Compound pendulum are made of A rod of length `l` suspended through a point located at distance `l // 4` from centre of rod. Find the time period in each case.

Explanation

To find the time period of a compound pendulum made of a rod of length \( l \) suspended through a point located at a distance \( \frac{l}{4} \) from the center of the rod, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Formula for Time Period**:
The time period \( T \) of a compound pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{I_{cm} + m \cdot d^2}{m \cdot g \cdot d}} \]
where:
- \( I_{cm} \) is the moment of inertia about the center of mass,
- \( m \) is the mass of the pendulum,
- \( g \) is the acceleration due to gravity,
- \( d \) is the distance from the center of mass to the point of suspension.

2. **Calculating Moment of Inertia for the Rod**:
The moment of inertia \( I_{cm} \) of a uniform rod about its center is given by:
\[ I_{cm} = \frac{1}{12} m l^2 \]

3. **Finding the Distance \( d \)**:
The distance \( d \) from the center of mass to the point of suspension is given as \( \frac{l}{4} \).

4. **Substituting Values into the Formula**:
Now substituting \( I_{cm} \) and \( d \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{\frac{1}{12} m l^2 + m \left(\frac{l}{4}\right)^2}{m g \cdot \frac{l}{4}}} \]

5. **Simplifying the Expression**:
Calculate \( m \left(\frac{l}{4}\right)^2 \):
\[ m \left(\frac{l}{4}\right)^2 = m \cdot \frac{l^2}{16} \]
Therefore:
\[ T = 2\pi \sqrt{\frac{\frac{1}{12} m l^2 + \frac{1}{16} m l^2}{m g \cdot \frac{l}{4}}} \]

6. **Finding a Common Denominator**:
The common denominator for \( \frac{1}{12} \) and \( \frac{1}{16} \) is \( 48 \):
\[ \frac{1}{12} = \frac{4}{48}, \quad \frac{1}{16} = \frac{3}{48} \]
Thus:
\[ \frac{1}{12} m l^2 + \frac{1}{16} m l^2 = \frac{4 m l^2 + 3 m l^2}{48} = \frac{7 m l^2}{48} \]

7. **Substituting Back into the Time Period Formula**:
Now substituting back:
\[ T = 2\pi \sqrt{\frac{\frac{7 m l^2}{48}}{m g \cdot \frac{l}{4}}} \]

8. **Simplifying Further**:
The \( m \) cancels out:
\[ T = 2\pi \sqrt{\frac{7 l^2}{48 \cdot g \cdot \frac{l}{4}}} \]
Simplifying the denominator:
\[ g \cdot \frac{l}{4} = \frac{g l}{4} \]
So:
\[ T = 2\pi \sqrt{\frac{7 l^2}{48 \cdot \frac{g l}{4}}} = 2\pi \sqrt{\frac{7 \cdot 4 l}{48 g}} = 2\pi \sqrt{\frac{28 l}{48 g}} = 2\pi \sqrt{\frac{7 l}{12 g}} \]

9. **Final Result**:
Thus, the time period \( T \) of the compound pendulum is:
\[ T = 2\pi \sqrt{\frac{7 l}{12 g}} \]