Question

Two stationary sources each emit wave of Wavelength $\lambda$. An observer moves from one source to another with velocity u. Then the number of beats to another with velocity u. Then number of beats

Answer 1

Let the two sources $A$ & $B$ emit sound of frequency $v_o=\frac{v}{\lambda }$

When observer moves $B\to A$

It receives compressed waves from $A$ whose frequency is:

$v_A=v_o\left(\frac{u+v}{u}\right)$ [From Doppler's Effect]

It recieves elongated waves from $B$, given by:

$v_B=v_o\left(\frac{u-v}{u}\right)$

Thus frequency of beats observed by him:

$v_{beat}=|v_A-v_B|=v\left(\frac{u+v-u+v}{u}\right)$

$\Rightarrow v_{beat}=\frac{2v^2}{u\lambda }$ $(as{v}_o=\frac{v}{\lambda })$