1. We are given the function $x(t)=\sin^2(\omega t).$

2. First, check whether it is periodic.

   Using the identity, $\sin^2\theta=\frac{1-\cos 2\theta}{2},$ we get $x(t)=\frac{1-\cos(2\omega t)}{2}.$

3. Now determine its period.

   Since $\cos(2\omega t)$ has angular frequency $2\omega$, its period is $T=\frac{2\pi}{2\omega}=\frac{\pi}{\omega}.$

   Hence, $\sin^2(\omega t)$ is a periodic function with period $\boxed{\frac{\pi}{\omega}}.$

4. Now check whether it represents SHM.

   For SHM, displacement must be of the form $x=A\sin(\Omega t+\phi) \quad \text{or} \quad x=A\cos(\Omega t+\phi),$ i.e. a pure sine/cosine function about a fixed mean position.

   But here, $x(t)=\frac{1}{2}-\frac{1}{2}\cos(2\omega t),$ which has a constant offset $\frac{1}{2}$ and is not directly of the standard SHM form as written.

   Also, $\sin^2(\omega t)$ is always non-negative, so it does not oscillate symmetrically about zero.

   Therefore, this function is periodic but not SHM.

5. Evaluate options:

   • A: periodic, but not SHM with period $\frac{\pi}{\omega}$ — Correct
   • B: periodic, but not SHM with period $\frac{2\pi}{\omega}$ — Incorrect period
   • C: SHM with period $\frac{\pi}{\omega}$ — Not SHM
   • D: SHM with period $\frac{2\pi}{\omega}$ — Not SHM and wrong period

Therefore, the correct answer is $\boxed{A}.$