The function `sin^(2) (omegat)` represents.

Options

A. a periodic motion, but not SHM with a period `2pi//omega`
B. a periodic motion, but not SHM with a period `pi//omega`
C. an SHM with a period `2pi//omega`
D. an SHM with a period `pi//omega`

Explanation

To solve the problem regarding the function \( \sin^2(\omega t) \), we will analyze its periodicity and whether it represents simple harmonic motion (SHM).

### Step-by-Step Solution:

1. **Identify the Function**: The function given is \( \sin^2(\omega t) \).

2. **Check for Simple Harmonic Motion (SHM)**:
- SHM is typically represented by functions of the form \( x = A \sin(\omega t) \) or \( x = A \cos(\omega t) \).
- The function \( \sin^2(\omega t) \) does not fit this form directly, so it is not a simple harmonic motion.

3. **Determine Periodicity**:
- The sine function \( \sin(\omega t) \) is periodic with a period of \( T = \frac{2\pi}{\omega} \).
- Since \( \sin^2(\omega t) \) is derived from the sine function, it is also periodic.

4. **Find the Period of \( \sin^2(\omega t) \)**:
- We can use the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \).
- Applying this to our function:
\[ \sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2} \]
- The cosine function \( \cos(2\omega t) \) has a period of \( \frac{2\pi}{2\omega} = \frac{\pi}{\omega} \).
- Therefore, the period of \( \sin^2(\omega t) \) is \( T = \frac{\pi}{\omega} \).

5. **Conclusion**:
- The function \( \sin^2(\omega t) \) is periodic and has a period of \( \frac{\pi}{\omega} \).
- It does not represent simple harmonic motion.

### Final Answer:
The function \( \sin^2(\omega t) \) represents a periodic motion but not simple harmonic motion, with a period of \( \frac{\pi}{\omega} \).