A simple pendulm has a length L and a bob of mass M. The bob is vibrating with amplitude a .What is the maximum tension in the string?

Options

A. `(mg)/(2)`
B. `(mg)/(2) [1+ 2 ((A)/(L))^(2)]`
C. `mg [1+ ((A)/(L))^(2)]`
D. `(mg)/(2) [1 + ((A)/(L))^(2)]`

Explanation

To find the maximum tension in the string of a simple pendulum with a length \( L \) and a bob of mass \( M \) vibrating with amplitude \( A \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Bob
At any point in the pendulum's swing, two main forces act on the bob:
- The gravitational force \( mg \) acting downward.
- The tension \( T \) in the string acting along the string towards the pivot.

### Step 2: Identify the Components of the Gravitational Force
When the bob is at an angle \( \theta \) from the vertical, the gravitational force can be resolved into two components:
- \( mg \cos \theta \) (along the string)
- \( mg \sin \theta \) (perpendicular to the string)

### Step 3: Apply Newton's Second Law
Using Newton's second law in the radial direction (along the string), we have:
\[ T - mg \cos \theta = \frac{mv^2}{L} \]
where \( v \) is the speed of the bob at that position.

### Step 4: Determine the Maximum Tension
The maximum tension will occur when the bob is at the lowest point of its swing (mean position), where \( \theta = 0 \). Thus, we can write:
\[ T_{\text{max}} = mg + \frac{mv^2}{L} \]

### Step 5: Calculate the Velocity at the Lowest Point
To find \( v^2 \), we can use the conservation of energy. At the highest point (amplitude \( A \)), the potential energy is maximum and kinetic energy is zero. At the lowest point, potential energy is minimum and kinetic energy is maximum.

The change in height \( h \) when the bob swings from the highest point to the lowest point is:
\[ h = L - L \cos \theta_0 = L(1 - \cos \theta_0) \]
Using the small angle approximation, \( \cos \theta \approx 1 - \frac{\theta^2}{2} \), we can express \( h \) in terms of amplitude \( A \):
\[ h \approx L \left(1 - \left(1 - \frac{A^2}{2L^2}\right)\right) = \frac{A^2}{2L} \]

Using conservation of energy:
\[ mgh = \frac{1}{2} mv^2 \]
Substituting for \( h \):
\[ mg \cdot \frac{A^2}{2L} = \frac{1}{2} mv^2 \]
This simplifies to:
\[ v^2 = gA \]

### Step 6: Substitute \( v^2 \) Back into the Tension Equation
Now, substituting \( v^2 \) back into the tension equation:
\[ T_{\text{max}} = mg + \frac{m(gA)}{L} \]
This simplifies to:
\[ T_{\text{max}} = mg + \frac{mgA}{L} = mg \left(1 + \frac{A}{L}\right) \]

### Final Expression
Thus, the maximum tension in the string is given by:
\[ T_{\text{max}} = mg \left(1 + \frac{A}{L}\right) \]