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Question
A small source of sound oscillates in simple harmonic motion with an amplitude of 17 cm. A detector is placed along the line of motion of the source. The source emits a sound of frequency 800 Hz which travels at a speed of 340 m s−1. If the width of the frequency band detected by the detector is 8 Hz, find the time period of the source.
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Solution
Given:
Amplitude r = 17 cm = \[\frac{17}{100}\]= 0.17 m
Frequency of sound emitted by source f = 800 Hz
Velocity of sound \[v\]= 340 m/s
Frequency band = f2\[-\]f1= 8 Hz
Here,
\[f_2\] and \[f_1\]correspond to the maximum and minimum apparent frequencies (Both will be at the mean position because the velocity is maximum).
\[\text { Now }, f_1 = \left( \frac{340}{340 + v_s} \right)f\]
\[ \text { and } f_2 = \left( \frac{340}{340 - v_s} \right)f\]
\[ \therefore f_2 - f_1 = 8\]
\[ \Rightarrow 8 = \left( \frac{340}{340 - v_s} \right)f - \left( \frac{340}{340 + v_s} \right)f\]
\[ \Rightarrow 8 = 340f\left[ \frac{1}{340 - v_s} - \frac{1}{340 + v_s} \right]\]
\[ \Rightarrow 8 = 340 \times 800 \times \left[ \frac{2 v_s}{{340}^2 - {v_s}^2} \right]\]
\[ \Rightarrow \frac{2 v_s}{{340}^2 - v_s^2} = \frac{8}{340 \times 800}\]
\[ \Rightarrow {340}^2 - {v_s}^2 = 68000 v_s \]
Solving for vs, we get:
\[v_s\]= 1.695 m/s
For SHM:
\[v_s = r\omega\]
\[ \Rightarrow \omega = \left( \frac{1 . 695}{0 . 17} \right) = 10\]
\[ \therefore T = \frac{2\pi}{w} = \frac{\pi}{5} = 0 . 63 \sec\]
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