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 find the points of equilibrium for the given potential energy function \( U(x) = \left(\frac{x^3}{3} - \frac{3x^2}{2} + 2x\right) \). We will follow these steps:

### Step 1: Find the Force
The force \( F \) acting on the particle is given by the negative gradient of the potential energy:
\[ F = -\frac{dU}{dx} \]
We need to differentiate \( U(x) \) with respect to \( x \).

### Step 2: Differentiate the Potential Energy
Differentiating \( U(x) \):
\[ \frac{dU}{dx} = \frac{d}{dx}\left(\frac{x^3}{3} - \frac{3x^2}{2} + 2x\right) \]
Calculating the derivative:
\[ \frac{dU}{dx} = x^2 - 3x + 2 \]

### Step 3: Set the Force to Zero
At equilibrium, the force is zero:
\[ F = -\frac{dU}{dx} = 0 \implies x^2 - 3x + 2 = 0 \]

### Step 4: Solve the Quadratic Equation
Now, we solve the quadratic equation \( x^2 - 3x + 2 = 0 \):
Factoring gives:
\[ (x - 1)(x - 2) = 0 \]
Thus, the solutions are:
\[ x = 1 \quad \text{and} \quad x = 2 \]

### Step 5: Determine the Nature of Equilibrium
To determine the nature of the equilibrium points, we need to find the second derivative of the potential energy:
\[ \frac{d^2U}{dx^2} = \frac{d}{dx}(x^2 - 3x + 2) = 2x - 3 \]

### Step 6: Evaluate the Second Derivative at the Equilibrium Points
Now we evaluate \( \frac{d^2U}{dx^2} \) at the points \( x = 1 \) and \( x = 2 \):
1. For \( x = 1 \):
\[ \frac{d^2U}{dx^2} = 2(1) - 3 = 2 - 3 = -1 \quad (\text{unstable equilibrium}) \]
2. For \( x = 2 \):
\[ \frac{d^2U}{dx^2} = 2(2) - 3 = 4 - 3 = 1 \quad (\text{stable equilibrium}) \]

### Conclusion
- The point \( x = 1 \) is an unstable equilibrium.
- The point \( x = 2 \) is a stable equilibrium.

### Final Answer
The correct option is that the point \( x = 2 \) is a point of stable equilibrium.

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