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 $340m{s}^{-1}$. If the width of the frequency band detected by the detector is 8 Hz, find the time period of the source.

Answer 1

Given that r =0.17 m, f=800 Hz,

u =340 m/s

Frequency band $=f_1+f_2=8Hz$

Where $f_1$ and $f_2$ correspond to the maximum and minimum apparent frequencies. (Both will be at the mean position, because the velocity is maximum).

Now, $f_1=\left(\frac{340}{340+v_s}\right)f$ and $f_2=\left(\frac{340}{340-v_s}\right)f$

$\therefore f_1-f_2=8$

\)=340 f\left [ \frac{1}{340-v_s}-\frac{1}{340+v_s} \right ]\)

=8

$\Rightarrow \frac{2vs}{{340}^2-v_s^2}=\frac{8}{340\times 800}$

$\Rightarrow {340}^{2}0v{s}^2=68000vs$

Solving for $v_s$, we get

vs=1.695 m/s

For SHM,

\(v_s =rw

$\Rightarrow \omega =\left(\frac{1.695}{0.17}\right)=10$

So, $T=\frac{2\pi }{\omega }=\frac{\pi }{5}=0.63sec$