A simple harmonic oscillator has an amplitude $$A$$ and time period $$6\pi$$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $$x = A$$ to $$x = \frac{\sqrt{3}}{2}A$$ will be $$\frac{\pi}{x}$$ s, where $$x =$$ ______.

Given time period

T=6π

So angular frequency

$$\omega=\frac{2\pi}{T}$$

$$=\frac{2\pi}{6\pi}$$

$$=\frac{1}{3}$$

Since oscillation starts from mean position,

$$x=A\sin\omega t$$

We need time taken from

$$x=A$$

to

$$x=\frac{\sqrt{3}}{2}A$$

At

x=A

$$\sin\omega t_1=1$$

$$t_1=\frac{\pi}{2}ω$$

At

$$x=\frac{\sqrt{3}}{2}A$$

while returning from extreme point,

$$\sin\omega t_2=\frac{\sqrt{3}}{2}$$

after $$\pi/2$$, corresponding angle is

$$\omega t_2=\frac{2\pi}{3}$$

So required time is

$$\Delta t=\frac{\frac{2\pi}{3}-\frac{\pi}{2}}{\omega}$$

$$=\frac{\pi/6}{1/3}$$

$$=\frac{\pi}{2}$$

Given

$$\Delta t=\frac{\pi}{x}$$

so x=2