A pendulum consisting of a small sphere of mass m, suspended by a inextensible and massless string of length 1 , is made to swing in a vertical plane. If the breaking strength of the string is 2 mg, then the maximum angular amplitude of the displacement from the vertical can be

Options

A. `0^@`
B. `30^@`
C. `60^@`
D. `90^@`

Answer

To find the maximum angular amplitude of a pendulum that can swing without breaking the string, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Forces**:
The pendulum consists of a mass \( m \) hanging from a string of length \( l \). When the pendulum swings, the forces acting on the mass are the gravitational force \( mg \) acting downward and the tension \( T \) in the string acting upward.

2. **Breaking Strength of the String**:
The breaking strength of the string is given as \( 2mg \). This means that the maximum tension \( T_{max} \) in the string can be \( 2mg \).

3. **Applying Energy Conservation**:
As the pendulum swings to an angle \( \theta \), it gains potential energy and loses kinetic energy. The height \( h \) gained when the pendulum is at angle \( \theta \) can be expressed as:
\[ h = l(1 - \cos \theta) \]
The loss in kinetic energy equals the gain in potential energy:
\[ \frac{1}{2} mv^2 = mgh \]
Substituting for \( h \):
\[ \frac{1}{2} mv^2 = mg \cdot l(1 - \cos \theta) \]
Simplifying gives:
\[ v^2 = 2g l(1 - \cos \theta) \]

4. **Finding the Tension at the Lowest Point**:
At the lowest point of the swing, the tension in the string can be expressed using Newton's second law:
\[ T - mg = \frac{mv^2}{l} \]
Rearranging gives:
\[ T = mg + \frac{mv^2}{l} \]
Substituting \( v^2 \) from the previous step:
\[ T = mg + \frac{m(2g l(1 - \cos \theta))}{l} \]
Simplifying, we find:
\[ T = mg + 2mg(1 - \cos \theta) \]
\[ T = mg(1 + 2(1 - \cos \theta)) = mg(3 - 2\cos \theta) \]

5. **Setting the Maximum Tension Equal to the Breaking Strength**:
We set the maximum tension equal to the breaking strength:
\[ 2mg = mg(3 - 2\cos \theta) \]
Dividing both sides by \( mg \) (assuming \( m \neq 0 \)):
\[ 2 = 3 - 2\cos \theta \]
Rearranging gives:
\[ 2\cos \theta = 1 \quad \Rightarrow \quad \cos \theta = \frac{1}{2} \]

6. **Finding the Angle**:
The angle \( \theta \) for which \( \cos \theta = \frac{1}{2} \) is:
\[ \theta = 60^\circ \]

### Final Answer:
Thus, the maximum angular amplitude of the displacement from the vertical is \( \theta = 60^\circ \).