MHT CET · Physics · Waves and Sound
An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency?
- A Zero
- B \(0.5 \%\)
- C \(5 \%\)
- D \(20 \%\)
Answer & Solution
Correct Answer
(D) \(20 \%\)
Step-by-step Solution
Detailed explanation
Given : \(v_{0}=\frac{v}{5} \Rightarrow v_{0}=\frac{320}{5}=64 \mathrm{~m} / \mathrm{s}\)
When observer moves towards the stationary source, then
\(
n^{\prime}=\left(\frac{v+v_{o}}{v}\right) n
\)
\(
\begin{array}{l}
n^{\prime}=\left(\frac{320+64}{320}\right) n \\
n^{\prime}=\left(\frac{384}{320}\right) n \\
\frac{n^{\prime}}{n}=\frac{384}{320}
\end{array}
\)
Hence, percentage increase
\(
\begin{aligned}
\left(\frac{n^{\prime}-n}{n}\right) &=\left(\frac{384-320}{320} \times 100\right) \% \\
&=\left(\frac{64}{320} \times 100\right) \%=20 \%
\end{aligned}
\)
When observer moves towards the stationary source, then
\(
n^{\prime}=\left(\frac{v+v_{o}}{v}\right) n
\)
\(
\begin{array}{l}
n^{\prime}=\left(\frac{320+64}{320}\right) n \\
n^{\prime}=\left(\frac{384}{320}\right) n \\
\frac{n^{\prime}}{n}=\frac{384}{320}
\end{array}
\)
Hence, percentage increase
\(
\begin{aligned}
\left(\frac{n^{\prime}-n}{n}\right) &=\left(\frac{384-320}{320} \times 100\right) \% \\
&=\left(\frac{64}{320} \times 100\right) \%=20 \%
\end{aligned}
\)
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