To identify the nature of the motion and its frequency, we simplify the given displacement equation using the trigonometric identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$.

Given: $x = a \sin^2 \omega t$ Substituting the identity: $x = a \left( \frac{1 - \cos 2\omega t}{2} \right)$ $x = \frac{a}{2} - \frac{a}{2} \cos(2\omega t)$

This equation describes a Simple Harmonic Motion (SHM) where the particle oscillates about a shifted equilibrium position ($x_0 = a/2$). The oscillating term is $\cos(2\omega t)$.

The angular frequency of this motion is the coefficient of $t$ in the cosine term: $\omega' = 2\omega$

The relationship between frequency ($f$ or $\nu$) and angular frequency is given by $f = \frac{\omega'}{2\pi}$ . Substituting $\omega'$: $f = \frac{2\omega}{2\pi} = \frac{\omega}{\pi}$

Therefore, the motion is simple harmonic with a frequency of $\omega/\pi$.