Question

An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency?
\left( A \right)0 \\\ \left( B \right)0.5\% \\\ \left( C \right)5\% \\\ \left( D \right)20\% \\\

Answer 1

In order to solve this question, we are going to use the Doppler’s effect, on the situation of the frequencies of the sound waves as given. On finding the velocity when the observer is moving towards the stationary source, we can easily find the percentage increase in the frequency.
From the Doppler’s effect, the velocity of the observer is
$v' = v\left[ {\dfrac{{{v_{sound}} + {v_{observer}}}}{{{v_{sound}}}}} \right]$

Complete step by step solution:
According to the Doppler’s effect, the Doppler’s effect is the change in the frequency of the wave in relation to an observer who is moving relative to the wave source. The pitch of the sound also seems different due to movement of source and observer.
Let the velocity of sound in air be $v$
This implies that according to the question, the velocity of observer is $\dfrac{v}{5}$
Let $v$ be the original frequency of the sound source and ${v'}$ be the apparent frequency heard by the moving observer.
Using Doppler Effect when observer is moving towards the stationary source:
$v' = v\left[ {\dfrac{{{v_{sound}} + {v_{observer}}}}{{{v_{sound}}}}} \right] = v\left[ {\dfrac{{v + \dfrac{v}{5}}}{v}} \right] = \dfrac{6}{5}v$
Percentage increase in frequency is
$\dfrac{{v' - v}}{v} \times 100 = \dfrac{{\dfrac{6}{5}v - v}}{v} \times 100 = 20\%$

Note:
Due to the Doppler’s effect, there is a change in the pitch of the sounds approaching us when a vehicle comes near to us and drifts away. Similarly, in this case, the speed of the observer has a huge impact on the frequency of the sound that is being received by the receiver and that is being transmitted.