A train blowing a whistle of frequency 320 Hz approaches an observer standing on the platform at a speed of 66 m/s. The frequency observed by the observer will be (given speed of sound = 330 ms⁻¹) __________ Hz.

Solution

Physics sir · 3 steps

The source is moving towards you, so the sound waves bunch up: subtract the train's speed in the denominator.

1. Who is moving. The train carries the whistle, so the source moves. The observer stands still on the platform. A moving source squeezes the wavelength in front of it, so the observer hears a higher frequency.

   f' = f v/(v - vₛ)

2. Put in numbers. Here f = 320 Hz, v = 330 m s⁻¹ and vₛ = 66 m s⁻¹. The denominator is 330 - 66 = 264, so f' = 320 × 330/264 = 320 × 1.25.
3. So. The observer hears f' = **400** Hz.

The common slip: 384. Treating the train as a moving observer, 320 × (330 + 66)/330, gives 384, not 400. The train carries the whistle, so it is the source.

The idea to keep. In Doppler problems, first decide who moves. A moving source changes the denominator, a moving observer changes the numerator, and approaching always raises the pitch.

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