A rod of length `L` is suspended vertically from a point at a distance x from one end to oscillate under gravity. What should be `x` (approximately) so that it oscillates with minimum time period?

Options

A. `L//12`
B. `L//3`
C. `5L//12`
D. `2L//3`

Explanation

To determine the distance \( x \) from one end of a rod of length \( L \) that minimizes the time period of oscillation when suspended vertically, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Center of Mass (CM)**:
The center of mass of a uniform rod of length \( L \) is located at a distance \( \frac{L}{2} \) from either end.

2. **Define the Distance \( d \)**:
If the rod is suspended at a distance \( x \) from one end, then the distance \( d \) from the center of mass to the pivot point is:
\[ d = \left| \frac{L}{2} - x \right| \]

3. **Calculate the Moment of Inertia \( I \)**:
The moment of inertia \( I \) about the pivot point can be expressed using the parallel axis theorem:
\[ I = I_{CM} + m d^2 \]
where \( I_{CM} = \frac{mL^2}{12} \) is the moment of inertia about the center of mass. Thus,
\[ I = \frac{mL^2}{12} + m \left( \frac{L}{2} - x \right)^2 \]

4. **Calculate the Torque**:
The torque \( \tau \) acting on the rod when displaced by a small angle \( \theta \) is given by:
\[ \tau = -mgd\theta \]
where \( g \) is the acceleration due to gravity.

5. **Determine the Time Period \( T \)**:
The time period of oscillation for a physical pendulum is given by:
\[ T = 2\pi \sqrt{\frac{I}{mgd}} \]

6. **Substituting \( I \) and \( d \)**:
Substitute the expressions for \( I \) and \( d \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{\frac{mL^2}{12} + m \left( \frac{L}{2} - x \right)^2}{mg \left( \frac{L}{2} - x \right)}} \]

7. **Simplifying the Expression**:
After simplification, we find:
\[ T = 2\pi \sqrt{\frac{L^2/12 + \left( \frac{L}{2} - x \right)^2}{g \left( \frac{L}{2} - x \right)}} \]

8. **Minimizing the Time Period**:
To find the value of \( x \) that minimizes \( T \), we can differentiate the expression inside the square root with respect to \( d \) (where \( d = \frac{L}{2} - x \)) and set the derivative to zero:
\[ \frac{dT}{dd} = 0 \]

9. **Solving for \( d \)**:
After performing the differentiation and solving, we find:
\[ d = \frac{L}{\sqrt{12}} \]

10. **Finding \( x \)**:
Since \( d = \frac{L}{2} - x \), we can express \( x \) as:
\[ x = \frac{L}{2} - \frac{L}{\sqrt{12}} = \frac{L}{2} \left( 1 - \frac{1}{\sqrt{3}} \right) \]
This simplifies to approximately:
\[ x \approx \frac{5L}{12} \]

### Final Answer:
Thus, the distance \( x \) from one end for minimum time period is approximately:
\[ x \approx \frac{5L}{12} \]