A thin uniform rod of length l is pivoted at its upper end. It is free to swing in a vertical plane. Its time period for oscillation of small amplitude is

Options

A. `2pisqrt((l)/(g))`
B. `2pisqrt((2l)/(3g))`
C. `2pisqrt((3l)/(2g))`
D. `2pisqrt((l)/(3g))`

Explanation

To find the time period of oscillation for a thin uniform rod of length \( l \) pivoted at its upper end, we can follow these steps:

### Step 1: Understand the System
The rod is pivoted at one end and can swing in a vertical plane. When it swings, it will experience torque due to gravity acting on its center of mass.

### Step 2: Identify the Center of Mass
For a uniform rod of length \( l \), the center of mass is located at a distance of \( \frac{l}{2} \) from the pivot point.

### Step 3: Calculate the Torque
The force acting on the center of mass is the weight of the rod, which is \( Mg \) (where \( M \) is the mass of the rod and \( g \) is the acceleration due to gravity). The torque \( \tau \) about the pivot point due to this force is given by:
\[ \tau = r \times F = \left(\frac{l}{2}\right) \times Mg = \frac{Mg \cdot l}{2} \]

### Step 4: Moment of Inertia
The moment of inertia \( I \) of a thin rod about an end is given by:
\[ I = \frac{1}{3}Ml^2 \]

### Step 5: Use the Formula for Time Period
The formula for the time period \( T \) of a physical pendulum is:
\[ T = 2\pi \sqrt{\frac{I}{\tau}} \]
Substituting the values of \( I \) and \( \tau \):
\[ T = 2\pi \sqrt{\frac{\frac{1}{3}Ml^2}{\frac{Mg \cdot l}{2}}} \]

### Step 6: Simplify the Expression
Now simplify the expression:
\[ T = 2\pi \sqrt{\frac{\frac{1}{3}Ml^2}{\frac{Mg \cdot l}{2}}} = 2\pi \sqrt{\frac{2l}{3g}} \]

### Final Result
Thus, the time period \( T \) for small amplitude oscillations of the rod is:
\[ T = 2\pi \sqrt{\frac{2l}{3g}} \]