A particle moving on x - axis has potential energy `U = 2 - 20x + 5x^(2)` joule along x - axis. The particle is relesed at `x = -3`. The maximum value of `x` will be (`x` is in metre)

Options

A. `5m`
B. `3 m`
C. `7 m`
D. `8 m`

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Write down the potential energy function
The potential energy \( U \) of the particle is given by:
\[ U = 2 - 20x + 5x^2 \]

### Step 2: Find the force acting on the particle
The force \( F \) acting on the particle can be found using the relation:
\[ F = -\frac{dU}{dx} \]
We need to differentiate the potential energy function \( U \) with respect to \( x \).

### Step 3: Differentiate the potential energy function
Differentiating \( U \):
\[ \frac{dU}{dx} = -20 + 10x \]
Thus, the force becomes:
\[ F = -(-20 + 10x) = 20 - 10x \]

### Step 4: Set the force equal to zero for equilibrium
For the particle to be in equilibrium, the force must be zero:
\[ 20 - 10x = 0 \]
Solving for \( x \):
\[ 10x = 20 \implies x = 2 \text{ meters} \]

### Step 5: Determine the amplitude of motion
The particle is released at \( x = -3 \) meters. The maximum displacement from the equilibrium position (which we found to be \( x = 2 \) meters) can be calculated as follows:
\[ \text{Amplitude} = |x_{\text{max}} - x_{\text{initial}}| = |2 - (-3)| = |2 + 3| = 5 \text{ meters} \]

### Step 6: Calculate the maximum value of \( x \)
The maximum value of \( x \) will be the equilibrium position plus the amplitude:
\[ x_{\text{max}} = x_{\text{equilibrium}} + \text{Amplitude} = 2 + 5 = 7 \text{ meters} \]

### Final Answer
The maximum value of \( x \) is:
\[ \boxed{7 \text{ meters}} \]

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