A source and a detector move away from each other in absence of wind with a speed of 20 m/s with respect to the ground. If the detector detects a frequency of 1800 Hz of the sound coming from the source, then the original frequency of source considering speed of sound in air 340 m/s will be ............... Hz.

Solution

Physics sir · 4 steps

Both are running away, so the detector's term and the source's term each pull the frequency down.

1. Set the signs. The detector moves away from the source, so it meets fewer wavefronts: use v - v_d on top. The source moves away from the detector, so it stretches the waves: use v + vₛ below. Both speeds are measured from the ground, which is also the frame of the still air.

   f' = f (v - v_d)/(v + vₛ)

2. Put in numbers. With v = 340 m s⁻¹ and v_d = vₛ = 20 m s⁻¹: f' = f × 320/360 = 8/9f.
3. Solve for f. 1800 = 8/9f, so f = (1800 × 9)/8 = 2025 Hz. The source frequency is higher than the detected one, as it should be when the two move apart.
4. So. The original frequency of the source is 2025 Hz.

The common slip: 1600. Multiplying 1800 by 8/9 instead of dividing by it gives 1600. That makes the source frequency lower than the detected one, which cannot happen when the two move apart.

The idea to keep. Fix the signs by physics, not by memory: motion apart lowers the frequency, motion together raises it. Then check that your answer moves the right way.

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