The maximum value attained by the tension in the string of a swinging pendulum is four times the minimum value it attains. There is no slack in the string. The angular amplitude of the pendulum is

Options

A. `90^(@)`
B. `60^(@)`
C. `45^(@)`
D. `30^(@)`

Explanation

To solve the problem, we need to determine the angular amplitude of a swinging pendulum, given that the maximum tension in the string is four times the minimum tension.

### Step-by-Step Solution:

1. **Understanding the Tension in the Pendulum**:
- The maximum tension \( T_1 \) occurs at the lowest point of the swing.
- The minimum tension \( T_2 \) occurs at the highest point of the swing.

2. **Setting Up the Relationship**:
- According to the problem, we have:
\[ T_1 = 4T_2 \]

3. **Analyzing Forces at the Minimum Tension Point**:
- At the highest point of the swing (where tension is minimum), the forces acting on the pendulum bob are:
- Weight \( mg \) acting downward.
- Tension \( T_2 \) acting along the string.
- The centripetal force required for circular motion at this point is provided by the component of the weight acting along the string:
\[ T_2 = mg \cos(\theta) \]

4. **Analyzing Forces at the Maximum Tension Point**:
- At the lowest point of the swing (where tension is maximum), the forces acting are:
- Weight \( mg \) acting downward.
- Tension \( T_1 \) acting upward.
- The net force provides the centripetal acceleration:
\[ T_1 - mg = \frac{mv^2}{r} \]
- Rearranging gives:
\[ T_1 = mg + \frac{mv^2}{r} \]

5. **Using the Relationship Between Tensions**:
- Substitute \( T_1 \) and \( T_2 \) into the equation \( T_1 = 4T_2 \):
\[ mg + \frac{mv^2}{r} = 4(mg \cos(\theta)) \]

6. **Using Conservation of Energy**:
- The potential energy at the highest point is converted into kinetic energy at the lowest point:
\[ mgh = \frac{1}{2} mv^2 \]
- The height \( h \) can be expressed as:
\[ h = r - r \cos(\theta) = r(1 - \cos(\theta)) \]
- Thus, we have:
\[ mg(r(1 - \cos(\theta))) = \frac{1}{2} mv^2 \]
- Simplifying gives:
\[ v^2 = 2gr(1 - \cos(\theta)) \]

7. **Substituting \( v^2 \) into the Tension Equation**:
- Substitute \( v^2 \) back into the tension equation:
\[ mg + \frac{m(2gr(1 - \cos(\theta)))}{r} = 4mg \cos(\theta) \]
- Simplifying gives:
\[ mg + 2mg(1 - \cos(\theta)) = 4mg \cos(\theta) \]
- Rearranging leads to:
\[ mg(1 + 2 - 2\cos(\theta)) = 4mg \cos(\theta) \]
- This simplifies to:
\[ 3 = 6\cos(\theta) \]
- Therefore:
\[ \cos(\theta) = \frac{1}{2} \]

8. **Finding the Angle**:
- The angle \( \theta \) corresponding to \( \cos(\theta) = \frac{1}{2} \) is:
\[ \theta = 60^\circ \]

### Final Answer:
The angular amplitude of the pendulum is \( \theta = 60^\circ \).