Question

A police bike with its siren on moves with a velocity ${v_s}$ towards a man standing at a bus stop. As the bike approaches the man hears the sound at a frequency of ${f_d}$ . Take the speed of the sound to be $v$. Calculate the wavelength of the sound produced by the siren.
A) $\dfrac{{v - {v_s}}}{{v{f_d}}}$
B) $\dfrac{{{v^2}}}{{\left( {v + {v_s}} \right){f_d}}}$
C) $\dfrac{{{v^2}}}{{\left( {v - {v_s}} \right){f_d}}}$
D) $\dfrac{{{v^2}{f_d}}}{{\left( {v - {v_s}} \right)}}$

Answer 1

Here the police bike approaching the man with its siren on is the source of the sound. The man at the bus stop is the observer of this sound. The situation demands the doppler effect be invoked. The Doppler effect describes the concept of change in frequency of sound when the source or the observer or both are in motion. The frequency and wavelength of a wave are related to each other.

Formulas used:
The apparent frequency heard by an observer is given by, ${f_o} = {f_s}\left( {\dfrac{{v + {v_s}}}{{v + {v_o}}}} \right)$ where ${f_s}$ is the frequency of the sound wave, $v$ is the speed of sound, ${v_s}$ is the velocity of the source and ${v_o}$ is the velocity of the observer.
The wavelength of a wave is given by, $\lambda = \dfrac{v}{f}$ where $v$ is the velocity of the wave and $f$ is the frequency of the wave.

Complete step by step answer:
Step 1: List the parameters given in the question.
In the given problem, the police bike is the source of the sound produced by the siren and the man at the bus stop is the observer.
As the bike is approaching the man the frequency of the sound heard by the observer will be different from the original frequency of the sound wave. This is referred to as the doppler effect.
The velocity of the bike is given to be ${v_s}$ .
The velocity of sound is given to be $v$ .
The apparent frequency heard by the man is given to be ${f_o} = {f_d}$ .
Let ${f_s}$ be the frequency of the sound produced by the siren.
Step 2: Based on the doppler effect, express the relation for the apparent frequency heard by the man.
The apparent frequency heard by the stationary man as the bike is in motion can be expressed as ${f_d} = {f_s}\left( {\dfrac{{v + {v_o}}}{{v + {v_s}}}} \right) = {f_s}\left( {\dfrac{v}{{v + {v_s}}}} \right)$
$\Rightarrow {f_s} = \dfrac{{{f_d}}}{{\left( {\dfrac{v}{{v + {v_s}}}} \right)}} = {f_d}\dfrac{{\left( {v + {v_s}} \right)}}{v}$
But as the police bike is moving towards the man, ${v_s}$ will be $- {v_s}$.
So we have the frequency of the sound wave as ${f_s} = {f_d}\dfrac{{\left( {v - {v_s}} \right)}}{v}$ -------- (1)
Step 3: Using equation (1) express the wavelength of the sound wave.
The wavelength of the sound wave can be expressed as $\lambda = \dfrac{v}{{{f_s}}}$ --------- (2)
Substituting equation (1) in equation (2) we get,
$\Rightarrow \lambda = \dfrac{v}{{{f_d}\dfrac{{\left( {v - {v_s}} \right)}}{v}}} = \dfrac{{{v^2}}}{{{f_d}\left( {v - {v_s}} \right)}}$
Thus the wavelength of the sound produced by the siren is obtained as $\lambda = \dfrac{{{v^2}}}{{{f_d}\left( {v - {v_s}} \right)}}$ .

Therefore, the correct option is C.

Note:
The frequency and the wavelength are inversely proportional to each other. So a sound wave of high frequency will have a short wavelength and a sound wave of low frequency sound will have a longer wavelength. The velocity of the source is taken to be negative if the source moves towards the observer and is taken to be positive if it moves away from the observer. Here the man is stationary and so the velocity of the man will be zero i.e., ${v_o} = 0$ .