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.55 Hz
- B.80 Hz
- C.33 Hz
- D.10 Hz
Show answer and solution
Correct option: A
Solution
For train A (approaching), the observed frequency is:
fA = 300 × 330/(330 - 30) = 300 × 330/300 = 330 Hz
For train B (leaving/receding), the observed frequency is:
fB = 300 × 330/(330 + 30) = 300 × 330/360 = 275 Hz
Difference = 330 - 275 = 55 Hz.
The Doppler effect gives higher frequency for an approaching source (denominator decreases) and lower for a receding source (denominator increases), so the difference between the two perceived frequencies is 55 Hz.
fA = 300 × 330/(330 - 30) = 300 × 330/300 = 330 Hz
For train B (leaving/receding), the observed frequency is:
fB = 300 × 330/(330 + 30) = 300 × 330/360 = 275 Hz
Difference = 330 - 275 = 55 Hz.
The Doppler effect gives higher frequency for an approaching source (denominator decreases) and lower for a receding source (denominator increases), so the difference between the two perceived frequencies is 55 Hz.
[JEE Main 2023 · 29 Jan · Shift 1]