A source of sound of frequency f₁ is placed on the ground. A detector placed at a height is released from rest on this source. The observed frequency f(Hz) is plotted against time t(sec). The speed of sound in air is 300 m/s. Find $f_{1} (g = 10 m / s^{2})$

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$0.2 \times 10^{3} Hz$

$0.25 \times 10^{3}$

$1 \times 10^{3} Hz$

$0.5 \times 10^{3} Hz$

answer is B.

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Detailed Solution

$\begin{array}{l} f = (\frac{v + v_{0}}{v}) f_{1} = f_{1} + f_{1} \frac{v_{0}}{v} \\ v_{0} = gt \\ So, f = f_{1} + (\frac{f_{1} g}{v}) t \end{array}$

$\begin{array}{l} Slope of graph = \frac{f_{1} g}{v} = \frac{2 \times 10^{3} - f_{1}}{30} = \frac{(f_{1}) (10)}{300} \\ or f_{1} = 10^{3} Hz \end{array}$

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