Question

A train has just completed a U-curve in a track which is a semicircle. The engine is at the forward end of the semicircular part of the track while the last carriage is at the rear end of the semicircular track. The driver blows a whistle of frequency 200 Hz. Velocity of sound is 340 m/sec. Then the apparent frequency as observed by a passenger in the middle of a train when the speed of the train is 30 m/sec is
A. 209 Hz
B. 288 Hz
C. 200 Hz
D. 181 Hz

Answer 1

Hint:The source of the sound, i.e. the engine is moving as well as the observer in the middle of the train is also moving. So, there will be a change in the frequency of the sound which can be caused 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:

Figure: The train in semi-circular track

Let the engine of the train be at A, the end of the train is at B and the observer is at the middle of the train, i.e. at C. The speed of the train is given as 30 m/s. So, the speed of every point on the train will have the same speed. The speed of sound is 340 m/s. Velocity of source of sound (engine) in the direction of the line joining the engine and the middle of the train is,
\[{v_s} = 30\,m/s\cos 45^\circ = \dfrac{{30}}{{\sqrt 2 }}m/s\]

Velocity of observer (middle of the train) in the direction of the line joining the engine and the middle of the train is,
\[{v_0} = 30\,m/s\cos 45^\circ = \dfrac{{30}}{{\sqrt 2 }}\,m/s\]
The source of the sound is moving away from the observer and the observer is moving towards the source of the sound. So, using the Doppler’s effect, the apparent frequency of the whistle detected by the observer at the middle of the train will be,
\[f' = {f_o}\left( {\dfrac{{v + {v_o}}}{{v + {v_s}}}} \right) \\ \]
\[{f_o} = 200\,Hz\] is the original frequency of the whistle.
\[v = 340\,m/s\] is the speed of sound in air

Putting the values in the Doppler’s effect formula, we get
\[f' = \left( {200Hz} \right)\left( {\dfrac{{340 + \dfrac{{30}}{{\sqrt 2 }}}}{{340 + \dfrac{{30}}{{\sqrt 2 }}}}} \right) \\ \]
\[\Rightarrow f' = \left( {200Hz} \right)\left( {\dfrac{{340 + 21.21}}{{340 + 21.21}}} \right) \\ \]
\[\Rightarrow f' = \left( {200Hz} \right) \times 1 \\ \]
\[\therefore f' = 200\,Hz\]

Therefore, the correct option is C.

Note: We take the velocity components along the line joining the source of the sound and the observer. The velocity component along the direction perpendicular to the line joining the source of the sound and the observer has no effect on the change in frequency due to Doppler’s effect.