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.

Answer 1

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).

$$\begin{gathered} \mathrm{Now},f_1=\left(\frac{340}{340+v_s}\right)f \\ and{f}_2=\left(\frac{340}{340-v_s}\right)f \\ \therefore f_2-f_1=8 \\ \Rightarrow 8=\left(\frac{340}{340-v_s}\right)f-\left(\frac{340}{340+v_s}\right)f \\ \Rightarrow 8=340f\left[\frac{1}{340-v_s}-\frac{1}{340+v_s}\right] \\ \Rightarrow 8=340\times 800\times \left[\frac{2v_s}{{340}^2-{v_s}^2}\right] \\ \Rightarrow \frac{2v_s}{{340}^2-v_s^2}=\frac{8}{340\times 800} \\ \Rightarrow {340}^2-{v_s}^2=68000v_s \end{gathered}$$

Solving for v_s, we get:
$v_s$ = 1.695 m/s

For SHM:

$$\begin{gathered} v_s=r\omega \\ \Rightarrow \omega =\left(\frac{1.695}{0.17}\right)=10 \\ \therefore T=\frac{2\mathrm{\pi }}{w}=\frac{\mathrm{\pi }}{5}=0.63\sec \end{gathered}$$