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 verticle plane. If the breaking strength of the string is 2 mg, then the maximum angular amplitude of the displacement from the verticle can be

Options

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

Answer

To solve the problem, we need to determine the maximum angular amplitude (θ) of a pendulum when the breaking strength of the string is given. Here’s a step-by-step solution:

### Step 1: Understand the forces acting on the pendulum
When the pendulum swings to an angle θ from the vertical, two main forces act on the mass m:
1. The gravitational force (weight) acting downwards: \( mg \)
2. The tension (T) in the string acting upwards and at an angle.

### Step 2: Resolve the forces
At the angle θ, the gravitational force can be resolved into two components:
- The component acting along the direction of the string: \( mg \cos \theta \)
- The component acting perpendicular to the string: \( mg \sin \theta \)

### Step 3: Write the equation for tension
The net force acting towards the center of the circular path of the pendulum is provided by the tension minus the component of the weight acting along the string. Therefore, we can write:
\[ T - mg \cos \theta = \frac{mv^2}{L} \]
Where:
- \( T \) is the tension in the string,
- \( v \) is the speed of the mass at angle θ,
- \( L \) is the length of the string (which is given as 1 meter).

### Step 4: Use the breaking strength of the string
The breaking strength of the string is given as \( 2mg \). At the maximum angle, the tension in the string reaches this breaking point:
\[ T = 2mg \]

### Step 5: Substitute the tension into the equation
Substituting \( T = 2mg \) into the tension equation gives:
\[ 2mg - mg \cos \theta = \frac{mv^2}{L} \]

### Step 6: Simplify the equation
Since \( L = 1 \), the equation simplifies to:
\[ 2mg - mg \cos \theta = mv^2 \]
Dividing through by \( m \) (assuming \( m \neq 0 \)):
\[ 2g - g \cos \theta = v^2 \]

### Step 7: Apply conservation of energy
At the maximum height (angle θ), the potential energy is maximum and kinetic energy is zero. The height (h) can be expressed as:
\[ h = L(1 - \cos \theta) \]
The potential energy at this height is:
\[ PE = mgh = mgL(1 - \cos \theta) = mg(1 - \cos \theta) \]

At the lowest point, all this potential energy converts to kinetic energy:
\[ KE = \frac{1}{2} mv^2 \]

Setting the potential energy equal to the kinetic energy gives:
\[ mg(1 - \cos \theta) = \frac{1}{2} mv^2 \]

### Step 8: Substitute for v^2
From the previous step, we have \( v^2 = 2g(1 - \cos \theta) \). Substitute this into the tension equation:
\[ 2g - g \cos \theta = 2g(1 - \cos \theta) \]

### Step 9: Solve for cos θ
Rearranging gives:
\[ 2g - g \cos \theta = 2g - 2g \cos \theta \]
This simplifies to:
\[ g \cos \theta = 2g \cos \theta \]
Which leads to:
\[ g \cos \theta = g \]
Thus:
\[ \cos \theta = \frac{1}{2} \]

### Step 10: Find θ
Taking the inverse cosine gives:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

### Final Answer
The maximum angular amplitude of the displacement from the vertical is \( 60^\circ \).