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
A. 120
B. 240
C. 60
D. 360

Hint:To solve this question you have to use the relation between the phase difference and path difference. The phase difference is defined as the difference in the phase angle of the two waves and the Path difference is defined as the difference in the path travelled by the two waves. Hence there is a direct relation between phase difference and path difference is. Both are directly proportional to each other.

Formula used:
In any two waves with the same frequency, the relation between Phase Difference and Path Difference is given as -
\[\Delta \phi = \dfrac{{2\pi }}{\lambda }\Delta x\]
Where \[\Delta x\] is the path difference between the two waves and \[\Delta \phi \] is the phase difference between the two waves.

Complete step by step solution:
Given: Phase difference, \[\Delta \phi = \dfrac{{1080}}{{180}}\pi = 6\pi \]
Wave velocity, \[v = 480\,m{s^{ - 1}}\]
Separation distance, \[\Delta x = 12\,m\]
As we know that
\[\Delta \phi = \dfrac{{2\pi }}{\lambda }\Delta x\]
\[\Rightarrow \lambda = \dfrac{{2\pi }}{{\Delta \phi }}\Delta x\]
Substituting the values, we have
\[\lambda = \dfrac{{2\pi }}{{6\pi }} \times 12\]
\[\Rightarrow \lambda = 4m\]
Now the number of waves passing can be,
\[n = \dfrac{v}{\lambda }\]
On substituting the values,
\[n = \dfrac{{480}}{4}\]
\[\therefore n = 120\]
Therefore, the number of waves passing across a point in 1 sec is 120.

Hence option A is the correct answer.

Note: There is a direct relation between Phase Difference and Path as they are directly proportional to each other. Phase difference is the difference between phase angles between two waves. On the other hand, Path difference refers to the difference in the path travelled by the two waves.