Question

A train moving at a speed of \[220m{s^{ - 1}}\]towards a stationary object, emits a sound of frequency 1000Hz. Some of the sound reaching the object gets reflected back to train as an echo. The frequency of the echo as detected by the driver of the train is (speed of sound in the air is\[330m{s^{ - 1}}\])
A. 4000Hz
B. 5000Hz
C. 3000Hz
D. 3500Hz

Answer 1

Hint: In this question, the speed of the train and the frequency of the sound by the train is given so to detect the frequency of the echo as observed by the driver of the train by using the Doppler effect.
Doppler Effect refers to the change in wave frequency during the relative motion between a wave source and its observer, which is given by the formula
\[f' = f\left( {\dfrac{{v + {v_s}}}{{v - {v_s}}}} \right)\]
Where, \[f\]is the frequency by the source, \[{v_s}\]is the speed of the source and \[f'\]is the change in frequency.

Complete step by step answer:
Frequency of sound emitted by a train \[f = 1000Hz\]
Speed of the moving train which is the source \[{v_s} = 220m{s^{ - 1}}\]
Speed of sound in the air\[v = 330m{s^{ - 1}}\]
We know that the relative motion between a wave source and its observer changes the frequency which is calculated by the formula
\[f' = f\left( {\dfrac{{v + {v_s}}}{{v - {v_s}}}} \right) - - (i)\]
Now we substitute the values in the given equation
\[
  f' = f\left( {\dfrac{{v + {v_s}}}{{v - {v_s}}}} \right) \\
   = 1000\left( {\dfrac{{330 + 220}}{{330 - 220}}} \right) \\
 \]
Now by further solving we get
\[
  f' = 1000\left( {\dfrac{{330 + 220}}{{330 - 220}}} \right) \\
   = 1000 \times \dfrac{{550}}{{110}} \\
   = 5000Hz \\
 \]
Hence the frequency of the echo as detected by the driver of the train \[ = 5000Hz\].

So, the correct answer is “Option B”.

Note:
Students must note that when a sound object moves towards the observer then the frequency of the sound wave observed by the observer increases and when the when a sound object moves away from the observer then frequency of the sound wave decreases.