Question

A tour boat approaches the shoreline, which is essentially a vertical cliff. When the boat's whistle is sounded, the passengers hear beating between the sound reaching them directly from the whistle and the sound reflected from the shore
If the frequency of the whistle is $100Hz$ and the beat frequency is $1Hz$, how fast is the boat going?
Take $v=330m/s$

Answer 1

Given $v=330m/s$ and ${\nu }_b=1Hz$

Also the velocity of the boat be $v_s$ and the frequency of the whistle be $f_s=100Hz$.

Assume that $f^{\prime }$ is frequency that reaches the shore and $f^{\prime \prime }$ be the frequency reflected from the shore.

So, $f^{\prime }=f_s\left(\frac{v}{v-vs}\right)$

Also, $f^{\prime \prime }=f^{\prime }\left(\frac{v+v_s}{v}\right)$

$f"=f_s\left(\frac{v+v_s}{v-v_s}\right)$

hence the beat frequecy ${\nu }_b$ is

${\nu }_b=f^{\prime \prime }-f_S⟹{\nu }_b=f_s\left(\frac{v+v_s-v+v_s}{v-v_s}\right)$

$⟹{\nu }_b=f_s\left(\frac{2v_s}{v-v_s}\right)$

$⟹1=100\left(\frac{2v_s}{330-v_s}\right)$

$⟹330-v_s=200v_s$

$v_s=\frac{330}{201}=1.64m{s}^{-1}$