A car P travelling at 20 ms⁻¹ sounds its horn at a frequency of 400 Hz. Another car Q is travelling behind the first car in the same direction with a velocity 40 ms⁻¹. The frequency heard by the passenger of the car Q is approximately [Take, velocity of sound =360 ms⁻¹ ]

Solution

Physics sir · 4 steps

Picture who moves toward whom: Q chases the sound, P runs away from it, and both signs follow.

1. Set up directions. P is ahead, so the horn's sound reaching Q travels backward, from P toward Q. Q moves forward at 40 m s⁻¹, toward P, so the observer moves toward the source. P moves forward at 20 m s⁻¹, away from Q, so the source moves away from the observer.

   f' = f (v + vₒ)/(v + vₛ)

2. Choose the signs. An observer moving toward the source meets more wavefronts each second, so vₒ adds in the numerator. A source moving away stretches the wavelength, so vₛ adds in the denominator.
3. Put in numbers. f' = 400 × (360 + 40)/(360 + 20) = 400 × 400/380 = 421.05 Hz.
4. So. The passenger in Q hears about 421 Hz, which is option (c).

The idea to keep. In Doppler problems, first decide whether the observer moves toward or away from the source, then do the same for the source. Toward raises the frequency, away lowers it. Put that sign on each speed separately.

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