A simple pendulum of length `l` and bob of mass `m` is displaced from its equilibrium position `O` to a position `P` so that hight of `P` above `O` is `h` . If is then released. What is the tension in the string when the bob passes through the through the equilibrium position `O` ? Neglect friction. `v` is the velocity of the bob at `O`

Options

A. `m(g+(v^(2))/(l))`
B. `(2mgh)/(l)`
C. `mg(1+(h)/(l))`
D. `mg(1+(2h)/(l))`

Explanation

To solve the problem of finding the tension in the string when the bob of a pendulum passes through the equilibrium position \( O \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Bob
When the bob is at the equilibrium position \( O \), two main forces act on it:
1. The gravitational force \( mg \) acting downwards.
2. The tension \( T \) in the string acting upwards.

### Step 2: Apply the Concept of Circular Motion
At the equilibrium position, the bob is in circular motion. The net force acting on the bob provides the necessary centripetal force for this motion. The equation for centripetal force is given by:
\[ F_{\text{net}} = T - mg = \frac{mv^2}{l} \]
where \( v \) is the velocity of the bob at the equilibrium position and \( l \) is the length of the pendulum.

### Step 3: Determine the Velocity \( v \) at Position \( O \)
To find the velocity \( v \) of the bob at the equilibrium position, we can use the conservation of energy principle. The potential energy at height \( h \) is converted into kinetic energy as the bob descends to position \( O \).

The potential energy at height \( h \) is given by:
\[ PE = mgh \]
The kinetic energy at the equilibrium position \( O \) is given by:
\[ KE = \frac{1}{2} mv^2 \]
Setting the potential energy equal to the kinetic energy:
\[ mgh = \frac{1}{2} mv^2 \]
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{1}{2} v^2 \]
Rearranging gives:
\[ v^2 = 2gh \]

### Step 4: Substitute \( v^2 \) into the Force Equation
Now we can substitute \( v^2 \) back into the centripetal force equation:
\[ T - mg = \frac{m(2gh)}{l} \]
This simplifies to:
\[ T = mg + \frac{2mgh}{l} \]

### Step 5: Factor Out Common Terms
We can factor out \( mg \) from the equation:
\[ T = mg \left(1 + \frac{2h}{l}\right) \]

### Final Result
Thus, the tension in the string when the bob passes through the equilibrium position \( O \) is:
\[ T = mg \left(1 + \frac{2h}{l}\right) \]