The angular amplitude of a simple pendulum is `theta_(0)`. The maximum tension in its string will be

Options

A. `mg(1-theta_(0))`
B. `mg(1+theta_(0))`
C. `mg(1-theta_(0)^(2))`
D. `mg(1+theta_(0)^(2))`

Answer

To find the maximum tension in the string of a simple pendulum with an angular amplitude of \( \theta_0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Forces**:
- When the pendulum swings to its maximum displacement (at angle \( \theta_0 \)), the forces acting on the mass \( m \) are the tension \( T \) in the string acting upwards and the weight \( mg \) acting downwards.

2. **Applying Newton's Second Law**:
- At the maximum displacement, the net force acting on the mass is given by:
\[ T - mg = m a \]
- Here, \( a \) is the centripetal acceleration at the maximum displacement.

3. **Finding Centripetal Acceleration**:
- The centripetal acceleration \( a \) can be expressed in terms of angular frequency \( \omega \) and the radius \( l \) (length of the pendulum):
\[ a = \omega^2 l \]

4. **Relating Angular Frequency to Angular Amplitude**:
- For a simple pendulum, the angular frequency \( \omega \) can be related to the gravitational acceleration \( g \) and the length \( l \) as:
\[ \omega = \sqrt{\frac{g}{l}} \sin(\theta_0) \]
- At maximum displacement, \( \sin(\theta_0) \approx \theta_0 \) (for small angles), so:
\[ \omega \approx \sqrt{\frac{g}{l}} \theta_0 \]

5. **Substituting for Acceleration**:
- Substitute \( \omega \) back into the equation for \( a \):
\[ a = \left(\sqrt{\frac{g}{l}} \theta_0\right)^2 l = \frac{g}{l} \theta_0^2 l = g \theta_0^2 \]

6. **Substituting Back into the Force Equation**:
- Now substitute \( a \) back into the force equation:
\[ T - mg = mg \theta_0^2 \]
- Rearranging gives:
\[ T = mg + mg \theta_0^2 \]
- Factoring out \( mg \):
\[ T = mg(1 + \theta_0^2) \]

### Final Answer:
The maximum tension in the string of the pendulum is:
\[ T_{\text{max}} = mg(1 + \theta_0^2) \]