Physics Circle home Student Login

Home›Chapters›Wave Motion PYQs›Doppler Effect›Question 16

MHT CET 2025 (20 Apr, Shift 1) · Doppler Effect

Doppler Effect question from MHT CET 2025 (20 Apr, Shift 1)

Q16MHT CET 2025 · 20 APR · SHIFT 1Medium

Doppler Effect

When an observer moves towards a stationary source with velocity ' V1V_1 ', the apparent frequency of emitted note is ' F1F_1 '. When observer moves away from stationary source with velocity ' V1V_1 ' the apparent frequency is ' F2F_2 '. If ' v ' is velocity of sound in air and F1F2=2\frac{F_1}{F_2} = 2, then VV1\frac{V}{V_1} is equal to

Show answer and solution

Answer: (c)

  1. Observer moving: F1=F V+V1VF_1 = F\,\dfrac{V + V_1}{V}, F2=F V−V1VF_2 = F\,\dfrac{V - V_1}{V}.
  2. V+V1V−V1=2⇒V=3V1\dfrac{V + V_1}{V - V_1} = 2 \Rightarrow V = 3V_1
  3. So the answer is (c) 3.
Report a mistake

More questions on Doppler Effect