A simple pendulum has a bob of mass `m` and swings with an angular amplitude `phi`. The tension in thread is `T`. At a certain time the string makes an angle `theta` with the vertical `(theta le phi)`

Options

A. T= mg cos `theta`, all values of `theta`
B. T= mg cos `theta`, only for `theta=phi`.
C. T =mg , for `theta= cos ^(-1) [1/3(2cos phi+1)]`
D. T will larger for smaller values of `theta`.

Answer

To solve the problem of finding the tension in the string of a simple pendulum at an angle θ with the vertical, we can follow these steps:

### Step 1: Identify the forces acting on the pendulum bob
At any position of the pendulum, two main forces act on the bob:
1. The weight of the bob (mg) acting vertically downward.
2. The tension (T) in the string acting along the string towards the pivot.

### Step 2: Break down the weight into components
When the pendulum makes an angle θ with the vertical, the weight can be resolved into two components:
- A component along the direction of the string: \( mg \cos(\theta) \)
- A component perpendicular to the string: \( mg \sin(\theta) \)

### Step 3: Apply Newton's second law
At an angle θ, the pendulum bob is not in free fall; it has a radial acceleration due to its circular motion. The net force acting along the direction of the string is given by:
\[ T - mg \cos(\theta) = \frac{mv^2}{L} \]
Where:
- \( T \) is the tension in the string.
- \( mg \cos(\theta) \) is the component of the weight acting along the string.
- \( \frac{mv^2}{L} \) is the centripetal force required for circular motion, with \( v \) being the velocity of the bob and \( L \) being the length of the string.

### Step 4: Find the relationship between potential energy and kinetic energy
As the pendulum swings from the maximum height (angle φ) to angle θ, potential energy is converted into kinetic energy. The change in height (h) can be expressed as:
\[ h = L \cos(\phi) - L \cos(\theta) \]
The change in potential energy (PE) is:
\[ \Delta PE = mg(L \cos(\phi) - L \cos(\theta)) = mgL(\cos(\phi) - \cos(\theta)) \]

This change in potential energy is equal to the kinetic energy (KE) gained:
\[ \Delta KE = \frac{1}{2} mv^2 \]

### Step 5: Set the potential energy equal to kinetic energy
Equating the change in potential energy to the change in kinetic energy gives:
\[ mgL(\cos(\phi) - \cos(\theta)) = \frac{1}{2} mv^2 \]

### Step 6: Solve for \( v^2 \)
From the equation above, we can solve for \( v^2 \):
\[ v^2 = 2gL(\cos(\phi) - \cos(\theta)) \]

### Step 7: Substitute \( v^2 \) back into the tension equation
Substituting \( v^2 \) into the tension equation gives:
\[ T - mg \cos(\theta) = \frac{m(2gL(\cos(\phi) - \cos(\theta)))}{L} \]
This simplifies to:
\[ T - mg \cos(\theta) = 2mg(\cos(\phi) - \cos(\theta)) \]

### Step 8: Solve for T
Rearranging the equation to isolate T:
\[ T = mg \cos(\theta) + 2mg(\cos(\phi) - \cos(\theta)) \]
\[ T = mg \cos(\theta) + 2mg \cos(\phi) - 2mg \cos(\theta) \]
\[ T = mg(2 \cos(\phi) - mg \cos(\theta)) \]

### Final Answer
Thus, the tension in the string at angle θ is given by:
\[ T = mg(2 \cos(\phi) - \cos(\theta)) \]