The bob of a simple pendulum is displaced from its equilibrium position 'O' to a position 'Q' which is at a height 'h' above 'O' and the bob is then released. Assuming the mass of the bob to be 'm' and time period of oscillation to be `2.0s`, the tension in the string when the bob passes through 'O' is

Options

A. `m(g +2pi^(2)h)`
B. `m(g +pi^(2) h)`
C. `m(g+(pi^(2))/(2)h)`
D. `m(g+(pi^(2))/(3)h)`

Explanation

To find the tension in the string when the bob of the simple pendulum passes through the equilibrium position 'O', we can follow these steps:

### Step 1: Understand the System
The bob is displaced to a height 'h' above the equilibrium position 'O' and then released. As it swings down, it will convert its potential energy at height 'h' into kinetic energy at the lowest point 'O'.

### Step 2: Calculate the Velocity at Point 'O'
Using the conservation of energy, we can equate the potential energy at height 'h' to the kinetic energy at the lowest point 'O':
\[ PE = KE \]
\[ mgh = \frac{1}{2} mv^2 \]
From this, we can solve for \(v^2\):
\[ v^2 = 2gh \]

### Step 3: Apply Newton's Second Law at Point 'O'
At the lowest point 'O', the forces acting on the bob are:
1. The gravitational force downward: \(mg\)
2. The tension in the string upward: \(T\)

Using Newton's second law, we can write:
\[ T - mg = \frac{mv^2}{L} \]
Where \(L\) is the length of the pendulum (which is the radius of circular motion).

### Step 4: Substitute the Expression for Velocity
Substituting \(v^2\) from Step 2 into the equation:
\[ T - mg = \frac{m(2gh)}{L} \]
Rearranging gives:
\[ T = mg + \frac{2mgh}{L} \]

### Step 5: Relate Length \(L\) to Time Period
The time period \(T\) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
Given that the time period is \(2.0\) seconds, we can set up the equation:
\[ 2 = 2\pi \sqrt{\frac{L}{g}} \]
Squaring both sides:
\[ 1 = \pi^2 \frac{L}{g} \implies L = \frac{g}{\pi^2} \]

### Step 6: Substitute \(L\) Back into the Tension Equation
Now substitute \(L\) into the tension equation:
\[ T = mg + \frac{2mgh}{\frac{g}{\pi^2}} = mg + \frac{2mgh \pi^2}{g} \]
This simplifies to:
\[ T = mg + \frac{2mgh \pi^2}{g} \]

### Final Expression for Tension
Thus, the final expression for the tension in the string when the bob passes through 'O' is:
\[ T = mg + 2mh\frac{\pi^2}{g} \]