When an observer moves towards a stationary source with velocity ' V ₁ ', the appearent frequency of emitted note is ' F ₁ '. When observer moves away from stationary source with velocity ' V ₁ ' the appearent frequency is ' F₂ '. If ' v ' is velocity of sound in air and F₁ F₂ =2 , then V V₁ is equal to

Step-by-step solution

The Doppler effect for sound waves gives the apparent frequency F' when an observer moves relative to a stationary source: F' = F ( v V_o v ) where F is the source frequency, v the speed of sound, and V_o the observer's speed. When the observer approaches the source, the apparent frequency is F₁ = F ( v + V₁ v ) . When moving away, F₂ = F ( v - V₁ v ) . Given F₁ F₂ = 2 , substitute the expressions: v + V₁ v - V₁ = 2 Solving for v V₁ : 2(v - V₁) = v + V₁ 2v - 2V₁ = v + V₁ v = 3V₁ Thus v V₁ = 3 , corresponding to opt