A ball suspended by a thread swing in a vertical plane that its acceleration values in the lowest possition and the extreme postition are equal . Find the thread deffection angle in the extreme possition.

Options

A. `2 tan^(-1)``(1)/(2)`
B. `tan^(-1)``(1)/(2)`
C. `tan^(-1)sqrt(2)`
D. `tan^(-1)2`

Explanation

To solve the problem of finding the deflection angle \( \theta \) in the extreme position of a ball suspended by a thread swinging in a vertical plane, we can follow these steps:

### Step 1: Understand the System
The ball is suspended by a thread of length \( l \) and swings in a vertical plane. At the lowest point of the swing, the ball has maximum speed, and at the extreme position, it has maximum height and minimum speed.

### Step 2: Identify the Forces and Accelerations
At the lowest point, the centripetal acceleration \( a_c \) is given by:
\[ a_c = \frac{v^2}{l} \]
where \( v \) is the speed of the ball at the lowest point.

At the extreme position, the tangential acceleration \( a_t \) is given by:
\[ a_t = g \sin \theta \]
where \( g \) is the acceleration due to gravity and \( \theta \) is the angle of deflection from the vertical.

### Step 3: Set Up the Equation
According to the problem, the acceleration values at the lowest position and the extreme position are equal:
\[ \frac{v^2}{l} = g \sin \theta \]

### Step 4: Use Conservation of Energy
The potential energy at the extreme position is converted into kinetic energy at the lowest point. The height \( h \) of the ball at the extreme position can be expressed as:
\[ h = l - l \cos \theta = l(1 - \cos \theta) \]
The potential energy at the extreme position is:
\[ PE = mgh = mg(l(1 - \cos \theta)) \]
The kinetic energy at the lowest point is:
\[ KE = \frac{1}{2} mv^2 \]
By conservation of energy:
\[ mg(l(1 - \cos \theta)) = \frac{1}{2} mv^2 \]

### Step 5: Simplify the Energy Equation
We can cancel \( m \) from both sides:
\[ g(l(1 - \cos \theta)) = \frac{1}{2} v^2 \]
From the previous centripetal acceleration equation, we have \( v^2 = g l \sin \theta \). Substituting this into the energy equation gives:
\[ g(l(1 - \cos \theta)) = \frac{1}{2} g l \sin \theta \]
Dividing both sides by \( g l \) (assuming \( g \) and \( l \) are not zero):
\[ 1 - \cos \theta = \frac{1}{2} \sin \theta \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ 2(1 - \cos \theta) = \sin \theta \]
or
\[ 2 - 2 \cos \theta = \sin \theta \]

### Step 7: Use Trigonometric Identity
Using the identity \( \sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \) and \( 1 - \cos \theta = 2 \sin^2 \frac{\theta}{2} \):
\[ 2 - 4 \sin^2 \frac{\theta}{2} = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \]
Dividing by 2:
\[ 1 - 2 \sin^2 \frac{\theta}{2} = \sin \frac{\theta}{2} \cos \frac{\theta}{2} \]

### Step 8: Solve for \( \theta \)
This leads us to:
\[ 1 - 2 \sin^2 \frac{\theta}{2} = \frac{1}{2} \sin \theta \]
After some algebra, we find:
\[ \frac{1 - \cos \theta}{\sin \theta} = \frac{1}{2} \]
This simplifies to:
\[ \theta = 2 \tan^{-1} \left(\frac{1}{2}\right) \]

### Final Answer
Thus, the deflection angle \( \theta \) in the extreme position is:
\[ \theta = 2 \tan^{-1} \left(\frac{1}{2}\right) \]