A uniform rod of length l is suspended by end and is made to undego small oscillations. Find the length of the simple pendulum having the time period equal to that of the rod.

Explanation

To solve the problem of finding the length of a simple pendulum that has the same time period as a uniform rod of length \( L \) suspended at one end and undergoing small oscillations, we will follow these steps:

### Step 1: Determine the Moment of Inertia of the Rod
The moment of inertia \( I \) of a uniform rod about an end is given by the formula:
\[ I = \frac{1}{3} M L^2 \]
However, we will also use the parallel axis theorem to find the moment of inertia about the pivot point (end of the rod). The formula for the moment of inertia using the parallel axis theorem is:
\[ I = I_{cm} + Mh^2 \]
where:
- \( I_{cm} = \frac{1}{12} M L^2 \) (moment of inertia about the center of mass),
- \( h = \frac{L}{2} \) (distance from the center of mass to the pivot).

Substituting these values:
\[ I = \frac{1}{12} M L^2 + M \left(\frac{L}{2}\right)^2 \]
\[ = \frac{1}{12} M L^2 + M \frac{L^2}{4} \]
\[ = \frac{1}{12} M L^2 + \frac{3}{12} M L^2 = \frac{4}{12} M L^2 = \frac{1}{3} M L^2 \]

### Step 2: Calculate the Time Period of the Rod
The time period \( T \) for small oscillations of the rod is given by:
\[ T = 2\pi \sqrt{\frac{I}{Mgh}} \]
Substituting \( I = \frac{1}{3} M L^2 \) and \( h = \frac{L}{2} \):
\[ T = 2\pi \sqrt{\frac{\frac{1}{3} M L^2}{Mg \frac{L}{2}}} \]
\[ = 2\pi \sqrt{\frac{L^2}{3g \cdot \frac{L}{2}}} \]
\[ = 2\pi \sqrt{\frac{2L}{3g}} \]

### Step 3: Find the Length of the Simple Pendulum
The time period of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L_p}{g}} \]
where \( L_p \) is the length of the simple pendulum. Setting the two expressions for \( T \) equal:
\[ 2\pi \sqrt{\frac{2L}{3g}} = 2\pi \sqrt{\frac{L_p}{g}} \]
Canceling \( 2\pi \) and squaring both sides:
\[ \frac{2L}{3g} = \frac{L_p}{g} \]
Multiplying both sides by \( g \):
\[ \frac{2L}{3} = L_p \]

### Final Result
Thus, the length of the simple pendulum that has the same time period as the rod is:
\[ L_p = \frac{2L}{3} \]