When a car is approaching the observer, the frequency of horn is 100 Hz. After passing the observer, it is 50 Hz. If the observer moves with the car, the frequency will be x/3 Hz where x= ________________.

Solution

Physics sir · 4 steps

Riding with the car kills the Doppler shift, so you only need the true horn frequency f₀.

1. Two Doppler equations. Let the true frequency be f₀, the speed of sound v and the car's speed vₛ. Approaching gives 100 = f₀ v/(v - vₛ), and receding gives 50 = f₀ v/(v + vₛ).
2. Flip and add. Take reciprocals: 1/100 = (v - vₛ)/(f₀ v) and 1/50 = (v + vₛ)/(f₀ v). Adding them cancels vₛ.

   2/f₀ = 1/100 + 1/50 = 3/100 ⇒ f₀ = 200/3 Hz

3. Observer with car. When the observer moves along with the car, there is no relative motion between source and observer. So he hears the true frequency f₀ = 200/3 Hz. As a check, the ratio of the two equations gives vₛ = v/3, and 100 × 2/3 = 200/3.
4. So. x/3 = 200/3, so x = 200.

The common slip: 225. Taking the plain average of 100 and 50 gives 75 Hz and x = 225. The reciprocals add, not the frequencies themselves.

The idea to keep. For a source passing an observer, the true frequency is the harmonic mean of the two heard frequencies: 2/f₀ = 1/f₁ + 1/f₂. With no relative motion, the observer hears exactly f₀.

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