Given data

• Amplitude $= A$
• Time period $T = 6\pi\,\text{s}$
• Oscillation starts from mean position

Find angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{6\pi} = \frac{1}{3}\,\text{rad/s}$

Write displacement equation

Since the particle starts from mean position, we can write $x = A\sin(\omega t)$

Find the time when particle is at $x=A$ $A\sin(\omega t_1)=A$ $\sin(\omega t_1)=1$ $\omega t_1 = \frac{\pi}{2}$ $t_1 = \frac{\pi/2}{\omega} = \frac{\pi/2}{1/3} = \frac{3\pi}{2}$

Find the next time when particle is at $x=\frac{\sqrt{3}}{2}A$ after reaching $x=A$ $A\sin(\omega t_2)=\frac{\sqrt{3}}{2}A$ $\sin(\omega t_2)=\frac{\sqrt{3}}{2}$

The relevant solution after $\omega t=\frac{\pi}{2}$ is: $\omega t_2 = \frac{2\pi}{3}$

Therefore, $t_2 = \frac{2\pi/3}{1/3} = 2\pi$

Required time interval $\Delta t = t_2 - t_1 = 2\pi - \frac{3\pi}{2} = \frac{\pi}{2}\,\text{s}$

Compare with given form $\Delta t = \frac{\pi}{x}$ So, $\frac{\pi}{x} = \frac{\pi}{2} \Rightarrow x=2$

Final answer $\boxed{2}$