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. (a)`m(g+2hpi^(2))`
B. (b)`m(g+sqrt(hpi^(2)gh))`
C. (c)`m(g+sqrt((pi^(2))/(2)gh))`
D. (d)`m(g+sqrt((pi^(2))/(3)gh))`

Explanation

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

### Step 1: Understand the Forces Acting on the Bob
When the bob is at the lowest point (equilibrium position 'O'), two forces act on it:
1. The gravitational force (weight) acting downwards, which is \( mg \).
2. The tension in the string \( T \) acting upwards.

### Step 2: Apply Newton's Second Law
At the lowest point, the net force acting on the bob is given by:
\[ T - mg = ma \]
Where \( a \) is the centripetal acceleration of the bob at point 'O'.

### Step 3: Determine the Velocity at Point 'O'
When the bob is released from height 'h', it converts potential energy into kinetic energy. The potential energy at height 'h' is given by:
\[ PE = mgh \]
At the lowest point 'O', this potential energy is converted into kinetic energy (KE):
\[ KE = \frac{1}{2} mv^2 \]
Setting the potential energy equal to the kinetic energy:
\[ mgh = \frac{1}{2} mv^2 \]
From this, we can solve for \( v^2 \):
\[ v^2 = 2gh \]

### Step 4: Substitute \( v^2 \) into the Force Equation
Now, substituting \( v^2 \) into the equation from Step 2:
\[ T - mg = \frac{mv^2}{L} \]
Where \( L \) is the length of the pendulum. Substituting \( v^2 \):
\[ T - mg = \frac{m(2gh)}{L} \]

### Step 5: Solve for Tension \( T \)
Rearranging the equation gives:
\[ T = mg + \frac{m(2gh)}{L} \]
This can be simplified to:
\[ T = mg + \frac{2mgh}{L} \]

### Step 6: Using the Time Period to Find \( L \)
Given the time period \( T \) of the pendulum is \( 2.0s \), we can use the formula for the time period of a simple pendulum:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
Squaring both sides:
\[ T^2 = 4\pi^2 \frac{L}{g} \]
From this, we can express \( L \):
\[ L = \frac{gT^2}{4\pi^2} \]

### Step 7: Substitute \( L \) Back into the Tension Equation
Substituting \( L \) into the tension equation:
\[ T = mg + \frac{2mgh}{\left(\frac{gT^2}{4\pi^2}\right)} \]
This simplifies to:
\[ T = mg + \frac{8\pi^2 mh}{T^2} \]

### Step 8: Final Expression for Tension
Now, substituting \( T = 2.0s \):
\[ T = mg + \frac{8\pi^2 mh}{4} \]
\[ T = mg + 2\pi^2 mh \]

### Conclusion
The tension in the string when the bob passes through 'O' is:
\[ T = mg + 2\pi^2 mh \]