A particle of mass m is present in a region where the potential energy of the particle depends on the x-coordinate according to the expression `U=(a)/(x^2)-(b)/(x)`, where a and b are positive constant. The particle will perform.

Options

A. oscillatory motion but not simple harmonic motion about its mean position for small displacements
B. simple harmonic motion with time period `2pisqrt((8a^2m)/(b^4))` about its mean position for small displacements
C. neither simple harmonic motion nor oscillatory about its mean position for small displacements
D. none of the above

Explanation

To solve the problem, we will analyze the potential energy function given and determine the motion of the particle based on the derived force and acceleration.

### Step-by-Step Solution:

1. **Identify the Potential Energy Function**:
The potential energy \( U \) is given by:
\[ U = \frac{a}{x^2} - \frac{b}{x} \]
where \( a \) and \( b \) are positive constants.

2. **Calculate the Force**:
The force \( F \) acting on the particle can be derived from the potential energy using the relation:
\[ F = -\frac{dU}{dx} \]
First, we differentiate \( U \):
\[ \frac{dU}{dx} = -\frac{2a}{x^3} + \frac{b}{x^2} \]
Therefore, the force becomes:
\[ F = -\left(-\frac{2a}{x^3} + \frac{b}{x^2}\right) = \frac{2a}{x^3} - \frac{b}{x^2} \]

3. **Determine the Equilibrium Position**:
At equilibrium, the force is zero:
\[ 0 = \frac{2a}{x_0^3} - \frac{b}{x_0^2} \]
Rearranging gives:
\[ \frac{2a}{x_0^3} = \frac{b}{x_0^2} \]
Multiplying both sides by \( x_0^3 \):
\[ 2a = b x_0 \]
Thus, we find the equilibrium position:
\[ x_0 = \frac{2a}{b} \]

4. **Analyze Small Displacements**:
Let the particle be displaced by a small amount \( \delta x \). The new position is \( x = x_0 + \delta x \). The force can be expressed as:
\[ F = \frac{2a}{(x_0 + \delta x)^3} - \frac{b}{(x_0 + \delta x)^2} \]

5. **Expand Using Taylor Series**:
For small \( \delta x \), we can use Taylor expansion to approximate:
\[ F \approx -b \left(\frac{1}{x_0^2} + \frac{2\delta x}{x_0^3}\right) + 2a \left(\frac{1}{x_0^3} - \frac{3\delta x}{x_0^4}\right) \]
This leads to:
\[ F \approx -\left(\frac{b}{x_0^2} - \frac{2a}{x_0^3}\right) + \text{higher order terms} \]

6. **Substituting \( x_0 \)**:
Substitute \( x_0 = \frac{2a}{b} \) into the force equation. After simplification, we find:
\[ F \approx -\frac{b^4 \delta x}{8a^3} \]

7. **Relate Force to Acceleration**:
According to Newton's second law, \( F = m a \):
\[ m \frac{d^2 x}{dt^2} = -\frac{b^4 \delta x}{8a^3} \]
This indicates that the motion is simple harmonic motion (SHM) since it can be expressed as:
\[ \frac{d^2 x}{dt^2} = -\frac{b^4}{8ma^3} x \]

8. **Determine the Angular Frequency**:
The angular frequency \( \omega \) is given by:
\[ \omega^2 = \frac{b^4}{8ma^3} \]
Thus,
\[ \omega = \sqrt{\frac{b^4}{8ma^3}} \]

9. **Calculate the Time Period**:
The time period \( T \) of the oscillation is given by:
\[ T = 2\pi \sqrt{\frac{8ma^3}{b^4}} \]

10. **Conclusion**:
The particle will perform simple harmonic motion (SHM) with the derived time period.