Question

The source $S_1$ is at rest. The observer and the source $S_2$ are moving toward $S_1$ as shown in the figure. The beat frequency heard by the observer if both the source have the frequency of $120Hz$ is (speed of sound is $330m/s$)

Answer 1

The beat frequency is the mod of the difference of two apparent frequency.
$f_0=120Hz$
Let, the apparent frequency heard by the observer due to source $S_2$ be $f_2$
the apparent frequency heard by the observer due to source $S_1$ be $f_1$
$f_2=f_0\left[\frac{v\pm v_0}{v\pm v_s}\right]$
When both the object are coming close to each other then the apparent frequency will increase. So, we will take $+ive$ in the numerator and $-ive$ in the denominator.
Opposite in case when object are going far from each other
$v_o=10\cos {53}^0$
$v_s=30\cos {37}^0$
$f_2=120\left[\frac{330+10\cos {53}^0}{330-30\cos {37}^0}\right]$
$f_2=120\left[\frac{336}{306}\right]$
IIy, in case of $f_1$
$f_1=120\left[\frac{330+10}{330-0}\right]$
$f_1=120\left[\frac{34}{33}\right]$
In this case $v_s=0$ because source is stationary
$f_b=|f_2-f_1|=\left|120[\frac{334}{306}-\frac{34}{33}]\right|$
$\therefore f_b=8.128Hz$