A partical of mass `m` is located in a unidimensionnal potential field where potentical energy of the partical depends on the coordinates `x as U (x) = (A)/(x^(2)) - (B)/(x)` where `A` and `B` are positive constant.
Find the time period of small oscillation that the partical perform about equilibrium possition.

Explanation

To solve the problem, we need to find the time period of small oscillations of a particle in a given potential energy field. The potential energy \( U(x) \) is given by:

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

where \( A \) and \( B \) are positive constants.

### Step 1: Find the equilibrium position
To find the equilibrium position, we need to determine where the force acting on the particle is zero. The force \( F \) is related to the potential energy \( U(x) \) by:

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

Calculating the derivative of \( U(x) \):

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

Setting the force to zero for equilibrium:

\[ -\frac{dU}{dx} = 0 \implies -\left(-\frac{2A}{x^3} + \frac{B}{x^2}\right) = 0 \]

This simplifies to:

\[ \frac{2A}{x^3} = \frac{B}{x^2} \]

Multiplying both sides by \( x^3 \):

\[ 2A = Bx \]

Thus, the equilibrium position \( x_0 \) is given by:

\[ x_0 = \frac{2A}{B} \]

### Step 2: Determine the effective spring constant \( k \)
To find the time period of small oscillations, we need to find the effective spring constant \( k \). This can be done by calculating the second derivative of the potential energy at the equilibrium position:

\[ k = \left. \frac{d^2U}{dx^2} \right|_{x = x_0} \]

Calculating the second derivative:

\[ \frac{d^2U}{dx^2} = \frac{d}{dx}\left(-\frac{2A}{x^3} + \frac{B}{x^2}\right) = \frac{6A}{x^4} - \frac{2B}{x^3} \]

Now substituting \( x = x_0 = \frac{2A}{B} \):

\[ k = \left. \frac{6A}{x^4} - \frac{2B}{x^3} \right|_{x = \frac{2A}{B}} \]

Calculating \( k \):

1. Calculate \( x_0^4 \) and \( x_0^3 \):
- \( x_0^3 = \left(\frac{2A}{B}\right)^3 = \frac{8A^3}{B^3} \)
- \( x_0^4 = \left(\frac{2A}{B}\right)^4 = \frac{16A^4}{B^4} \)

2. Substitute into \( k \):
\[ k = \frac{6A \cdot B^4}{16A^4} - \frac{2B \cdot B^3}{8A^3} = \frac{3B^4}{8A^3} - \frac{B^4}{4A^3} = \frac{3B^4 - 2B^4}{8A^3} = \frac{B^4}{8A^3} \]

### Step 3: Calculate the time period \( T \)
The time period \( T \) of small oscillations is given by:

\[ T = 2\pi \sqrt{\frac{m}{k}} \]

Substituting \( k \):

\[ T = 2\pi \sqrt{\frac{m}{\frac{B^4}{8A^3}}} = 2\pi \sqrt{\frac{8mA^3}{B^4}} \]

### Final Answer
Thus, the time period of small oscillations about the equilibrium position is:

\[ T = 2\pi \sqrt{\frac{8mA^3}{B^4}} \]