There is a rod of length `l` and mass `m`. It is hinged at one end to the ceiling. The period of small oscillation is

Options

A. `2pi sqrt((mL)/(2g))`
B. `2pi sqrt((mL)/(g))`
C. `2pi sqrt((2L)/(3g))`
D. `2pi sqrt((m)/(gL))`

Answer

To find the period of small oscillations of a rod of length \( l \) and mass \( m \) that is hinged at one end to the ceiling, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the System**:
We have a uniform rod of length \( l \) and mass \( m \) hinged at one end (point O) to the ceiling. When the rod is displaced by a small angle \( \theta \), it will oscillate about the hinge.

2. **Determine the Forces**:
The weight of the rod acts at its center of mass, which is located at a distance \( \frac{l}{2} \) from the hinge. The force due to gravity is \( mg \), acting downward.

3. **Calculate the Torque**:
The torque \( \tau \) about the hinge due to the weight of the rod when displaced by angle \( \theta \) is given by:
\[ \tau = mg \sin(\theta) \cdot \frac{l}{2} \]
For small angles, we can approximate \( \sin(\theta) \approx \theta \):
\[ \tau \approx mg \cdot \frac{l}{2} \cdot \theta \]

4. **Relate Torque to Angular Acceleration**:
The torque is also related to the moment of inertia \( I \) and angular acceleration \( \alpha \) by the equation:
\[ \tau = I \alpha \]
For a rod hinged at one end, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{3} ml^2 \]
Therefore, we can write:
\[ mg \cdot \frac{l}{2} \cdot \theta = \frac{1}{3} ml^2 \cdot \alpha \]

5. **Express Angular Acceleration**:
Rearranging the equation gives:
\[ \alpha = \frac{3g}{2l} \theta \]

6. **Relate to Simple Harmonic Motion**:
In simple harmonic motion (SHM), the angular acceleration is related to the displacement by:
\[ \alpha = -\omega^2 \theta \]
Comparing this with our expression for \( \alpha \):
\[ -\omega^2 = \frac{3g}{2l} \]
Thus, we find:
\[ \omega = \sqrt{\frac{3g}{2l}} \]

7. **Calculate the Period**:
The period \( T \) of oscillation is given by:
\[ T = \frac{2\pi}{\omega} \]
Substituting for \( \omega \):
\[ T = 2\pi \sqrt{\frac{2l}{3g}} \]

### Final Result:
The period of small oscillations of the rod is:
\[ T = 2\pi \sqrt{\frac{2l}{3g}} \]