A simple pendulum is oscillating with a maximum angular displacement of `theta` radian. If `theta` is very small, the ratio of maximum tension to the minimum tension in the string during oscillations is

Options

A. `(1 + theta^(2)) : (1 - theta^(2))`
B. `(1 + theta^(2)) : (1 - (theta^(2))/(2))`
C. `(1 + theta^(2)) : 1`
D. `(1 + (theta^(2))/(2)) : (1 - theta^(2))`

Explanation

To find the ratio of maximum tension (T_max) to minimum tension (T_min) in a simple pendulum oscillating with a small angular displacement θ, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Tension in the Pendulum**:
- The tension in the string varies during the oscillation of the pendulum.
- The maximum tension (T_max) occurs when the pendulum is at the lowest point (mean position), while the minimum tension (T_min) occurs at the extreme position (maximum displacement).

2. **Calculating T_min**:
- At the extreme position, the forces acting on the pendulum bob are:
- Weight (mg) acting downwards.
- The component of weight acting along the direction of the string is mg cos(θ).
- Therefore, the minimum tension can be expressed as:
\[ T_{min} = mg \cos(\theta) \]

3. **Calculating T_max**:
- At the mean position, the forces acting on the pendulum bob are:
- Weight (mg) acting downwards.
- The centripetal force required for circular motion, which is provided by the tension in the string.
- Thus, the maximum tension can be expressed as:
\[ T_{max} = mg + m a_c \]
- Where \( a_c \) is the centripetal acceleration. For a pendulum, \( a_c = \frac{v^2}{L} \), where \( L \) is the length of the pendulum.

4. **Finding the Velocity (v)**:
- Using conservation of energy, the potential energy lost when the pendulum moves from the height to the lowest point is converted into kinetic energy.
- The height \( h \) when the pendulum is at an angle θ is given by:
\[ h = L(1 - \cos(\theta)) \]
- The loss in potential energy is:
\[ \Delta PE = mg h = mg L(1 - \cos(\theta)) \]
- The kinetic energy at the lowest point is:
\[ KE = \frac{1}{2} mv^2 \]
- Setting these equal gives:
\[ mg L(1 - \cos(\theta)) = \frac{1}{2} mv^2 \]
- Solving for \( v^2 \):
\[ v^2 = 2gL(1 - \cos(\theta)) \]

5. **Substituting for T_max**:
- Substitute \( v^2 \) back into the equation for \( T_{max} \):
\[ T_{max} = mg + m \left(\frac{v^2}{L}\right) = mg + m \left(\frac{2gL(1 - \cos(\theta))}{L}\right) \]
- This simplifies to:
\[ T_{max} = mg + 2mg(1 - \cos(\theta)) = mg(1 + 2(1 - \cos(\theta))) \]

6. **Finding the Ratio \( \frac{T_{max}}{T_{min}} \)**:
- Now, we can find the ratio:
\[ \frac{T_{max}}{T_{min}} = \frac{mg(1 + 2(1 - \cos(\theta)))}{mg \cos(\theta)} = \frac{1 + 2(1 - \cos(\theta))}{\cos(\theta)} \]
- For small angles, \( \cos(\theta) \approx 1 - \frac{\theta^2}{2} \), hence:
\[ 1 - \cos(\theta) \approx \frac{\theta^2}{2} \]
- Substituting this in gives:
\[ \frac{T_{max}}{T_{min}} \approx \frac{1 + 2 \left(\frac{\theta^2}{2}\right)}{1 - \frac{\theta^2}{2}} = \frac{1 + \theta^2}{1 - \frac{\theta^2}{2}} \]

7. **Final Simplification**:
- For very small θ, this can be approximated further to:
\[ \frac{T_{max}}{T_{min}} \approx \frac{1 + \theta^2}{1} = 1 + \theta^2 \]

### Final Answer:
The ratio of maximum tension to minimum tension in the string during oscillations is approximately \( 1 + \theta^2 \).