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Home » When the observer moves towards the stationary source with velocity, ‘’, the apparent frequency of emitted note is ‘’. When the observer moves away from the source with velocity ‘’, the apparent frequency is ‘’. If ‘’ is the velocity of sound in air and then A) 2 B) 3 C) 4 D) 5
When the observer moves towards the stationary source with velocity, ‘V_1’, the apparent frequency of 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
Physics
When the observer moves towards the stationary source with velocity, ‘V_1’, the apparent frequency of 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

Answer is (B)
\frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=2, \quad
Speed of approach = Speed of leaving
The apparent frequency of sound level by observer when it is approaching source is given by
f_{1}=\frac{V}{V-V_{1}} \cdot f_{0}
When observer is moving away from source
f_{2}=\frac{V}{V+V_{1}} \cdot f_{0}
Here f_{0} is the frequency of sound main
Taking ratios of (1) with (2) we have
\begin{array}{l} \frac{f_{1}}{f_{2}}=\frac{V}{V-V_{1}} \times \frac{V+V_{1}}{V}=\frac{V+V_{1}}{V-V_{1}} \\ \therefore \quad 2=\frac{V+V_{1}}{V-V_{1}} \\ \therefore \quad 2 V-2 V_{1}=V+V_{1} \\ \therefore \quad 2 V-V=3 V_{1} \\ \therefore \quad \frac{V}{V_{1}}=3 \end{array}

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