A pendulum is displaced to an angle `theta` from its equilibrium position, then it will pass through its mean position with a velocity v equal to

Options

A. `sqrt(2g l)`
B. `sqrt(2g l sin theta)`
C. `sqrt(2 g l cos theta)`
D. `sqrt(2gl(1-cos theta))`

Answer

To solve the problem of finding the velocity \( v \) of a pendulum as it passes through its mean position after being displaced to an angle \( \theta \), we can use the principle of conservation of mechanical energy. Here’s a step-by-step solution:

### Step 1: Understand the System
- A pendulum is displaced to an angle \( \theta \) from its equilibrium position (mean position).
- The pendulum swings back to the mean position where we want to find the velocity \( v \).

### Step 2: Identify the Points
- Let point \( P \) be the maximum displacement (at angle \( \theta \)).
- Let point \( B \) be the mean position (equilibrium position).

### Step 3: Calculate the Height \( h \)
- The height \( h \) of the pendulum at point \( P \) can be calculated using the length \( L \) of the pendulum and the angle \( \theta \):
\[ h = L - L \cos(\theta) = L(1 - \cos(\theta)) \]

### Step 4: Apply Conservation of Energy
- At point \( P \) (maximum displacement):
- Potential Energy (PE) = \( mgh \)
- Kinetic Energy (KE) = 0 (since it momentarily stops)

- At point \( B \) (mean position):
- Potential Energy (PE) = 0 (minimum)
- Kinetic Energy (KE) = \( \frac{1}{2} mv^2 \)

Using conservation of energy:
\[ \text{PE at P} = \text{KE at B} \]
\[ mgh = \frac{1}{2} mv^2 \]

### Step 5: Simplify the Equation
- The mass \( m \) cancels out from both sides:
\[ gh = \frac{1}{2} v^2 \]
- Rearranging gives:
\[ v^2 = 2gh \]

### Step 6: Substitute for \( h \)
- Substitute \( h = L(1 - \cos(\theta)) \) into the equation:
\[ v^2 = 2gL(1 - \cos(\theta)) \]
- Taking the square root to find \( v \):
\[ v = \sqrt{2gL(1 - \cos(\theta))} \]

### Final Result
- Thus, the velocity \( v \) of the pendulum as it passes through the mean position is:
\[ v = \sqrt{2gL(1 - \cos(\theta))} \]