A particle travels along the X-axis governed by the equation x =...

Given equation of motion is $x = x_0 \sin^2 \omega t$.

Using the trigonometric identity $\sin^2 \theta = \dfrac{1 - \cos 2\theta}{2}$, the equation can be rewritten as:

$x = \dfrac{x_0}{2} (1 - \cos 2\omega t)$

$x - \dfrac{x_0}{2} = - \dfrac{x_0}{2} \cos 2\omega t$

This represents simple harmonic motion about the mean position $x = \dfrac{x_0}{2}$.

The amplitude of the SHM is $A = \dfrac{x_0}{2}$.

The angular frequency of the SHM is $\omega' = 2\omega$.

The time period of the motion is $T = \dfrac{2\pi}{\omega'} = \dfrac{2\pi}{2\omega} = \dfrac{\pi}{\omega}$.