\({v_{t}=v}\)
\({f_{B}=2150 \mathrm{Hz}}\)
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,
\(\therefore \mathrm{f}_{\mathrm{A}}^{\prime}=\left(\frac{\mathrm{v}_{\mathrm{s}}+\mathrm{v}_{\mathrm{t}}}{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{t}}}\right) \mathrm{f}_{\mathrm{A}}\)
\(=\left(\begin{array}{l}{343+55.83} \\ {343-55.83}\end{array}\right) \times 1800\)
\(=2499.44 \approx 2500 \mathrm{Hz}\)
$(a)$ $\left(x^2-v t\right)^2$
$(b)$ $\log \left[\frac{(x+v t)}{x_0}\right]$
$(c)$ $e^{\left\{-\frac{(x+v t)}{x_0}\right\}^2}$
$(d)$ $\frac{1}{x+v t}$
$y_1=A \sin \left(k x-\omega t+\frac{\pi}{6}\right), \quad y_2=A \sin \left(k x-\omega t-\frac{\pi}{6}\right)$
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