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 find the time period \( T \) of a physical pendulum when the pivot points are separated by a distance \( L \), we can follow these steps:

### Step 1: Understand the Time Period of a Physical Pendulum
The time period \( T \) of a physical pendulum about a pivot point is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{mgd}} \]
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,
- \( d \) is the distance from the pivot point to the center of mass.

### Step 2: Define the Two Pivot Points
Let’s denote the first pivot point as \( O \) and the second pivot point as \( O' \). The distance between these two points is \( L \). The center of mass (CM) of the pendulum lies on the line joining these two pivot points.

### Step 3: Calculate the Moment of Inertia about Each Pivot Point
1. For the first pivot point \( O \):
- Let the distance from the center of mass to point \( O \) be \( x \).
- The moment of inertia about point \( O \) is given by:
\[ I_O = I_{CM} + mx^2 \]
where \( I_{CM} \) is the moment of inertia about the center of mass.

2. For the second pivot point \( O' \):
- The distance from the center of mass to point \( O' \) is \( L - x \).
- The moment of inertia about point \( O' \) is given by:
\[ I_{O'} = I_{CM} + m(L - x)^2 \]

### Step 4: Set Up the Time Period Equations
Using the time period formula for both pivot points:
1. For pivot point \( O \):
\[ T = 2\pi \sqrt{\frac{I_O}{mgx}} = 2\pi \sqrt{\frac{I_{CM} + mx^2}{mgx}} \]

2. For pivot point \( O' \):
\[ T' = 2\pi \sqrt{\frac{I_{O'}}{mg(L - x)}} = 2\pi \sqrt{\frac{I_{CM} + m(L - x)^2}{mg(L - x)}} \]

### Step 5: Equate the Time Periods
Since it is given that \( T = T' \):
\[ \sqrt{\frac{I_{CM} + mx^2}{mgx}} = \sqrt{\frac{I_{CM} + m(L - x)^2}{mg(L - x)}} \]

### Step 6: Square Both Sides and Simplify
Squaring both sides gives:
\[ \frac{I_{CM} + mx^2}{mgx} = \frac{I_{CM} + m(L - x)^2}{mg(L - x)} \]
Cross-multiplying and simplifying leads to:
\[ (I_{CM} + mx^2)(L - x) = (I_{CM} + m(L - x)^2)x \]

### Step 7: Solve for \( I_{CM} \)
Rearranging terms and solving for \( I_{CM} \) leads to:
\[ I_{CM} = mx(L - x) \]

### Step 8: Substitute \( I_{CM} \) Back into the Time Period Formula
Substituting \( I_{CM} \) back into the time period equation:
\[ T = 2\pi \sqrt{\frac{mx(L - x) + mx^2}{mgx}} = 2\pi \sqrt{\frac{mLx}{mgx}} = 2\pi \sqrt{\frac{L}{g}} \]

### Final Result
Thus, the time period \( T \) is:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]