Question

A train moving at a speed of $220\,ms^{-1}$ towards a stationary object, emits a sound of frequency $1000\, Hz$. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected by the driver of the train is (Speed of sound in air is $330\,ms^{-1}$)

• A. 3500 Hz
• B. 4000 Hz
• C. 5000 Hz
• D. 3000 Hz

Answer 1

The Correct Option is C

To solve this problem, we need to determine the frequency of the sound detected by the driver of the train after it is reflected by the stationary object. This situation involves a combination of the Doppler Effect as sound is emitted, reflected, and then received.

First, let's note down the given data: 

• Speed of the train, \( v_s = 220\, \text{ms}^{-1} \)
• Frequency of the sound emitted, \( f_0 = 1000\, \text{Hz} \)
• Speed of sound in air, \( v = 330\, \text{ms}^{-1} \)

The sequence of events is as follows:

1. The train emits a sound of frequency \( f_0 = 1000\, \text{Hz} \).
2. This sound reaches a stationary object and gets reflected.
3. The reflected sound is received back by the train.

In such problems, the Doppler effect formula is applied twice:

1. Sound from the train to the object: The observed frequency by the stationary object (\( f' \)) can be calculated using the formula for the Doppler effect when the source is moving towards a stationary listener:

 

\(f' = f_0 \left( \frac{v}{v - v_s} \right)\)

Substituting the known values:

\(f' = 1000 \left( \frac{330}{330 - 220} \right) = 1000 \left( \frac{330}{110} \right) = 3000\, \text{Hz}\)

1. Sound from the object back to the train: Now the stationary object acts as a new source emitting frequency \( f' = 3000\, \text{Hz} \) towards the moving train (which receives this sound). Again applying the Doppler effect formula when source is stationary and receiver is moving towards the source:

\(f_{\text{echo}} = f' \left( \frac{v + v_s}{v} \right)\)

Substituting \( f' = 3000 \, \text{Hz} \), we have:

\(f_{\text{echo}} = 3000 \left( \frac{330 + 220}{330} \right) = 3000 \left( \frac{550}{330} \right) = 5000\, \text{Hz}\)

Therefore, the frequency of the echo detected by the driver of the train is 5000 Hz.

This matches the correct option.