Compound pendulum are made of :
(a) A rod of length `l` suspended through a point located at distance `l // 4` from centre of rod.
(b) A ring of mass `m` and radius `r` suspended through a point on its periphery.
(c) A uniform square plate of edge a suspended through a corner.
(d) A uniform disc of mass `m` and radius `r` suspeneded through a point `r//2` away from the centre.
Find the time period in each case.

Explanation

To find the time period of a compound pendulum for the given cases, we will use the formula for the time period \( T \) of a compound pendulum:

\[ T = 2\pi \sqrt{\frac{I}{Mgh}} \]

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

Let's analyze each case step by step.

### Case (a): Rod of length \( l \) suspended through a point located at distance \( \frac{l}{4} \) from the center of the rod.

1. **Moment of Inertia**: For a rod of length \( l \) about its center, the moment of inertia \( I_{CM} \) is given by:
\[ I_{CM} = \frac{1}{12} Ml^2 \]
To find the moment of inertia about the pivot point, we use the parallel axis theorem:
\[ I = I_{CM} + Md^2 = \frac{1}{12} Ml^2 + M\left(\frac{l}{4}\right)^2 = \frac{1}{12} Ml^2 + \frac{Ml^2}{16} = \frac{1}{12} Ml^2 + \frac{3}{48} Ml^2 = \frac{7}{48} Ml^2 \]

2. **Distance to Center of Mass**: The distance \( h \) from the pivot to the center of mass is \( \frac{l}{4} \).

3. **Time Period Calculation**:
\[ T = 2\pi \sqrt{\frac{I}{Mgh}} = 2\pi \sqrt{\frac{\frac{7}{48} Ml^2}{Mg\left(\frac{l}{4}\right)}} \]
Simplifying:
\[ T = 2\pi \sqrt{\frac{7l}{12g}} \]

### Case (b): Ring of mass \( m \) and radius \( r \) suspended through a point on its periphery.

1. **Moment of Inertia**: For a ring about its center, the moment of inertia is:
\[ I_{CM} = mr^2 \]
Using the parallel axis theorem:
\[ I = I_{CM} + Md^2 = mr^2 + m(r)^2 = 2mr^2 \]

2. **Distance to Center of Mass**: The distance \( h \) from the pivot to the center of mass is \( r \).

3. **Time Period Calculation**:
\[ T = 2\pi \sqrt{\frac{I}{Mgh}} = 2\pi \sqrt{\frac{2mr^2}{mgr}} \]
Simplifying:
\[ T = 2\pi \sqrt{\frac{2r}{g}} \]

### Case (c): Uniform square plate of edge \( a \) suspended through a corner.

1. **Moment of Inertia**: The moment of inertia about the center is:
\[ I_{CM} = \frac{1}{6} Ma^2 \]
Using the parallel axis theorem:
\[ I = I_{CM} + Md^2 = \frac{1}{6} Ma^2 + M\left(\frac{a\sqrt{2}}{2}\right)^2 = \frac{1}{6} Ma^2 + \frac{1}{2} Ma^2 = \frac{2}{3} Ma^2 \]

2. **Distance to Center of Mass**: The distance \( h \) from the pivot to the center of mass is \( \frac{a}{2\sqrt{2}} \).

3. **Time Period Calculation**:
\[ T = 2\pi \sqrt{\frac{I}{Mgh}} = 2\pi \sqrt{\frac{\frac{2}{3} Ma^2}{Mg\left(\frac{a}{2\sqrt{2}}\right)}} \]
Simplifying:
\[ T = 2\pi \sqrt{\frac{4a}{3g}} \]

### Case (d): Uniform disc of mass \( m \) and radius \( r \) suspended through a point \( \frac{r}{2} \) away from the center.

1. **Moment of Inertia**: The moment of inertia about the center is:
\[ I_{CM} = \frac{1}{2} mr^2 \]
Using the parallel axis theorem:
\[ I = I_{CM} + Md^2 = \frac{1}{2} mr^2 + m\left(\frac{r}{2}\right)^2 = \frac{1}{2} mr^2 + \frac{1}{4} mr^2 = \frac{3}{4} mr^2 \]

2. **Distance to Center of Mass**: The distance \( h \) from the pivot to the center of mass is \( \frac{r}{2} \).

3. **Time Period Calculation**:
\[ T = 2\pi \sqrt{\frac{I}{Mgh}} = 2\pi \sqrt{\frac{\frac{3}{4} mr^2}{mg\left(\frac{r}{2}\right)}} \]
Simplifying:
\[ T = 2\pi \sqrt{\frac{3r}{2g}} \]

### Summary of Time Periods:
- Case (a): \( T = 2\pi \sqrt{\frac{7l}{12g}} \)
- Case (b): \( T = 2\pi \sqrt{\frac{2r}{g}} \)
- Case (c): \( T = 2\pi \sqrt{\frac{4a}{3g}} \)
- Case (d): \( T = 2\pi \sqrt{\frac{3r}{2g}} \)