Question 276: Waves

Question

When an observer moves towards a stationary source with velocity ' $\mathrm{V}_1$ ', the apparent frequency of emitted note is ' $\mathrm{F}_1$ '. When observer moves away from stationary source with velocity ' $\mathrm{V}_1$ ' the apparent frequency is ' $\mathrm{F}_2$ '. If ' v ' is velocity of sound in air and $\frac{\mathrm{F}_1}{\mathrm{~F}_2}=2$, then $\frac{\mathrm{V}}{\mathrm{V}_1}$ is equal to

Accepted Answer

Let:
Actual frequency of the source = $f$
Velocity of sound = $v$
Speed of observer = $V_1$
**Case 1: Observer moves *
towards
* stationary source**
The apparent frequency ($F_1$) is:
$ F_1 = f \left(\frac{v + V_1}{v}\right) $
**Case 2: Observer moves *
away from
* stationary source**
The apparent frequency ($F_2$) is:
$ F_2 = f \left(\frac{v - V_1}{v}\right) $
Given:
$ \frac{F_1}{F_2} = 2 $
Now, substitute the values of $F_1$ and $F_2$:
$ \frac{\frac{v + V_1}{v}f }{\frac{v - V_1}{v}f} = 2 $
$ \frac{v + V_1}{v - V_1} = 2 $
Now, solve for $\frac{v}{V_1}$:
Cross-multiply:
$ v + V_1 = 2(v - V_1) $
$ v + V_1 = 2v - 2V_1 $
$ v + V_1 - 2v + 2V_1 = 0 $
$ -v + 3V_1 = 0 $
$ v = 3V_1 $
So,
$ \frac{v}{V_1} = 3 $
Final Answer:
Option C: 3

Answer

6

Answer

5

Answer

3

Answer

4

Waves

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