A particle free to move along the (x - axis) has potential energy given by `U(x)= k[1 - exp(-x^2)] for -o o le x le + o o`, where (k) is a positive constant of appropriate dimensions. Then.

Options

A. at points away from the origin, the particle is in unstable equilibrium.
B. for any finite non-zero value of `x`, there is a force directed away from the origin
C. if its total mehanical energy is `k//2`, it has its minimum kinetic energy at the origin
D. for small displacements from `x = 0`, the motion is simple harmonic.

Explanation

To solve the problem, we need to analyze the given potential energy function and derive the force acting on the particle. We will also determine the nature of equilibrium based on the force derived from the potential energy.

### Step-by-Step Solution:

1. **Identify the Potential Energy Function**:
The potential energy \( U(x) \) is given by:
\[ U(x) = k \left( 1 - e^{-x^2} \right) \]
where \( k \) is a positive constant.

2. **Calculate the Force**:
The force \( F \) acting on the particle is related to the potential energy by the equation:
\[ F = -\frac{dU}{dx} \]
We need to differentiate \( U(x) \) with respect to \( x \).

3. **Differentiate the Potential Energy**:
To find \( \frac{dU}{dx} \):
\[ \frac{dU}{dx} = k \cdot \frac{d}{dx}(1 - e^{-x^2}) = k \cdot 0 - k \cdot \left(-2x e^{-x^2}\right) = 2kx e^{-x^2} \]
Thus, the force becomes:
\[ F = -\frac{dU}{dx} = -2kx e^{-x^2} \]

4. **Analyze the Force**:
The expression for the force is:
\[ F = -2kx e^{-x^2} \]
- For \( x > 0 \), \( F < 0 \) (force is directed towards the origin).
- For \( x < 0 \), \( F > 0 \) (force is also directed towards the origin).

5. **Determine Equilibrium**:
The equilibrium points occur where \( F = 0 \). This happens when:
\[ -2kx e^{-x^2} = 0 \]
This implies \( x = 0 \) is the only equilibrium point.

6. **Stability of Equilibrium**:
To determine if this equilibrium is stable or unstable, we can analyze the second derivative of \( U(x) \):
\[ \frac{d^2U}{dx^2} = \frac{d}{dx}(2kx e^{-x^2}) = 2k \left( e^{-x^2} - 2x^2 e^{-x^2} \right) = 2k e^{-x^2} (1 - 2x^2) \]
- At \( x = 0 \):
\[ \frac{d^2U}{dx^2} = 2k e^{0} (1 - 0) = 2k > 0 \]
Since the second derivative is positive, the equilibrium at \( x = 0 \) is stable.

7. **Conclusion**:
The particle will perform simple harmonic motion around the origin, and the equilibrium is stable.

### Final Answer:
The particle is in stable equilibrium at the origin.