In the compound pendulum, the minimum period of oscillation will be :

Options

A. `2 pi sqrt ((k)/(g))`
B. ` 2 pi sqrt ((l)/(g))`
C. ` 2 pi sqrt ((2l)/(g))`
D. `2 pi sqrt ((2k)/(g))`

Answer

To find the minimum period of oscillation for a compound pendulum, we can follow these steps:

### Step-by-Step Solution:

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

2. **Apply the Parallel Axis Theorem**:
According to the parallel axis theorem, the moment of inertia \( I_o \) can be expressed as:
\[ I_o = I_{cg} + md^2 \]
where \( I_{cg} \) is the moment of inertia about the center of mass.

3. **Substitute \( I_o \) into the Time Period Formula**:
Substituting \( I_o \) into the time period formula gives:
\[ T = 2\pi \sqrt{\frac{I_{cg} + md^2}{mgd}} \]

4. **Replace \( I_{cg} \) with \( mk^2 \)**:
Here, \( k \) is the radius of gyration, so we can write:
\[ I_{cg} = mk^2 \]
Thus, the time period becomes:
\[ T = 2\pi \sqrt{\frac{mk^2 + md^2}{mgd}} = 2\pi \sqrt{\frac{k^2 + d^2}{gd}} \]

5. **Minimize the Time Period**:
To find the minimum period, we need to minimize the expression:
\[ f(d) = \frac{k^2 + d^2}{d} \]
This can be rewritten as:
\[ f(d) = \frac{k^2}{d} + d \]

6. **Differentiate and Set to Zero**:
Differentiate \( f(d) \) with respect to \( d \):
\[ f'(d) = -\frac{k^2}{d^2} + 1 \]
Setting \( f'(d) = 0 \) gives:
\[ -\frac{k^2}{d^2} + 1 = 0 \implies \frac{k^2}{d^2} = 1 \implies d^2 = k^2 \implies d = k \]

7. **Substitute Back to Find Minimum Time Period**:
Substitute \( d = k \) back into the time period formula:
\[ T = 2\pi \sqrt{\frac{k^2 + k^2}{gk}} = 2\pi \sqrt{\frac{2k^2}{gk}} = 2\pi \sqrt{\frac{2k}{g}} \]

### Final Result:
The minimum period of oscillation for the compound pendulum is:
\[ T = 2\pi \sqrt{\frac{2k}{g}} \]