The wave velocity of a progressive wave is $480m{s}^{-1}$ and the phase difference between the two particles separated by a distance of $12m$ is ${1080}^0.$ The number of waves passing across a point in $1\sec$ is

Given ,

Velocity $v=480m/s$

Phase difference $\varphi ={1080}^{\circ }$

                                  $=1080\times \frac{\text{π}}{180}=6\text{π}$

Distance $x=12m$

Time $t=1\sec$

Number of waves passing $n=?$

Phase difference $\varphi =\frac{2\text{π}}{\lambda }.x$

                            $6\text{π}=\frac{2\text{π}}{\lambda }\times 12$

                       $\Rightarrow \lambda =4m$

As    $v=n\lambda$

     $480=n\times 4$

$$ $\therefore n=120\text{Hz}$