The frequency of a car horn encountered a change from 400 Hz to 500 Hz, when the car approaches a vertical wall. If the speed of sound is 330 m/s. Then the speed of car is ___________ km/h.

Solution

Physics sir · 4 steps

The sound hits the wall and comes back, so the Doppler shift acts twice, once each way.

1. Wall as listener. The car is a source moving towards a still wall, so the wall receives f' = fv/(v-u), where u is the car's speed.
2. Wall as source. The wall reflects f' back. Now the car is a listener moving towards a still source, so the driver hears f'' = f'(v+u)/v.

   f''/f = (v+u)/(v-u) ⇒ 500/400 = (330+u)/(330-u)

3. Solve for u. Cross-multiplying gives 5(330-u) = 4(330+u), so 1650 - 5u = 1320 + 4u. Then 9u = 330 and u = 110/3 m s⁻¹ ≈ 36.7 m s⁻¹.
4. So. In km/h, u = 110/3 × 18/5 = 132 km/h. The answer is 132.

The common slip: 237.6. Using the one-way formula fv/(v-u) gives u = 66 m s⁻¹, which is 237.6 km/h. That misses the second shift when the echo reaches the moving car.

The idea to keep. For an echo from a wall, treat the wall first as a listener, then as a source. For an approaching car the two shifts combine into (v+u)/(v-u).

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