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

Options

A. a. Find the equilibrium position of the body
B. b. Show that oscilliations of the body about this equilibrium positon in sample harmonic motion and find its time period
C. c. Find the amplitude of oscilliations if speed of the body at equilibrium position is `2sqrt6 m//s`
D. d. Find the amplitude of oscillators if speed of the body at equilibrium position is `2sqrt(6) m//s`.

Answer

To solve the problem step by step, we will break down the question into parts and provide a clear explanation for each step.

### Given:
The potential energy \( U \) of a unit mass body in a one-dimensional force field is given by:
\[ U = x^2 - 4x + 3 \]

### Step 1: Find the Equilibrium Position
The equilibrium position occurs where the net force acting on the body is zero. The force \( F \) can be derived from the potential energy using the relation:
\[ F = -\frac{dU}{dx} \]

First, we need to differentiate \( U \):
1. Differentiate \( U \):
\[ U = x^2 - 4x + 3 \]
\[ \frac{dU}{dx} = 2x - 4 \]

2. Set the force to zero:
\[ F = -\frac{dU}{dx} = 0 \implies - (2x - 4) = 0 \]
\[ 2x - 4 = 0 \implies 2x = 4 \implies x = 2 \]

**Equilibrium Position**: \( x = 2 \)

### Step 2: Show that the Oscillation is Simple Harmonic Motion (SHM)
To show that the oscillation is SHM, we need to express the force in the form:
\[ F = -k(x - x_0) \]
where \( k \) is a positive constant and \( x_0 \) is the equilibrium position.

From our earlier calculation:
\[ F = -\frac{dU}{dx} = - (2x - 4) = -2(x - 2) \]

This shows that:
\[ F = -2(x - 2) \]
Here, \( k = 2 \) and \( x_0 = 2 \). This confirms that the motion is SHM since the force is proportional to the displacement from the equilibrium position.

### Step 3: Calculate the Time Period
The time period \( T \) of SHM can be calculated using the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Given that the mass \( m = 1 \) (unit mass) and \( k = 2 \):
\[ T = 2\pi \sqrt{\frac{1}{2}} = 2\pi \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \cdot 2\pi \]

**Time Period**: \( T = \sqrt{2} \cdot 2\pi \)

### Step 4: Calculate the Amplitude of Oscillation
Given the speed at equilibrium \( v = 2\sqrt{6} \) m/s, we know that at the equilibrium position, the speed is the maximum speed \( v_{\text{max}} \):
\[ v_{\text{max}} = A \omega \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

From the previous calculations, we know:
\[ \omega = \sqrt{k/m} = \sqrt{2} \]

Now, substituting the values:
\[ 2\sqrt{6} = A \cdot \sqrt{2} \]
Solving for \( A \):
\[ A = \frac{2\sqrt{6}}{\sqrt{2}} = 2\sqrt{3} \]

**Amplitude**: \( A = 2\sqrt{3} \)

### Summary of Results:
1. **Equilibrium Position**: \( x = 2 \)
2. **Oscillation Type**: Simple Harmonic Motion (SHM)
3. **Time Period**: \( T = \sqrt{2} \cdot 2\pi \)
4. **Amplitude**: \( A = 2\sqrt{3} \)