A simple harmonic oscillator has an amplitude A and time period 6π...

From phasor diagram particle has to move from $P$ to $Q$ in a circle of radius equal to amplitude of $SHM$.

$cosϕ = \dfrac{\dfrac{\sqrt{3}A}{2}}{A} = \dfrac{\sqrt{3}}{2}$

$⇒ ϕ = \dfrac{\pi}{6}$

Now, $\dfrac{\pi}{6} = \omegat$

$⇒ \dfrac{\pi}{6} = \dfrac{2\pi}{T}t$

$⇒ \dfrac{\pi}{6} = \dfrac{2\pi}{6\pi}t$

$⇒ t = \dfrac{\pi}{2}$

So, $x = 2$