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A source of sound A emitting waves of frequency $1800 \mathrm{~Hz}$ is falling towards ground with a terminal speed $\mathrm{v}$. The observer $\mathrm{B}$ on the ground directly beneath the source receives waves of frequency $2150 \mathrm{~Hz}$. The source A receives waves, reflected from ground of frequency nearly: (Speed of sound $=343 \mathrm{~m} / \mathrm{s}$ )
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The correct answer is:
$2500 \mathrm{~Hz}$
$2500 \mathrm{~Hz}$
Given $\mathrm{f}_{\mathrm{A}}=1800 \mathrm{~Hz}$
$$
\begin{aligned}
&\mathrm{v}_{\mathrm{t}}=\mathrm{v} \\
&\mathrm{f}_{\mathrm{B}}=2150 \mathrm{~Hz}
\end{aligned}
$$
Reflected wave frequency received by $\mathrm{A}$, $\mathrm{f}_{\mathrm{A}}{ }^{\prime}=$ ?
Applying doppler's effect of sound, $\mathrm{f}^{\prime}=\frac{\mathrm{v}_{\mathrm{s}} \mathrm{f}}{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{t}}}$
Here, $\mathrm{v}_{\mathrm{t}}=\mathrm{v}_{\mathrm{s}}\left(1-\frac{\mathrm{f}_{\mathrm{A}}}{\mathrm{f}_{\mathrm{B}}}\right)=343\left(1-\frac{1800}{2150}\right)$ $\mathrm{v}_{\mathrm{t}}=55.8372 \mathrm{~m} / \mathrm{s}$
Now, for the reflected wave,
$$
\begin{aligned}
&\therefore \mathrm{f}_{\mathrm{A}}{ }^{\prime}=\left(\frac{\mathrm{v}_{\mathrm{s}}+\mathrm{v}_{\mathrm{t}}}{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{t}}}\right) \mathrm{f}_{\mathrm{A}} \\
&=\left(\frac{343+55.83}{343-55.83}\right) \times 1800 \\
&=2499.44 \approx 2500 \mathrm{~Hz}
\end{aligned}
$$
$$
\begin{aligned}
&\mathrm{v}_{\mathrm{t}}=\mathrm{v} \\
&\mathrm{f}_{\mathrm{B}}=2150 \mathrm{~Hz}
\end{aligned}
$$
Reflected wave frequency received by $\mathrm{A}$, $\mathrm{f}_{\mathrm{A}}{ }^{\prime}=$ ?
Applying doppler's effect of sound, $\mathrm{f}^{\prime}=\frac{\mathrm{v}_{\mathrm{s}} \mathrm{f}}{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{t}}}$
Here, $\mathrm{v}_{\mathrm{t}}=\mathrm{v}_{\mathrm{s}}\left(1-\frac{\mathrm{f}_{\mathrm{A}}}{\mathrm{f}_{\mathrm{B}}}\right)=343\left(1-\frac{1800}{2150}\right)$ $\mathrm{v}_{\mathrm{t}}=55.8372 \mathrm{~m} / \mathrm{s}$
Now, for the reflected wave,
$$
\begin{aligned}
&\therefore \mathrm{f}_{\mathrm{A}}{ }^{\prime}=\left(\frac{\mathrm{v}_{\mathrm{s}}+\mathrm{v}_{\mathrm{t}}}{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{t}}}\right) \mathrm{f}_{\mathrm{A}} \\
&=\left(\frac{343+55.83}{343-55.83}\right) \times 1800 \\
&=2499.44 \approx 2500 \mathrm{~Hz}
\end{aligned}
$$
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