a
From phasor diagram particle has to move from \(\mathrm{P}\) to \(Q\) in a circle of radius equal to amplitude of \(SHM.\)

\( \cos \phi=\frac{\frac{\sqrt{3} \mathrm{~A}}{2}}{\mathrm{~A}}=\frac{\sqrt{3}}{2} \)

\( \phi=\frac{\pi}{6}\)

Now, \(\frac{\pi}{6}=\omega t\)

\(\frac{\pi}{6}=\frac{2 \pi}{T} t\)

\(\frac{\pi}{6}=\frac{2 \pi}{6 \pi} t\)

\(\mathrm{t}=\frac{\pi}{2}\)

So, \(\mathrm{x}=2\)