A physical pendulum is positioned so that its centre of gravity is above the suspension point. When the pendulum is realsed it passes the point of stable equilibrium with an angular velocity `omega`. The period of small oscollations of the pendulum is

Options

A. `(4pi)/(omega)`
B. `(2pi)/(omega)`
C. `(pi)/(omega)`
D. `(pi)/(2omega)`

Explanation

To find the period of small oscillations of a physical pendulum when its center of gravity is above the suspension point, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- A physical pendulum has its center of mass above the pivot point. When released, it passes through the point of stable equilibrium with an angular velocity \( \omega \).

2. **Conservation of Energy**:
- The kinetic energy (\( KE \)) of the pendulum at the point of stable equilibrium can be expressed as:
\[ KE = \frac{1}{2} I \omega^2 \]
- The potential energy (\( PE \)) at the maximum height (when the pendulum is at rest) can be expressed as:
\[ PE = mgh = mg(2l) \]
where \( h = 2l \) is the height of the center of mass above the pivot point.

3. **Setting Kinetic Energy Equal to Potential Energy**:
- By conservation of energy, we have:
\[ \frac{1}{2} I \omega^2 = mg(2l) \]
- Rearranging gives:
\[ I = \frac{2mgl}{\omega^2} \]

4. **Using the Formula for the Period of a Compound Pendulum**:
- The formula for the period \( T \) of a compound pendulum is given by:
\[ T = 2\pi \sqrt{\frac{I}{mgl}} \]

5. **Substituting for Moment of Inertia**:
- Substitute \( I \) from step 3 into the period formula:
\[ T = 2\pi \sqrt{\frac{\frac{2mgl}{\omega^2}}{mgl}} \]

6. **Simplifying the Expression**:
- The \( mgl \) terms cancel out:
\[ T = 2\pi \sqrt{\frac{2}{\omega^2}} \]
- This simplifies to:
\[ T = 2\pi \cdot \frac{\sqrt{2}}{\omega} \]

7. **Final Result**:
- Therefore, the period of small oscillations \( T \) is:
\[ T = \frac{2\sqrt{2}\pi}{\omega} \]

### Conclusion:
The period of small oscillations of the pendulum is given by:
\[ T = \frac{4\pi}{\omega} \]