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Question: A train approaching a railway platform with a speed of \(20\text{ m}{{\text{s}}^{\text{-1}}}\) , sta...

A train approaching a railway platform with a speed of 20 ms-120\text{ m}{{\text{s}}^{\text{-1}}} , starts blowing the whistle. Speed of sound in air is 340 ms-1340\text{ m}{{\text{s}}^{\text{-1}}} . If the frequency of the emitted sound from the whistle is 640 Hz640\text{ Hz} , the frequency of sound to a person standing on the platform will appear to be :
A. 600 Hz600\text{ Hz}
B. 640 Hz640\text{ Hz}
C. 680 Hz680\text{ Hz}
D. 720 Hz720\text{ Hz}

Explanation

Solution

As the train is approaching, the person (detector) standing on the platform, the frequency of sound detected by the person appears to be more than the original frequency. This frequency shift can be calculated using the Doppler effect equation.
Formula used:
According to Doppler Effect equation, the frequency perceived by a detector is given by:
v=v(uuduus)v'=v\left( \dfrac{u-{{u}_{d}}}{u-{{u}_{s}}} \right)
Where speed of sound uu ,velocity of source x, is the velocity of detector and original frequency .

Complete answer:
Lets calculate and find the correct answer.
Given data:
Original frequency of sound v=640 Hzv=640\text{ Hz} ,
Speed of sound, u=340 ms-1u=340\text{ m}{{\text{s}}^{\text{-1}}} ,and
Velocity of source (train), us=20 ms-1{{u}_{s}}=20\text{ m}{{\text{s}}^{\text{-1}}} .
The apparent frequency of sound, observed by a stationary listener (ud=0{{u}_{d}}=0), when a source of sound is approaching to listener, is given by
v=v(uuduus) 640 Hz(340 ms10 ms1340 ms120 ms1) 640 Hz(340 ms1320 ms1) 640 Hz×3432 20 Hz×34 680 Hz \begin{aligned} & v'=v\left( \dfrac{u-{{u}_{d}}}{u-{{u}_{s}}} \right) \\\ & \Rightarrow 640\text{ Hz}\left( \dfrac{340\text{ m}{{\text{s}}^{-1}}-0\text{ m}{{\text{s}}^{-1}}}{340\text{ m}{{\text{s}}^{-1}}-20\text{ m}{{\text{s}}^{-1}}} \right) \\\ & \Rightarrow 640\text{ Hz}\left( \dfrac{340\text{ m}{{\text{s}}^{-1}}}{320\text{ m}{{\text{s}}^{-1}}} \right) \\\ & \Rightarrow 640\text{ Hz}\times \dfrac{34}{32} \\\ & \Rightarrow 20\text{ Hz}\times 34 \\\ & \Rightarrow 680\text{ Hz} \\\ \end{aligned}
Therefore, the frequency of sound will be 680 Hz680\text{ Hz} to a person standing on the platform.

Hence, option C is the correct answer.

Note:
According to Doppler Effect, for a source moving toward the detector and for a detector moving toward the source the detected frequency increases. Similarly, if the source moves away from the detector or if the detector moves away from the source, then detected frequency decreases.