Question

A receiver and a source of sonic oscillations of frequency $2000\text{Hz}$ are located on the x-axis. The source swings harmonically along x-axis with circular frequency $\omega$ and amplitude $50\text{cm}$. At what approximate value of $\omega$ will the frequency bandwidth (difference of maximum and minimum frequency) registered by stationary receiver be equal to $200\text{Hz}$? (The velocity of sound in air is $340\text{m/s}$ and assume that the velocity of source is small compared to the velocity of sound)

• A. $20\text{rad/s}$
• B. $34\text{rad/s}$
• C. $48\text{rad/s}$
• D. $62\text{rad/s}$

Answer 1

The correct option is B $34\text{rad/s}$

We know
$f_{app}=f_0\left(\frac{v\pm v_0}{v\pm v_s}\right)$ $\begin{array}{cc}{f}_{max}-f_{min}& =f_0\left(\frac{v}{v-v_s}\right)-f_0\left(\frac{v}{v+v_s}\right)\\ & =f_{0}v\left(\frac{v+v_s-v+v_s}{v^2-v_s^2}\right)\end{array}$
$\because v>>v_s⟹f_{max}-f_{min}=f_o\left(\frac{2v_s}{v}\right)$
$\begin{array}{cc}200& =\frac{2v_s}{v}{f}_0\\ & =\frac{2\times \frac{1}{2}\times w}{340}\times 2000\end{array}$
Hence $\omega =34\frac{rad}{s}$