The potential energy (U) of a body of unit mass moving in a one-dimension force field is given by
`U=(x^(2)-4x+3)` . All units are in S.L

Explanation

To solve the problem, we will follow these steps:

### Step 1: Find the Force from Potential Energy
The potential energy \( U \) is given by:
\[ U = x^2 - 4x + 3 \]
To find the force \( F \), we use the relation:
\[ F = -\frac{dU}{dx} \]
Calculating the derivative:
\[ \frac{dU}{dx} = 2x - 4 \]
Thus, the force is:
\[ F = - (2x - 4) = -2x + 4 \]

### Step 2: Determine the Equilibrium Position
At equilibrium, the force \( F \) is zero:
\[ -2x + 4 = 0 \]
Solving for \( x \):
\[ 2x = 4 \implies x = 2 \, \text{m} \]

### Step 3: Check for Simple Harmonic Motion (SHM)
For the motion to be simple harmonic, the force must be proportional to the displacement from the equilibrium position. We can express the force as:
\[ F = -2(x - 2) \]
This indicates that the motion about \( x = 2 \) is simple harmonic, with the force constant \( k = 2 \, \text{N/m} \).

### Step 4: Calculate the Time Period of Oscillation
The time period \( T \) for simple harmonic motion is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Here, \( m = 1 \, \text{kg} \) (unit mass) and \( k = 2 \, \text{N/m} \):
\[ T = 2\pi \sqrt{\frac{1}{2}} = \pi \sqrt{2} \, \text{s} \]

### Step 5: Maximum Velocity at Equilibrium Position
At the equilibrium position, the velocity is maximum. We know that:
\[ v_{\text{max}} = A \omega \]
Where \( \omega = \sqrt{\frac{k}{m}} = \sqrt{2} \). If the maximum speed is given as \( 2\sqrt{6} \):
\[ 2\sqrt{6} = A \sqrt{2} \]
Solving for amplitude \( A \):
\[ A = \frac{2\sqrt{6}}{\sqrt{2}} = 2\sqrt{3} \, \text{m} \]

### Summary of Results
1. The equilibrium position is at \( x = 2 \, \text{m} \).
2. The system exhibits simple harmonic motion with \( k = 2 \, \text{N/m} \).
3. The time period of oscillation is \( T = \pi \sqrt{2} \, \text{s} \).
4. The amplitude of oscillation is \( A = 2\sqrt{3} \, \text{m} \).