Question

A sound source of frequency $f_0$ moves with speed $u_S$ relative to still air toward a receiver who is moving with speed $u_{[}r]$ relative to still air away from the source.
Use the result that $(1-x{)}^{-1}=1+x$ to show that is both $u_S$ and $u_r$ are small compared to $v$, then the received frequency is approximately
$f^{\prime }=(1+\frac{u_S-u_r}{v})f_0=(1+\frac{u_{rel}}{v})f_0$
where $u_{rel}$ is the relative velocity of the source and the receiver.

Answer 1

Using the Doppler effect equation, here both source and observer are moving.

where $u_s$ and $u_r$ are the velocity of the source and observer respectively.

Therefore, the revieced frequency

$f^{\prime }=f_0\left(\frac{v-u_r}{v-u_s}\right)$

Or, $f^{\prime }=f_0\left(\frac{1-\frac{u_r}{v}}{1-\frac{u_s}{v}}\right)...............using$ $(1-x{)}^{-1}=1+x$ as $\frac{u_s}{v}<<<1$

$⟹f^{\prime }=f_0(1-\frac{u_r}{v})(1+\frac{u_s}{v})$

$f^{\prime }=f_0(1+\frac{u_s}{v}-\frac{u_r}{v}+\frac{u_s{u}_r}{v^2})$

Or$f^{\prime }=f_0(1+\frac{u_s-u_r}{v}).............since$ $\frac{u_r{u}_s}{v^2}<<<1$

Or, $f^{\prime }=f_0(1+\frac{u_{rel}}{v})$