Two cars X and Y are approaching each other with velocities 36 km/h and 72 km/h respectively. The frequency of a whistle sound as emitted by a passenger in car X, heard by the passenger in car Y is 1320 Hz. If the velocity of sound in air is 340 m/s, the actual frequency of the whistle sound produced is .................. Hz.

Solution

Physics sir · 4 steps

Both cars approach, so the pitch heard in Y is raised. Undo that shift to find the real frequency.

1. Convert speeds. Car X carries the source: 36 km/h = 10 m s⁻¹. Car Y carries the listener: 72 km/h = 20 m s⁻¹.
2. Pick the signs. A listener moving towards the source meets more wavefronts, so vₒ is added in the numerator. A source moving towards the listener squeezes the waves, so vₛ is subtracted in the denominator.

   f' = f (v + vₒ)/(v - vₛ) = f (340 + 20)/(340 - 10) = f 360/330

3. Solve for f. f = 1320 × 330/360 = 1320 × 11/12 = 1210 Hz.
4. So. The whistle actually sounds at 1210 Hz. The 1320 Hz heard in car Y is higher because the cars are approaching each other.

The common slip: 1440. Treating 1320 Hz as the emitted frequency gives 1320 × 360/330 = 1440. But 1320 Hz is what car Y hears, so you must divide by the Doppler factor, not multiply.

The idea to keep. Write f' = f(v ± vₒ)/(v ∓ vₛ) and choose each sign so that approach raises the frequency. Always convert km/h to m/s before you substitute.

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