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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 ?
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Verified Answer
The correct answer is:
$20 \%$
Given : $v_o=\frac{v}{5} \Rightarrow v_o=\frac{320}{5}=64 \mathrm{~m} / \mathrm{s}$
When observer moves towards the stationary source, then
$\begin{aligned}
n^{\prime} & =\left(\frac{v+v_0}{v}\right) n \\
n^{\prime} & =\left(\frac{320+64}{320}\right) n \\
n^{\prime} & =\left(\frac{384}{320}\right) n \\
\frac{n^{\prime}}{n} & =\frac{384}{320}
\end{aligned}$
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
$\begin{aligned}
n^{\prime} & =\left(\frac{v+v_0}{v}\right) n \\
n^{\prime} & =\left(\frac{320+64}{320}\right) n \\
n^{\prime} & =\left(\frac{384}{320}\right) n \\
\frac{n^{\prime}}{n} & =\frac{384}{320}
\end{aligned}$
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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