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:

Solution

Physics sir · 4 steps

The source moves, so speed of sound minus source speed when it comes in, plus when it goes away.

1. Moving source. The person stands still and the trains move, so use the moving-source formula. Take the minus sign when the source approaches and the plus sign when it recedes.

   f' = f v/(v ∓ vₛ)

2. Train A approaching. f_A = 300 × 330/(330 - 30) = 300 × 330/300 = 330 Hz.
3. Train B leaving. f_B = 300 × 330/(330 + 30) = 300 × 330/360 = 275 Hz.
4. So. The difference is 330 - 275 = 55 Hz, which is option (b).

The idea to keep. For a moving source, the approaching pitch rises more than the receding pitch falls. Work out each frequency separately with v ∓ vₛ in the denominator, then subtract.

Answer from the NTA final answer key. Solution by PrepWiser.