_s)}$

$f^{\prime }=f\frac{\left(330\right)}{(330-v_3)}$

ii) When the car is separating from him,

Then , $f_2^{\prime }=f\frac{(v+o)}{v+v_s}$

Change in frequency observed $=f_e^{\prime }-f_1^{\prime }$

As, pitch drops by $10\%$

$\frac{△f}{f}=\frac{-1}{10}\Rightarrow △f=\frac{-f}{10}$

So, $f_2^{\prime }-f_1^{\prime }=\frac{-f}{10}$

$\Rightarrow f\frac{\left(v\right)}{v+v_s}-f\frac{\left(v\right)}{v-v_s}=\frac{-f}{10}$

$\Rightarrow \frac{-330\left[2v_s\right]}{(330{)}^2-v{s}^2}=\frac{-}{10}$

$\Rightarrow v_s^2-2\times 3300v_s-(330{)}^{2}20$

By solving, we get

$v_s=-16.458$ and $v_s=6616.45$

As, second roots is not much practical.

Option $D$ is correct as $|v_s|=16.4m/s$