The maximum tension in the string of an oscillating simple pendulum is 3% more than the minimum tension in the string. The angular amplitude of oscillations of the pendulum is

Options

A. `cos^(-1) ((100)/(101))`
B. `cos^(-1) ((100)/(103))`
C. `cos^(-1) ((2)/(3))`
D. `cos^(-1) ((97)/(100))`

Explanation

To solve the problem, we need to find the angular amplitude of the oscillating simple pendulum given that the maximum tension in the string is 3% more than the minimum tension. Let's go through the steps systematically.

### Step-by-Step Solution:

1. **Understanding Tension in the Pendulum**:
- Let \( T_{\text{max}} \) be the maximum tension in the string and \( T_{\text{min}} \) be the minimum tension.
- According to the problem, \( T_{\text{max}} = T_{\text{min}} + 0.03 T_{\text{min}} = 1.03 T_{\text{min}} \).

2. **Setting Up the Relationship**:
- We can express this relationship as:
\[ \frac{T_{\text{max}}}{T_{\text{min}}} = \frac{1.03 T_{\text{min}}}{T_{\text{min}}} = 1.03 \]

3. **Analyzing Forces at Maximum Displacement**:
- At maximum angular displacement \( \theta \), the forces acting on the pendulum bob are:
- The gravitational force \( mg \) acting downwards.
- The tension \( T_{\text{min}} \) acting along the string.
- The minimum tension occurs when the pendulum is at the maximum height, where the centripetal force is zero. Thus:
\[ T_{\text{min}} = mg \cos \theta \]

4. **Analyzing Forces at the Lowest Point**:
- At the lowest point (mean position), the forces are:
- The gravitational force \( mg \) acting downwards.
- The maximum tension \( T_{\text{max}} \) acting upwards.
- The equation for maximum tension can be expressed as:
\[ T_{\text{max}} = mg + \frac{mv^2}{L} \]
- Here, \( v \) is the speed of the pendulum at the lowest point, and \( L \) is the length of the pendulum.

5. **Using Energy Conservation**:
- The potential energy at the maximum height is converted to kinetic energy at the lowest point:
\[ mgh = \frac{1}{2} mv^2 \]
- The height \( h \) can be expressed in terms of \( L \) and \( \theta \):
\[ h = L(1 - \cos \theta) \]
- Thus, we have:
\[ mgL(1 - \cos \theta) = \frac{1}{2} mv^2 \]
- From this, we can find \( v^2 \):
\[ v^2 = 2gL(1 - \cos \theta) \]

6. **Substituting into Tension Equation**:
- Substitute \( v^2 \) into the equation for \( T_{\text{max}} \):
\[ T_{\text{max}} = mg + m \frac{2gL(1 - \cos \theta)}{L} = mg + 2mg(1 - \cos \theta) = mg(3 - 2\cos \theta) \]

7. **Setting Up the Ratio of Tensions**:
- Now we can set up the ratio of tensions:
\[ \frac{T_{\text{max}}}{T_{\text{min}}} = \frac{mg(3 - 2\cos \theta)}{mg \cos \theta} = \frac{3 - 2\cos \theta}{\cos \theta} \]
- Setting this equal to 1.03 gives:
\[ \frac{3 - 2\cos \theta}{\cos \theta} = 1.03 \]

8. **Solving for \( \cos \theta \)**:
- Rearranging gives:
\[ 3 - 2\cos \theta = 1.03 \cos \theta \]
\[ 3 = 1.03 \cos \theta + 2\cos \theta \]
\[ 3 = (1.03 + 2)\cos \theta \]
\[ 3 = 3.03 \cos \theta \]
\[ \cos \theta = \frac{3}{3.03} = \frac{300}{303} \]

9. **Finding the Angular Amplitude**:
- Finally, we find the angular amplitude \( \theta \):
\[ \theta = \cos^{-1}\left(\frac{300}{303}\right) \]

### Final Answer:
The angular amplitude of the oscillations of the pendulum is:
\[ \theta = \cos^{-1}\left(\frac{300}{303}\right) \]