Question

A boy with a radio, playing music at a frequency f is moving towards a wall with velocity ${v_b}$. A motorist is following the boy with speed ${v_m}$. The expression for the beat frequency heard by the motorist, if the speed of the sound $v$ is:
A) \[\left( {\dfrac{{v + {v_m}}}{{v + {v_b}}}} \right)f\].
B) \[\left( {\dfrac{{v + {v_m}}}{{v - {v_b}}}} \right)f\].
C) $\dfrac{{2{v_b}\left( {v + {v_m}} \right)}}{{{v^2} - {v_b}^2}}f$.
D) $\dfrac{{2{v_m}\left( {v + {v_b}} \right)}}{{{v^2} - {v_m}^2}}f$.

Answer 1

Hint: When the two sounds of different frequencies reach the ear then interference of sound and formation of constructive and destructive waves are formed. The Doppler effect is the change in the frequency of the observer which is moving relative to the source. The formula of the Doppler effect can be used to calculate the correct answer for this problem.

Formula used:
The formula for the Doppler’s effect is given by $f' = \left( {\dfrac{{v \pm {v_o}}}{{v \mp {v_s}}}} \right)f$ where $f'$ is the frequency of the observed $f$ is the source frequency $v$ is the velocity of the sound ${v_o}$ is the velocity of the observer and ${v_s}$ is the velocity of the source.

Complete step by step answer:
Case 1:
Let us suppose that the source is the boy and the observer is a motorist.
The observer is moving towards the source so the frequency heard by the motorist is given by,
$ \Rightarrow f' = \left( {\dfrac{{v + {v_m}}}{{v + {v_b}}}} \right)f$.........eq. (1)
Case 2:
And now suppose the case where the sound reflects from the wall so here the source is boy and the observer is wall therefore,
$ \Rightarrow f'' = \left( {\dfrac{v}{{v - {v_b}}}} \right)f$.........eq. (2)
Case 3:
Now when the sound reflects from the wall and hits the observer then the source will be a wall and the observer will be a motorist.
So the frequency heard by the motorist will be,
$ \Rightarrow f''' = f''\left( {\dfrac{{v + {v_m}}}{v}} \right)$.........eq. (3)
Let us replace the value of$f''$ from equation (2) into equation (3).
$ \Rightarrow f''' = f''\left( {\dfrac{{v + {v_m}}}{v}} \right)$
$ \Rightarrow f''' = \left( {\dfrac{v}{{v - {v_b}}}} \right)f \cdot \left( {\dfrac{{v + {v_m}}}{v}} \right)$
$ \Rightarrow f''' = \left( {\dfrac{{v + {v_m}}}{{v - {v_b}}}} \right)f$
Now let us calculate the beat frequency heard by the motorist. The beat frequency of the motorist is given by,
\[ \Rightarrow {\text{Beat frequency}} = f' - f'''\]
\[ \Rightarrow f' - f''' = \left( {\dfrac{{v + {v_m}}}{{v + {v_b}}}} \right)f - \left( {\dfrac{{v + {v_m}}}{{v - {v_b}}}} \right)f\]
\[ \Rightarrow f' - f''' = f \cdot \left( {v + {v_m}} \right)\left[ {\dfrac{{\left( {v - {v_b}} \right) - \left( {v + {v_b}} \right)}}{{\left( {{v^2} - {v_b}^2} \right)}}} \right]\]
\[ \Rightarrow f' - f''' = f \cdot \left( {v + {v_m}} \right) \cdot \left( {\dfrac{{ - 2{v_b}}}{{{v^2} - {v_b}^2}}} \right)\]
Let just consider the magnitude,
\[ \Rightarrow f' - f''' = f \cdot \left( {v + {v_m}} \right) \cdot \left( {\dfrac{{2{v_b}}}{{{v^2} - {v_b}^2}}} \right)\]
\[ \Rightarrow f' - f''' = \dfrac{{2{v_b}\left( {v + {v_m}} \right) \cdot f}}{{\left( {{v^2} - {v_b}^2} \right)}}\]

Hence, the correct option for this problem is C.

Note:
It is advisable to learn the concept and formula of the beat frequency as it is a little bit tricky. It is very important to remember the different signs for different conditions of velocity of observer and velocity of source in the formula of the beat frequency.