A simple pendulum has time period T. A uniform rod, whose length is the same as that of the pendulum, undergoes small oscillations about its upper end. Its time period of oscillation will be -

Options

A. `lt T`
B. T
C. `gtT`
D. may be (A), (B) or (C) depending on whether T is `lt` equal to or `gt2 `seconds

Answer

To find the time period of a uniform rod undergoing small oscillations about its upper end, we can follow these steps:

### Step 1: Understand the Problem
We need to find the time period of a uniform rod that is pivoted at one end and undergoes small oscillations. The length of the rod is the same as that of a simple pendulum, which has a known time period \( T \).

### Step 2: Recall the Time Period of a Simple Pendulum
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

### Step 3: Use the Formula for a Physical Pendulum
For a physical pendulum (like our uniform rod), the time period \( T_1 \) is given by:
\[ T_1 = 2\pi \sqrt{\frac{I}{mgh}} \]
where:
- \( I \) is the moment of inertia about the pivot point,
- \( m \) is the mass of the rod,
- \( h \) is the distance from the pivot to the center of mass.

### Step 4: Calculate the Moment of Inertia
For a uniform rod of length \( L \) pivoted at one end, the moment of inertia \( I \) is:
\[ I = \frac{1}{3} mL^2 \]

### Step 5: Find the Center of Mass
The center of mass of a uniform rod is located at a distance of \( \frac{L}{2} \) from the pivot.

### Step 6: Substitute Values into the Time Period Formula
Now substituting \( I \) and \( h \) into the time period formula:
\[ T_1 = 2\pi \sqrt{\frac{\frac{1}{3} mL^2}{mg \cdot \frac{L}{2}}} \]
This simplifies to:
\[ T_1 = 2\pi \sqrt{\frac{\frac{1}{3} L^2}{g \cdot \frac{L}{2}}} \]
\[ T_1 = 2\pi \sqrt{\frac{2L}{3g}} \]

### Step 7: Relate \( T_1 \) to \( T \)
From the time period of the simple pendulum:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
We can express \( T_1 \) in terms of \( T \):
\[ T_1 = \sqrt{\frac{2}{3}} T \]

### Conclusion
Thus, the time period of the uniform rod undergoing small oscillations about its upper end is:
\[ T_1 = \sqrt{\frac{2}{3}} T \]