d
Given:- Two points are located at a distance of \(10\,m\) and \(15\,m\)

The period of oscillation \(=0.05\,s\)

The velocity of the wave \(=300\,m / s\)

To find:- The phase difference between the oscillation of two points.

Solution:- We know that,

Path difference \(=\frac{2 \pi}{\lambda} \times\) path difference

Path difference between two points,

\(\Delta x =15-10=5\,m\)

Time period, \(T =0.5^{ s }\)

\(\Rightarrow\) frequency \(v =\frac{1}{ T }=\frac{1}{0.05}=20\,Hz\)

Velocity \(v =300\,m / s\)

\(\therefore\) Wavelenght \(\lambda=\frac{ v }{ n }=\frac{300}{20}=15\,m\)

Hence, phase difference

\(\Delta \phi=\frac{2 \pi}{\lambda} \times x =\frac{2 \pi}{15} \times 5=\frac{2 \pi}{3}\)