A simple pendulum of mass 'm' , swings with maximum angular displacement of `60^(@)` . When its angular displacement is `30^(@)` ,the tension in the string is

Options

A. Less than mg `cos 30^(@)`
B. Equal to `mg cos 30^(@)`
C. Greater than `mg cos (30^(@)`
D. Zero.

Answer

To find the tension in the string of a simple pendulum when it is at an angular displacement of \(30^\circ\), we can follow these steps:

### Step 1: Identify the forces acting on the pendulum bob
At an angle of \(30^\circ\), the forces acting on the bob of the pendulum are:
- The gravitational force (\(mg\)) acting downwards.
- The tension (\(T\)) in the string acting along the string towards the pivot.

### Step 2: Resolve the gravitational force
The gravitational force can be resolved into two components:
1. A component along the direction of the tension: \(mg \cos(30^\circ)\)
2. A component perpendicular to the tension: \(mg \sin(30^\circ)\)

### Step 3: Apply Newton's second law
In the radial direction (along the string), we can apply Newton's second law. The net force acting towards the center of the circular path is the difference between the tension and the component of the gravitational force acting along the string:
\[ T - mg \cos(30^\circ) = \frac{mv^2}{L} \]
where \(L\) is the length of the pendulum and \(v\) is the linear velocity of the bob at \(30^\circ\).

### Step 4: Find the velocity \(v\) at \(30^\circ\)
To find the velocity \(v\), we can use conservation of energy. The potential energy at the maximum height (when the pendulum is at \(60^\circ\)) is converted into kinetic energy and potential energy at \(30^\circ\):
\[ mgh_{60} = \frac{1}{2}mv^2 + mgh_{30} \]
The height \(h\) can be calculated using:
\[ h = L(1 - \cos(\theta)) \]
Thus,
\[ h_{60} = L(1 - \cos(60^\circ)) = L(1 - 0.5) = 0.5L \]
\[ h_{30} = L(1 - \cos(30^\circ)) = L(1 - \frac{\sqrt{3}}{2}) = L(\frac{2 - \sqrt{3}}{2}) \]
Substituting these heights into the energy equation:
\[ mg(0.5L) = \frac{1}{2}mv^2 + mg\left(L\frac{2 - \sqrt{3}}{2}\right) \]
Simplifying gives:
\[ 0.5mgL = \frac{1}{2}mv^2 + mg\left(\frac{2 - \sqrt{3}}{2}\right) \]

### Step 5: Solve for \(v^2\)
Rearranging the equation:
\[ \frac{1}{2}mv^2 = mg\left(0.5 - \frac{2 - \sqrt{3}}{2}\right) \]
\[ \frac{1}{2}mv^2 = mg\left(\frac{\sqrt{3} - 1}{2}\right) \]
Thus,
\[ v^2 = g(\sqrt{3} - 1) \]

### Step 6: Substitute \(v^2\) back into the tension equation
Now substituting \(v^2\) back into the tension equation:
\[ T = mg \cos(30^\circ) + \frac{m(\sqrt{3} - 1)g}{L} \]

### Step 7: Final expression for tension
The final expression for the tension in the string at \(30^\circ\) is:
\[ T = mg \cos(30^\circ) + mg \frac{(\sqrt{3} - 1)}{L} \]

### Conclusion
The tension in the string when the pendulum is at an angle of \(30^\circ\) is greater than \(mg \cos(30^\circ)\).