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

Explanation

To solve the problem, we need to analyze the potential energy function given and determine the conditions for simple harmonic motion (SHM). The potential energy function is given by:

\[ U(x) = k[1 - e^{-x^2}] \]

where \( k \) is a positive constant.

### Step 1: Analyze the Potential Energy Function
The potential energy function describes how the potential energy changes with position \( x \). At \( x = 0 \):

\[ U(0) = k[1 - e^{0}] = k[1 - 1] = 0 \]

As \( x \) moves away from zero, the term \( e^{-x^2} \) approaches zero, and thus:

\[ U(x) \to k \text{ as } |x| \to \infty \]

### Step 2: Determine the Force from Potential Energy
The force \( F \) acting on the particle can be derived from the potential energy using the relation:

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

Now, we differentiate \( U(x) \):

1. Differentiate \( U(x) \):

\[ \frac{dU}{dx} = k \cdot \frac{d}{dx} [1 - e^{-x^2}] = k \cdot (0 + 2x e^{-x^2}) = 2kx e^{-x^2} \]

2. Therefore, the force is:

\[ F = -\frac{dU}{dx} = -2kx e^{-x^2} \]

### Step 3: Analyze the Force for Small Displacements
For small displacements around \( x = 0 \), we can approximate \( e^{-x^2} \) using its Taylor series expansion:

\[ e^{-x^2} \approx 1 - x^2 \quad \text{(for small } x\text{)} \]

Substituting this approximation into the force equation:

\[ F \approx -2kx(1 - x^2) \approx -2kx \quad \text{(ignoring higher order terms)} \]

### Step 4: Identify Simple Harmonic Motion Condition
The force \( F \approx -2kx \) indicates that the force is proportional to \( -x \), which is the condition for simple harmonic motion. Thus, we can conclude that:

- The motion is simple harmonic for small displacements from \( x = 0 \).

### Step 5: Analyze the Total Mechanical Energy
The total mechanical energy \( E \) of the system is given as \( \frac{k}{2} \). At the equilibrium position \( x = 0 \), the kinetic energy is maximum, and potential energy is minimum (zero). Thus, the statement about the total mechanical energy having its minimum kinetic energy at the origin is incorrect.

### Step 6: Stability of Equilibrium
For any finite non-zero value of \( x \), the force \( F \) is directed towards the origin (not away from it), indicating that the equilibrium at \( x = 0 \) is stable.

### Conclusion
Based on the analysis, we can summarize the findings:

1. For small displacements from \( x = 0 \), the motion is simple harmonic.
2. The statement about minimum kinetic energy at the origin is incorrect; kinetic energy is maximum at the origin.
3. The force is always directed towards the origin, indicating stable equilibrium.

Thus, the correct answer is that for a small displacement from \( x = 0 \), the motion is simple harmonic.