MHT CET · Physics · Waves and Sound

When the observer moves towards the 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 velocity of sound in air and $\frac{F_1}{F_2}=2$ then $\frac{V}{V_1}=$ ?

A $2$
B $3$
C $4$
D $5$

Verified Solution

Answer & Solution

Correct Answer

(B) $3$

Step-by-step Solution

Detailed explanation

\frac{F_{1}}{F_{2}} = 2

, Speed of approach = Speed of leaving
The apparent frequency of sound observed by observer when it is approaching source is given by

F_{1} = \frac{V + V_{1}}{V} . f_{0}

....(i)
When observer is moving away from source

F_{2} = \frac{V - V_{1}}{V} f_{0}

....... (ii)
Here

f_{0}

is the frequency of sound wave
Taking ratio of (i) with (ii) we have

\frac{F_{1}}{F_{2}} = \frac{V}{V - V_{1}} \times \frac{V + V_{1}}{V} = \frac{V + V_{1}}{V - V_{1}}

∴ 2 = \frac{V + V_{1}}{V - V_{1}}

∴ 2 V - 2 V_{1} = V + V_{1}

∴ 2 V - V = 3 V_{1}

∴ \frac{V}{V_{1}} = 3

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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