Potential energy of a particle is related to x coordinate by equation `x^2 - 2x` . Particle will be in stable equilibrium at :

Options

A. `x = 0.5`
B. `x = 1`
C. `x = 2`
D. `x = 4`

Explanation

To determine the position of stable equilibrium for a particle whose potential energy \( U \) is given by the equation \( U(x) = x^2 - 2x \), we can follow these steps:

### Step 1: Understand the concept of stable equilibrium
Stable equilibrium occurs when the net force acting on the particle is zero, and the potential energy has a local minimum at that point.

### Step 2: Relate force to potential energy
The force \( F \) acting on the particle can be derived from the potential energy using the equation:
\[ F = -\frac{dU}{dx} \]

### Step 3: Differentiate the potential energy function
Given the potential energy function:
\[ U(x) = x^2 - 2x \]
we differentiate it with respect to \( x \):
\[ \frac{dU}{dx} = 2x - 2 \]

### Step 4: Set the force to zero for equilibrium
For stable equilibrium, we set the force equal to zero:
\[ F = -\frac{dU}{dx} = 0 \]
This implies:
\[ - (2x - 2) = 0 \]
or:
\[ 2x - 2 = 0 \]

### Step 5: Solve for \( x \)
Solving the equation \( 2x - 2 = 0 \):
\[ 2x = 2 \implies x = 1 \]

### Step 6: Confirm stability
To confirm that this point is indeed a point of stable equilibrium, we can check the second derivative of the potential energy:
\[ \frac{d^2U}{dx^2} = \frac{d}{dx}(2x - 2) = 2 \]
Since the second derivative is positive (\( \frac{d^2U}{dx^2} > 0 \)), this indicates that the potential energy has a local minimum at \( x = 1 \), confirming stable equilibrium.

### Final Answer
The particle will be in stable equilibrium at \( x = 1 \).

---