Question

A boy with a radio playing music at frequency ‘f’ is moving towards a wall with velocity ${{v}_{b}}$ . A motorist is following the boy with speed ${{v}_{m}}$. The expression for beat frequency heard by the motorist, if the speed of sound is v, is:
$\begin{aligned} & a)\left( \dfrac{v+{{v}_{m}}}{v+{{v}_{b}}} \right)f \\\ & b)\left( \dfrac{v+{{v}_{m}}}{v-{{v}_{b}}} \right)f \\\ & c)\dfrac{2{{v}_{b}}(v+{{v}_{m}})}{{{v}^{2}}-{{v}_{b}}^{2}}f \\\ & d)\dfrac{2{{v}_{m}}(v+{{v}_{b}})}{{{v}^{2}}-{{v}_{m}}^{2}}f \\\ \end{aligned}$

Answer 1

It is given in the question that the boy is moving towards the wall. Hence the motorist will hear two sounds i.e. one directly from the radio and one after the reflection of the sound wave from the wall. When there is a relative motion between the source and the observer the frequency of sound heard by the observer is different from the actual frequency of the source. Hence we will determine the apparent frequency heard by the motorist from the reflection from the wall and the radio and then determine the beat frequency i.e. the difference between the two frequencies.

Complete answer:
The apparent frequency heard by an observer, when there is relative distance between the source and the observer is given by,
${{f}^{'}}=\left( \dfrac{v-{{V}_{o}}}{v-{{V}_{s}}} \right)f$
Where f is the actual frequency of the source, v is the frequency of sound ${{V}_{o}}$is the velocity of the observer and ${{V}_{s}}$is the velocity of the source. The velocity of the observer is taken as negative when its approaching the observer and positive when moving apart. Similarly, the velocity of the source is taken as positive when its approaching the observer and negative when moving apart.
In the above case the motorist is moving towards the boy while the boy with the radio moves away from the motorist. Hence the apparent frequency heard by the motorist is,
$\begin{aligned} & {{f}_{1}}=\left( \dfrac{v-{{V}_{o}}}{v-{{V}_{s}}} \right)f \\\ & {{f}_{1}}=\left( \dfrac{v-\left( -{{V}_{m}} \right)}{v-\left( -{{V}_{b}} \right)} \right)f \\\ & {{f}_{1}}=\left( \dfrac{v+{{V}_{m}}}{v+{{V}_{b}}} \right)f \\\ \end{aligned}$
The sound waves from the radio hit the wall and reflect back to the motorist. The frequency of the sound wave hitting the wall keeps on changing as the boy is moving towards the wall. The apparent frequency of the sound wave that reaches the wall( is stationary) is given by,
$\begin{aligned} & {{f}_{2}}=\left( \dfrac{v-{{V}_{o}}}{v-{{V}_{s}}} \right)f \\\ & {{f}_{2}}=\left( \dfrac{v-0}{v-{{V}_{b}}} \right)f \\\ & {{f}_{2}}=\left( \dfrac{v}{v-{{V}_{b}}} \right)f \\\ \end{aligned}$
This frequency will then hit the wall and reach the observer i.e. the motorist. Hence the apparent frequency heard by the motorist after hitting the wall is given by,
$\begin{aligned} & {{f}_{3}}=\left( \dfrac{v-{{V}_{o}}}{v-{{V}_{s}}} \right){{f}_{2}} \\\ & {{f}_{3}}=\left( \dfrac{v-\left( -{{V}_{m}} \right)}{v-0} \right){{f}_{2}} \\\ & {{f}_{3}}=\left( \dfrac{v+{{V}_{m}}}{v} \right){{f}_{2}} \\\ \end{aligned}$
Substituting ${{f}_{2}}$ in the above equation we get,
$\begin{aligned} & {{f}_{3}}=\left( \dfrac{v+{{V}_{m}}}{v} \right){{f}_{2}} \\\ & {{f}_{3}}=\left( \dfrac{v+{{V}_{m}}}{v} \right)\left( \dfrac{v}{v-{{V}_{b}}} \right)f \\\ & {{f}_{3}}=\left( \dfrac{v+{{V}_{m}}}{v-{{V}_{b}}} \right)f \\\ \end{aligned}$
The beat frequency heard by an observer is the difference between the two frequencies i.e.
${{f}_{beat}}={{f}_{3}}-{{f}_{1}}$ hence substituting the above values of frequencies in the adjacent equation we get,
$\begin{aligned} & {{f}_{beat}}={{f}_{3}}-{{f}_{1}} \\\ & {{f}_{beat}}=\left( \dfrac{v+{{V}_{m}}}{v-{{V}_{b}}} \right)f-\left( \dfrac{v+{{V}_{m}}}{v+{{V}_{b}}} \right)f \\\ & {{f}_{beat}}=\dfrac{2{{V}_{b}}\left( v+{{V}_{m}} \right)}{{{v}^{2}}-{{V}_{b}}^{2}}f \\\ \end{aligned}$

So, the correct answer is “Option C”.

Note:
It is to be noted that the difference between the apparent frequency reflected from the wall and the apparent frequency heard by the motorist directly from the radio should not be more than 10. This is because our ears cannot detect two sounds if the time interval between them is 1/10. Hence if we analyze the above equation we can conclude that there will be a limit on the velocity of the boy and the velocity of the motorist.