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+pisqrt(2gh))`
B. `m(g+sqrt(pi^(2)gh))`
C. `m(g+sqrt((pi^(2))/2)gh)`
D. `m(g+sqrt((pi^(2))/3)gh)`

Answer

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

### Step 1: Understand the Forces Acting on the Bob
At the equilibrium position 'O', two forces act on the bob:
1. The gravitational force (weight) acting downward: \( F_g = mg \)
2. The tension in the string acting upward: \( T \)

### Step 2: Apply Newton's Second Law
When the bob passes through point 'O', it is moving with a certain velocity. According to Newton's second law, the net force acting on the bob is equal to the mass of the bob times its acceleration. The centripetal acceleration is provided by the tension in the string minus the weight of the bob.

The equation can be written as:
\[ T - mg = \frac{mv^2}{r} \]
Where:
- \( T \) is the tension in the string
- \( mg \) is the weight of the bob
- \( \frac{mv^2}{r} \) is the centripetal force required to keep the bob moving in a circular path

### Step 3: Solve for Tension
Rearranging the equation gives:
\[ T = mg + \frac{mv^2}{r} \]

### Step 4: Find the Velocity \( v \)
To find \( v \), we can use the conservation of energy principle. When the bob is at the height \( h \) (position 'Q'), all the energy is potential energy, and when it passes through 'O', all the potential energy converts into kinetic energy.

The potential energy at height \( h \) is:
\[ PE = mgh \]
The kinetic energy at point 'O' is:
\[ KE = \frac{1}{2} mv^2 \]

Setting the potential energy equal to the kinetic energy gives:
\[ mgh = \frac{1}{2} mv^2 \]
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{1}{2} v^2 \implies v^2 = 2gh \]

### Step 5: Find the Angular Frequency \( \omega \)
The angular frequency \( \omega \) can be calculated using the time period \( T \):
\[ \omega = \frac{2\pi}{T} \]
Given that the time period \( T = 2.0 \, s \):
\[ \omega = \frac{2\pi}{2} = \pi \, \text{rad/s} \]

### Step 6: Substitute \( v^2 \) into the Tension Equation
Now substituting \( v^2 = 2gh \) into the tension equation:
\[ T = mg + \frac{m(2gh)}{r} \]

### Step 7: Final Expression for Tension
Substituting \( v^2 \) into the tension equation gives:
\[ T = mg + \frac{m(2gh)}{r} \]
This can be simplified to:
\[ T = mg + 2mg = 3mg \]

### Conclusion
Thus, the tension in the string when the bob passes through point 'O' is:
\[ T = 3mg \]