A simple pendulum is oscillating with an angular amplitude `60^@`. If mass of bob is `50` gram, then the tension in the string at mean position is `(g = 10 ms^(-2))`

Options

A. 0.5 N
B. 1 N
C. 1.5 N
D. 2 N

Explanation

To find the tension in the string of a simple pendulum at the mean position, we can follow these steps:

### Step 1: Understand the Problem
We have a simple pendulum with:
- Angular amplitude \( \theta = 60^\circ \)
- Mass of the bob \( m = 50 \, \text{grams} = 0.05 \, \text{kg} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

We need to find the tension in the string at the mean position.

### Step 2: Calculate the Height at Maximum Displacement
At the maximum displacement (angular amplitude), the height \( h \) of the bob above the lowest point can be calculated using:
\[ h = L - L \cos(\theta) \]
Where \( L \) is the length of the pendulum. The height can be simplified as:
\[ h = L(1 - \cos(60^\circ)) = L(1 - 0.5) = 0.5L \]

### Step 3: Calculate Potential Energy at Maximum Displacement
The potential energy \( PE \) at the maximum displacement is given by:
\[ PE = mgh = mg(0.5L) = 0.05 \times 10 \times 0.5L = 0.25mL \]

### Step 4: Calculate Kinetic Energy at Mean Position
At the mean position, all potential energy converts to kinetic energy \( KE \):
\[ KE = \frac{1}{2}mv^2 \]
Since \( PE = KE \):
\[ 0.25mL = \frac{1}{2}mv^2 \]
Cancelling \( m \) from both sides:
\[ 0.25L = \frac{1}{2}v^2 \]
Rearranging gives:
\[ v^2 = 0.5L \]

### Step 5: Calculate the Tension in the String
At the mean position, the tension \( T \) in the string is the sum of the gravitational force and the centripetal force:
\[ T = mg + \frac{mv^2}{L} \]
Substituting \( v^2 \):
\[ T = mg + \frac{m(0.5L)}{L} = mg + 0.5mg = 1.5mg \]

### Step 6: Substitute Values
Now substituting \( m = 0.05 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \):
\[ T = 1.5 \times 0.05 \times 10 = 0.75 \, \text{N} \]

### Step 7: Conclusion
The tension in the string at the mean position is:
\[ T = 0.75 \, \text{N} \]