Question

A source of sound attached to the bob of a simple pendulum executes S.H.M. The difference between the apparent frequency of sound as received by an observer during its approach and recession at the mean position of the S.H.M. motion is 2% of the natural frequency of the source. The velocity of the source at the mean position is:(velocity of sound in the air is 340 m/s)
[Assume velocity of sound source << velocity of sound in air]
A. 1.4 m/s
B. 3.4 m/s
C. 1.7 m/s
D. 2.1 m/s

Answer 1

Hint: As the source of sound is attached to the bob, so the source of sound will be in motion with respect to the observer as the bob executes the simple harmonic motion about the mean position. So, there will be a change in frequency of the sound due to Doppler’s effect.

Formula used:
\[{f_{ap}} = {f_o}\left( {\dfrac{{v \pm {v_o}}}{{v \pm {v_s}}}} \right)\]
where \[{f_{ap}}\] is the apparent frequency heard by the listener moving with speed \[{v_o}\] with respect to the source which is moving with speed \[{v_s}\], \[{f_o}\] is the original frequency and v is the speed of sound in air.

Complete step by step solution:
Let the speed of the sound is v and the speed of the bob at the mean position is \[{v_s}\]. When the bob is executing simple harmonic motion, then at regular intervals it moves away from the observer at mean position and towards the observer at the mean position. The observed frequency by the observer will be maximum when the source of sound is moving towards the observer. The observed frequency by the observer will be minimum when the source of sound is moving away from the observer.

Let the natural frequency of the source of the sound is f. When the bob is moving away from the stationary observer then the frequency observed will be minimum. Using Doppler’s effect, the minimum frequency observed will be,
\[f{'_{\min }} = f\left( {\dfrac{v}{{v + {v_s}}}} \right)\]
When the bob is moving towards the stationary observer then the frequency observed will be maximum. Using Doppler’s effect, the maximum frequency observed will be,
\[f{'_{\max }} = f\left( {\dfrac{v}{{v - {v_s}}}} \right)\]

It is given that the difference between the apparent frequency of sound as received by an observer during its approach and recession at the mean position of the S.H.M. motion is 2% of the natural frequency of the source.
\[\dfrac{{f{'_{\max }} - f{'_{\min }}}}{f} = 2\% \\ \]
\[\Rightarrow \dfrac{{f\left( {\dfrac{v}{{v - {v_s}}}} \right) - f\left( {\dfrac{v}{{v + {v_s}}}} \right)}}{f} = 0.02 \\ \]
\[\Rightarrow \left( {\dfrac{v}{{v - {v_s}}}} \right) - \left( {\dfrac{v}{{v + {v_s}}}} \right) = 0.02 \\ \]
\[\Rightarrow \dfrac{{v\left( {v + {v_s}} \right) - v\left( {v - {v_s}} \right)}}{{{v^2} - v_s^2}} = 0.02 \\ \]
\[\Rightarrow \dfrac{{{v^2} + v{v_s} - {v^2} + v{v_s}}}{{{v^2} - v_s^2}} = 0.02 \\ \]
As the speed of the source is very small than the speed of the sound in air, so
\[{v^2} - v_s^2 \approx {v^s} \\ \]
\[\Rightarrow \dfrac{{2v{v_s}}}{{{v^2}}} = 0.02 \\ \]
\[\Rightarrow {v_s} = 0.01v \\ \]
\[\therefore {v_s} = 0.01 \times 340\,m/s = 3.4\,m/s\]
Hence, the speed of the source at the mean position is 3.4 m/s.

Therefore, the correct option is B.

Note: We should assume that the source is approaching or receding at the mean position parallel to the line joining the source and the observer. The difference of the frequency is instantaneous because the object passes through the mean position.