The time period of a physical pendulum about some pivot point is T. When we take another pivot point, opposite of the first one such that the centre of mass of the physical pendulum lies on the line joining these two pivot points, we obtain the same time period. If the two points are separated by a distance L, then the time period T is

Options

A. `2pisqrt((L)/(g))`
B. `pisqrt((L)/(g))`
C. `2pisqrt((L)/(2g))`
D. `pisqrt((2L)/(g))`

Answer

To solve the problem, we need to derive the expression for the time period \( T \) of a physical pendulum when pivoting about two different points that are separated by a distance \( L \).

### Step-by-Step Solution:

1. **Understanding the Physical Pendulum**:
A physical pendulum oscillates about a pivot point, and its time period \( T \) can be expressed using the formula:
\[ T = 2\pi \sqrt{\frac{I}{mgh}} \]
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, and \( h \) is the distance from the pivot to the center of mass.

2. **Identifying the Two Pivot Points**:
Let’s denote the two pivot points as \( P_1 \) and \( P_2 \). The distance between these two points is \( L \). The center of mass \( C \) of the pendulum lies on the line joining these two pivot points.

3. **Moment of Inertia Calculation**:
The moment of inertia \( I \) about the first pivot point \( P_1 \) can be expressed as:
\[ I_{P_1} = I_C + m d_1^2 \]
where \( I_C \) is the moment of inertia about the center of mass, \( m \) is the mass of the pendulum, and \( d_1 \) is the distance from the center of mass to the pivot point \( P_1 \).

Similarly, for the second pivot point \( P_2 \):
\[ I_{P_2} = I_C + m d_2^2 \]
where \( d_2 = L - d_1 \) (since the total distance between the two pivot points is \( L \)).

4. **Time Period for Both Pivot Points**:
The time period for the first pivot point \( P_1 \) is:
\[ T_1 = 2\pi \sqrt{\frac{I_{P_1}}{mgd_1}} = 2\pi \sqrt{\frac{I_C + m d_1^2}{mgd_1}} \]
The time period for the second pivot point \( P_2 \) is:
\[ T_2 = 2\pi \sqrt{\frac{I_{P_2}}{mgd_2}} = 2\pi \sqrt{\frac{I_C + m (L - d_1)^2}{mg(L - d_1)}} \]

5. **Equating the Time Periods**:
According to the problem, the time periods \( T_1 \) and \( T_2 \) are equal:
\[ T_1 = T_2 \]

6. **Simplifying the Expression**:
Since both expressions are equal, we can set them equal to each other and simplify:
\[ \sqrt{\frac{I_C + m d_1^2}{g d_1}} = \sqrt{\frac{I_C + m (L - d_1)^2}{g (L - d_1)}} \]

7. **Finding the Time Period**:
After simplifying the above equation, we can derive the expression for the time period \( T \) in terms of the distance \( L \):
\[ T = 2\pi \sqrt{\frac{L}{g}} \]

### Final Answer:
The time period \( T \) of the physical pendulum is:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]