A person driving car at a constant speed of 15 m / s is approaching a vertical wall. The person notices a change of 40 Hz in the frequency of his car's horn upon reflection from the wall. The frequency of horn is _______________ Hz.
(Given: Speed of sound : 330 m / s )

Solution

Physics sir · 4 steps

The wall acts as a still source, so apply Doppler twice: once going to the wall and once coming back.

1. Two Doppler shifts. The horn is a source moving towards the wall at u = 15 m s⁻¹, so the wall receives a raised frequency. The wall reflects that frequency back as a stationary source, and the driver moves towards it as an observer, so the frequency rises again.

   f' = f(v/(v-u))((v+u)/v) = f (v+u)/(v-u)

2. Put in numbers. f' = f (330+15)/(330-15) = f 345/315. The change is f' - f = f (345-315)/315 = f 30/315.
3. Solve for f. Set the change equal to 40 Hz: f 30/315 = 40, so f = (40 × 315)/30 = 420 Hz.
4. So. The frequency of the horn is 420 Hz.

The common slip: 840. Applying only one Doppler shift, f 330/315 - f = 40, gives 840 Hz instead of 420.

The idea to keep. For echoes from a wall, treat the wall as an observer first and then as a stationary source. A moving car gives f' = f (v+u)/(v-u), because both shifts point the same way.

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