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(B) According to the Doppler effect, when an observer moves with velocity $V_O$ relative to a stationary source, the apparent frequency $F'$ is given

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) According to the Doppler effect, when an observer moves with velocity $V_O$ relative to a stationary source, the apparent frequency $F'$ is given by $F' = F_0 \left( \frac{V \pm V_O}{V} \right)$, where $F_0$ is the actual frequency and $V$ is the speed of sound.For an observer moving towards the source with velocity $V_1$: $F_1 = F_0 \left( \frac{V + V_1}{V} \right)$.For an observer moving away from the source with velocity $V_1$: $F_2 = F_0 \left( \frac{V - V_1}{V} \right)$.Given the ratio $F_1/F_2 = 2$, we have: $\frac{F_0(V + V_1)/V}{F_0(V - V_1)/V} = 2$.This simplifies to $\frac{V + V_1}{V - V_1} = 2$.Cross-multiplying gives $V + V_1 = 2(V - V_1)$, which is $V + V_1 = 2V - 2V_1$.Rearranging the terms, we get $3V_1 = V$, which implies $V/V_1 = 3$.

When the observer moves towards a stationary source with velocity $V_1$, the apparent frequency of the emitted note is $F_1$. When the observer moves away from the source with velocity $V_1$, the apparent frequency is $F_2$. If $V$ is the speed of sound in air and $F_1/F_2 = 2$, then $V/V_1 =$

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