Two particles P and Q are executing SHM across the same straight line whose equations are…

Two particles P and Q are executing SHM across the same straight line whose equations are given as

\(y_{p}=A \sin \left[\omega t+\Phi_{1}\right]\) and \(V_{\mathbf{q}}=A \cos \left[\omega t+\Phi_{2}\right]\).

\(\overleftarrow{\mathrm{C}} \quad \mathrm{O} \quad \overrightarrow{\mathrm{~B}}\)

An observer, at t = 0, observes the particle P at a distance A/√2 moving to the right from the mean position O while Q at \(\frac{\sqrt{3}}{2} A\) moving to the left from the mean position O as shown. Then, \(\left[\Phi_{2}-\Phi_{1}\right]\) is