The function `sin^(2)(omega t)` represents:

Options

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

Explanation

To determine whether the function \( \sin^2(\omega t) \) represents simple harmonic motion (SHM) or periodic motion, we can follow these steps:

### Step 1: Understand the basic form of SHM
The standard equations for simple harmonic motion are:
\[ X = A \sin(\omega t) \quad \text{or} \quad X = A \cos(\omega t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

### Step 2: Analyze the function \( \sin^2(\omega t) \)
The function \( \sin^2(\omega t) \) can be expressed as the square of the sine function. This means that it will always yield non-negative values (since squaring any real number results in a non-negative number).

### Step 3: Determine if it is periodic
To check if \( \sin^2(\omega t) \) is periodic, we need to see if it repeats its values over time. The sine function itself is periodic with a period of \( 2\pi \). Therefore, \( \sin^2(\omega t) \) will also be periodic, but with a different period.

### Step 4: Find the period of \( \sin^2(\omega t) \)
Using the double angle identity, we can rewrite \( \sin^2(\theta) \):
\[ \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \]
In our case, substituting \( \theta = \omega t \):
\[ \sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2} \]
This shows that \( \sin^2(\omega t) \) is periodic with a period related to \( \cos(2\omega t) \).

### Step 5: Determine the period of \( \cos(2\omega t) \)
The period of \( \cos(2\omega t) \) is given by:
\[ T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega} \]
Thus, the period of \( \sin^2(\omega t) \) is \( \frac{\pi}{\omega} \).

### Conclusion
1. The function \( \sin^2(\omega t) \) is **not** simple harmonic motion because it does not fit the standard form of SHM.
2. However, it is **periodic** with a period of \( \frac{\pi}{\omega} \).

### Final Answer
The function \( \sin^2(\omega t) \) represents periodic motion with a period of \( \frac{\pi}{\omega} \).

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