The driver of a bus approaching a big wall notices that the frequency of his bus's horn changes from 420 Hz to 490 Hz when he hears it after it gets reflected from the wall. Find the speed of the bus if speed of the sound is 330 ms –1 .

Solution

Physics sir · 4 steps

The wall acts as a listener first and then as a source, so apply Doppler twice and multiply.

1. Bus to wall. The bus is a source moving at speed u toward the wall, which is at rest. The wall receives f₁ = f v/(v-u), where f = 420 Hz.
2. Wall to driver. The wall now sends out f₁ as a stationary source. The driver moves toward it at u, so he hears f' = f₁ (v+u)/v. Putting the two stages together, the v cancels.

   f' = f (v+u)/(v-u) ⇒ 490/420 = 7/6 = (330+u)/(330-u)

3. Solve for u. Cross-multiplying gives 6(330+u) = 7(330-u), so 13u = 330 and u = 25.38 m s⁻¹.
4. So. In km/h, u = 25.38 × 3.6 = 91.4 km h⁻¹ ≈ 91 km/h, which is option (a).

The idea to keep. For an echo heard by a source moving toward a wall, the wall is a stationary listener and then a stationary source. Both shifts combine to f' = f (v+u)/(v-u).

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