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 can follow these steps:

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

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

where:
- \( I \) is the moment of inertia about the point of suspension,
- \( M \) is the mass of the pendulum,
- \( g \) is the acceleration due to gravity,
- \( d \) is the distance from the point of suspension to the center of mass.

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

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

The center of mass of the rod is located at \( \frac{d}{2} \) from the pivot point. Thus, we can substitute \( I \) into the period formula:

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

### Step 2: Determine the period of Pendulum B (Simple Pendulum)
For a simple pendulum, the period \( T_B \) is given by the formula:

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

where \( L \) is the length of the pendulum. In this case, \( L = d \):

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

### Step 3: Calculate the ratio \( \frac{T_A}{T_B} \)
Now, we can find the ratio of the periods:

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

The \( 2\pi \) and \( \sqrt{d} \) terms cancel out:

\[ \frac{T_A}{T_B} = \frac{\sqrt{\frac{1}{3}}}{\sqrt{1}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \approx 0.577 \]

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

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