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).

\[\text { Now },    f_1  = \left( \frac{340}{340 + v_s} \right)f\] 

\[  \text { and }    f_2  = \left( \frac{340}{340 - v_s} \right)f\] 

\[     \therefore    f_2  -  f_1  = 8\] 

\[ \Rightarrow   8   = \left( \frac{340}{340  - v_s} \right)f - \left( \frac{340}{340  + v_s} \right)f\] 

\[ \Rightarrow   8   =   340f\left[ \frac{1}{340 - v_s} - \frac{1}{340 + v_s} \right]\] 

\[ \Rightarrow   8   = 340 \times 800 \times \left[ \frac{2 v_s}{{340}^2 - {v_s}^2} \right]\] 

\[ \Rightarrow   \frac{2 v_s}{{340}^2 - v_s^2} = \frac{8}{340 \times 800}\] 

\[ \Rightarrow    {340}^2  -  {v_s}^2  = 68000   v_s \]

Solving for v_s, we get:

\[v_s\]= 1.695 m/s

For SHM:

\[v_s  = r\omega\] 

\[ \Rightarrow   \omega = \left( \frac{1 . 695}{0 . 17} \right) = 10\] 

\[ \therefore   T = \frac{2\pi}{w} = \frac{\pi}{5} = 0 . 63  \sec\]