Pendulum `A` is a physical pendulum made from a thin, rigid and uniform rod whose length is `d`. One end of this rod is attached to the ceiling by a frictionless hinge, so that the rod is free to swing back and forth. Pendulum `B` is a simple pendulum whose length is also `d`. Obtain the ratio `(T_(A))/(T_(B))` of their periods for small angle oscillations.

Explanation

To find the ratio of the periods of Pendulum A (a physical pendulum) and Pendulum B (a simple pendulum), we will follow these steps:

### Step 1: Determine the period of Pendulum A (Physical Pendulum)
The formula for the period \( T_A \) of a physical pendulum is given by:

\[ T_A = 2\pi \sqrt{\frac{I}{mgh}} \]

Where:
- \( I \) is the moment of inertia about the pivot point,
- \( m \) is the mass of the pendulum,
- \( g \) is the acceleration due to gravity,
- \( h \) is the distance from the pivot to the center of mass.

For a uniform rod of length \( d \), the moment of inertia \( I \) about one end is:

\[ I = \frac{1}{3} m d^2 \]

The center of mass of the rod is located at a distance \( \frac{d}{2} \) from the pivot. Thus, \( h = \frac{d}{2} \).

Substituting these values into the formula for \( T_A \):

\[ T_A = 2\pi \sqrt{\frac{\frac{1}{3} m d^2}{mg \cdot \frac{d}{2}}} \]

### Step 2: Simplify the expression for \( T_A \)
Now we simplify the expression:

\[ T_A = 2\pi \sqrt{\frac{\frac{1}{3} d^2}{g \cdot \frac{d}{2}}} \]

This simplifies to:

\[ T_A = 2\pi \sqrt{\frac{2d}{3g}} \]

### Step 3: Determine the period of Pendulum B (Simple Pendulum)
The formula for the period \( T_B \) of a simple pendulum is given by:

\[ T_B = 2\pi \sqrt{\frac{L}{g}} \]

For Pendulum B, the length \( L \) is equal to \( d \):

\[ T_B = 2\pi \sqrt{\frac{d}{g}} \]

### Step 4: Find the ratio \( \frac{T_A}{T_B} \)
Now, we can find the ratio of the periods:

\[ \frac{T_A}{T_B} = \frac{2\pi \sqrt{\frac{2d}{3g}}}{2\pi \sqrt{\frac{d}{g}}} \]

The \( 2\pi \) cancels out:

\[ \frac{T_A}{T_B} = \frac{\sqrt{\frac{2d}{3g}}}{\sqrt{\frac{d}{g}}} \]

### Step 5: Simplify the ratio
This simplifies to:

\[ \frac{T_A}{T_B} = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}} \]

### Final Result
Thus, the ratio of the periods of Pendulum A to Pendulum B is:

\[ \frac{T_A}{T_B} = \sqrt{\frac{2}{3}} \approx 0.816 \]