Question

A person observes two moving trains, ‘A’ reaching the station and ‘B’ leaving the station with equal speed of 30 m/s. If both trains emit sounds with frequency 300 Hz, (Speed of sound: 330 m/s) approximate difference of frequencies heard by the person will be:

• A. 33 Hz
• B. 10 Hz
• C. 55 Hz
• D. 80 Hz

Hint:

For Doppler Effect calculations:

• Use \[ f' = f \frac{v \pm v_o}{v \mp v_s}, \] where \(v\) is the speed of sound, \(v_o\) is the observer speed, and \(v_s\) is the source speed.
• Add velocities for approaching sources, and subtract for receding sources.

Answer 1

The Correct Option is C

To solve this problem, we will use the Doppler effect formula. The Doppler effect describes the changes in frequency of a wave in relation to an observer moving relative to the wave source. The formula for the observed frequency when the source is moving towards or away from the observer is given by:

f' = f \left(\frac{v + v_o}{v + v_s}\right)

where:

• f' is the observed frequency.
• f is the emitted frequency (300 Hz in this case).
• v is the speed of sound (330 m/s given).
• v_o is the speed of the observer (0 m/s, as the observer is stationary).
• v_s is the speed of the source (30 m/s for both trains A and B).

For train A, which is approaching the observer:

f'_{\text{approach}} = 300 \times \frac{330 + 0}{330 - 30}

f'_{\text{approach}} = 300 \times \frac{330}{300} = 300 \times 1.1 = 330 \text{ Hz}

For train B, which is moving away from the observer:

f'_{\text{recede}} = 300 \times \frac{330 + 0}{330 + 30}

f'_{\text{recede}} = 300 \times \frac{330}{360} = 300 \times 0.9167 \approx 275 \text{ Hz}

The difference in frequency heard by the person is:

\Delta f = f'_{\text{approach}} - f'_{\text{recede}} = 330 - 275 = 55 \text{ Hz}

Thus, the approximate difference in frequencies heard by the person is 55 Hz, which matches the correct answer.