MHT CET · Physics · Waves and Sound
When the listener moves towards stationary source with velocity ' \(\mathrm{V}_1\) ', the apparent frequency of emitted note is ' \(F_1\) '. When observer moves away from the source with velocity ' \(\mathrm{V}_1\) ', apparent frequency is ' \(\mathrm{F}_2\) '. If' V is the velocity of sound in air and \(\frac{F_1}{F_2}=2\) then \(\frac{V}{V_1}\) is
- A 2
- B 3
- C 4
- D 5
Answer & Solution
Correct Answer
(B) 3
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& F_1=F\left(\frac{V+V_1}{V}\right) \\
F_2 & =F\left(\frac{V-V_1}{V}\right) \\
\therefore \quad & \frac{F_1}{F_2}=\left(\frac{V+V_1}{V-V_1}\right)=2 \\
\therefore \quad & \frac{V}{V_1}=3
\end{aligned}\)
...(given)
& F_1=F\left(\frac{V+V_1}{V}\right) \\
F_2 & =F\left(\frac{V-V_1}{V}\right) \\
\therefore \quad & \frac{F_1}{F_2}=\left(\frac{V+V_1}{V-V_1}\right)=2 \\
\therefore \quad & \frac{V}{V_1}=3
\end{aligned}\)
...(given)
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