The potential energy of a particle varies with position X according to the relation`U(x) = [(X^3/3)-(3X^2/2)-2X]` then

Options

A. The point x = 1 is point of stable equilibrium
B. The point x = 2 is point of stable equilibrium
C. Both x = 1 and x = 2 are points of unstable equilibrium
D. The point x = 1 is point of neutral equilibrium

Answer

To solve the problem, we need to analyze the potential energy function given by:

\[ U(x) = \frac{x^3}{3} - \frac{3x^2}{2} - 2x \]

We will find the equilibrium points and determine their stability.

### Step 1: Find the force acting on the particle
The force \( F \) acting on the particle is given by the negative gradient of the potential energy:

\[ F = -\frac{dU}{dx} \]

### Step 2: Differentiate the potential energy function
We need to differentiate \( U(x) \) with respect to \( x \):

\[ \frac{dU}{dx} = \frac{d}{dx} \left( \frac{x^3}{3} - \frac{3x^2}{2} - 2x \right) \]

Calculating the derivative:

\[ \frac{dU}{dx} = x^2 - 3x - 2 \]

### Step 3: Set the force to zero for equilibrium
For equilibrium, we set the force \( F \) to zero:

\[ -\left( x^2 - 3x - 2 \right) = 0 \]

This simplifies to:

\[ x^2 - 3x - 2 = 0 \]

### Step 4: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = -3 \), and \( c = -2 \):

\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2} \]

This gives us two equilibrium points:

1. \( x_1 = \frac{3 + \sqrt{17}}{2} \)
2. \( x_2 = \frac{3 - \sqrt{17}}{2} \)

### Step 5: Determine the stability of the equilibrium points
To determine the stability, we need to find the second derivative of \( U(x) \):

\[ \frac{d^2U}{dx^2} = \frac{d}{dx}(x^2 - 3x - 2) = 2x - 3 \]

### Step 6: Evaluate the second derivative at the equilibrium points
1. For \( x_1 = \frac{3 + \sqrt{17}}{2} \):

\[ \frac{d^2U}{dx^2}\bigg|_{x_1} = 2\left(\frac{3 + \sqrt{17}}{2}\right) - 3 = \sqrt{17} > 0 \]
- This indicates a stable equilibrium.

2. For \( x_2 = \frac{3 - \sqrt{17}}{2} \):

\[ \frac{d^2U}{dx^2}\bigg|_{x_2} = 2\left(\frac{3 - \sqrt{17}}{2}\right) - 3 = -\sqrt{17} < 0 \]
- This indicates an unstable equilibrium.

### Conclusion
The equilibrium points are:
- \( x_1 = \frac{3 + \sqrt{17}}{2} \) (stable)
- \( x_2 = \frac{3 - \sqrt{17}}{2} \) (unstable)