An employee of a factory moving away from his workplace by a car listens to the siren of the factory. He drives the car at the speed of 72 kmh⁻¹ in the direction of wind which is blowing at 72 kmh⁻¹ speed. Frequency of siren is 720 Hz. The employee hears an apparent frequency of ____________ Hz.
(Assume speed of sound to be 340 ms⁻¹)

Solution

Physics sir · 4 steps

The wind carries the sound with it, so make sound 360 m/s first, then apply Doppler for a receding listener.

1. Convert speeds. 72 km h⁻¹ = 72 × 5/18 = 20 m s⁻¹. So the car moves at vₒ = 20 m s⁻¹ and the wind blows at w = 20 m s⁻¹.
2. Include the wind. The wind blows from the factory towards the car, the same way the sound travels. Relative to the ground, sound now moves at 340 + 20 = 360 m s⁻¹.

   f' = f ((v + w) - vₒ)/(v + w) = 720 × (360 - 20)/360

3. Observer moving away. The siren is at rest and the employee drives away from it, so his speed is subtracted in the numerator. This gives f' = 720 × 340/360 = 680 Hz.
4. So. The employee hears an apparent frequency of 680 Hz.

The common slip: 760. Adding the car's speed in the numerator, as if he were driving towards the siren, gives 720 × 380/360 = 760. Moving away lowers the frequency.

The idea to keep. Wind gives no Doppler shift by itself. It only changes the speed of sound: use v + w when it blows from source to observer and v - w when it blows the other way.

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