Question

In the figure shown a source of sound of frequency 510 Hz moves with constant velocity \[{v_s} = 20m/s\] in the direction shown. The wind is blowing at a constant velocity of \[{v_w} = 20m/s\] towards an observer who is at rest at point B. The frequency detected by the observer corresponding to the sound emitted by the source at initial position A will be (speed of sound relative to air = 330 m/s)


A. 485 Hz
B. 500 Hz
C. 512 Hz
D. 525 Hz

Answer 1

Hint: As the source of sound is in motion with respect to the observer so there will be change in the frequency of the sound observed by the observer due to Doppler’s effect. To find the frequency detected by the observer we use the 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:
The original frequency of the sound emitted from the source is 510 Hz. It is given that the source is moving with constant velocity \[{v_s} = 20\,m/s\] in the direction making 60° with the line joining the source and the stationary observer. The wind is blowing at a constant velocity of \[{v_w} = 20\,m/s\] towards an observer.

The speed of sound in still air is given as 330 m/s. So, the effective speed of the sound in wind will be the resultant of the velocity of sound and the velocity of wind,
\[v = \left( {330 + 20} \right)\,m/s\]
The component of velocity of the source along the line joining the source and the object is \[{v_s}\sin 30^\circ \].

Using Doppler’s effect, the apparent frequency detected by the observer will be,
\[f' = {f_o}\left[ {\dfrac{{\left( {330 + 20} \right)}}{{\left( {330 + 20} \right) - {v_s}\sin 30^\circ }}} \right] \\ \]
\[\Rightarrow f' = {f_o}\left[ {\dfrac{{\left( {330 + 20} \right)}}{{\left( {330 + 20} \right) - {v_s}\sin 30^\circ }}} \right] \\ \]
\[\Rightarrow f' = \left( {510Hz} \right)\left[ {\dfrac{{\left( {330 + 20} \right)}}{{\left( {330 + 20} \right) - 20 \times \sin 30^\circ }}} \right] \\ \]
\[\Rightarrow f' = \left( {510Hz} \right) \times \left( {\dfrac{{350}}{{340}}} \right) \\ \]
\[\therefore f' = 525\,Hz\]
So, the frequency detected by the observer is 525 Hz.

Therefore, the correct option is D.

Note: The velocity component along the line joining the source and the observer is used for the Doppler’s effect formula. The perpendicular component of the velocity doesn’t have an effect on the change in frequency due to Doppler’s effect.