When the listener moves towards stationary source with velocity ' V₁ ', the apparent frequency of emitted note is ' F₁ '. When observer moves away from the source with velocity ' V₁ ', apparent frequency is ' F₂ '. If V is the velocity of sound in air and (F₁)/(F₂) = 2 then V/V₁ is (a) ,2 (b) 3 (c) 4 (d) 5

Answer: (b). Observer moving: F₁ = F (V + V₁)/(V), F₂ = F (V - V₁)/(V). (V + V₁)/(V - V₁) = 2 ⇒ V = 3V₁ So the answer is (b) 3.

When the listener moves towards stationary source with velocity ' V₁… | MHT CET 2024 (10 May, Shift 1)

Q23MHT CET 2024 · 10 MAY · SHIFT 1Medium

Doppler Effect

When the listener moves towards stationary source with velocity ' $V_1$ ', the apparent frequency of emitted note is ' $F_1$ '. When observer moves away from the source with velocity ' $V_1$ ', apparent frequency is ' $F_2$ '. If V is the velocity of sound in air and $\frac{F_1}{F_2} = 2$ then $\frac{V}{V_1}$ is

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Answer: (b)

Observer moving: $F_1 = F\,\dfrac{V + V_1}{V}$ , $F_2 = F\,\dfrac{V - V_1}{V}$ .
$\dfrac{V + V_1}{V - V_1} = 2 \Rightarrow V = 3V_1$
So the answer is (b) 3.

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Doppler Effect question from MHT CET 2024 (10 May, Shift 1)

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