Question

When the observer moves towards the stationary source with velocity, 'V₁', the apparent frequency of the emitted note is 'F₁'. When the observer moves away from the source with velocity 'V₁', the apparent frequency is 'F₂'. If 'V' is the velocity of sound in air and `"F"_1/"F"_2 = 4` then `"V"/"V"_1 = ?`

Options

• 2

• `5/3`

• `3/5`

• 5

Answer 1

`underline(5/3)`

Explanation:

Apparent frequency is given by,

F' = `[(V ± V_0)/(V ± V_s)]F`

∵ source is stationary,

∴ V_s = 0; V₀ = V₁ 

∴ F₁ = `[(V + V_1)/V]F`

F₂ = `[(V - V_1)/V]F`

∴ `F_1/F_2 = (V + V_1)/(V - V_1)`

∴ 4 = `(V + V_1)/(V - V_1)`

∴ 4V - 4V₁ = V + V₁

∴ V = `5/3V_1`

∴ `V/V_1 = 5/3`