A source of sound $A$ emitting waves of frequency $1800\,Hz$ is falling towards ground with a terminal speed $v.$ The observer $B$ on the ground directly beneath the source receives waves of frequency $2150\,Hz.$ The source $A$ receives waves, reflected from ground of frequency nearly $..... Hz ($Speed of sound $= 343\,m/s )$
JEE MAIN 2014, Diffcult
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Given $f_{A}=1800 \mathrm{Hz}$${v_{t}=v}$
${f_{B}=2150 \mathrm{Hz}}$
Reflected wave frequency received by $A.$
$f_{A}^{\prime}=?$
Applying doppler's effect of sound,
$f^{\prime}=\frac{v_{s} f}{v_{s}-v_{t}}$
Here, $v_{t}=v_{s}\left(1-\frac{f_{A}}{f_{B}}\right)=343\left(1-\frac{1800}{2150}\right)$
$v_{t}=55.8372 \mathrm{m} / \mathrm{s}$
Now, for the reflected wave,
$\therefore \mathrm{f}_{\mathrm{A}}^{\prime}=\left(\frac{\mathrm{v}_{\mathrm{s}}+\mathrm{v}_{\mathrm{t}}}{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{t}}}\right) \mathrm{f}_{\mathrm{A}}$
$=\left(\begin{array}{l}{343+55.83} \\ {343-55.83}\end{array}\right) \times 1800$
$=2499.44 \approx 2500 \mathrm{Hz}$
${f_{B}=2150 \mathrm{Hz}}$
Reflected wave frequency received by $A.$
$f_{A}^{\prime}=?$
Applying doppler's effect of sound,
$f^{\prime}=\frac{v_{s} f}{v_{s}-v_{t}}$
Here, $v_{t}=v_{s}\left(1-\frac{f_{A}}{f_{B}}\right)=343\left(1-\frac{1800}{2150}\right)$
$v_{t}=55.8372 \mathrm{m} / \mathrm{s}$
Now, for the reflected wave,
$\therefore \mathrm{f}_{\mathrm{A}}^{\prime}=\left(\frac{\mathrm{v}_{\mathrm{s}}+\mathrm{v}_{\mathrm{t}}}{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{t}}}\right) \mathrm{f}_{\mathrm{A}}$
$=\left(\begin{array}{l}{343+55.83} \\ {343-55.83}\end{array}\right) \times 1800$
$=2499.44 \approx 2500 \mathrm{Hz}$
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