The potential energy of a `1 kg` particle free to move along the x- axis is given by `V(x) = ((x^(4))/(4) - x^(2)/(2)) J`
The total mechainical energy of the particle is `2 J` . Then , the maximum speed (in m//s) is

Options

A. (a)`1/(sqrt2)`
B. (b)`2`
C. (c)`3/(sqrt2)`
D. (d)`sqrt(2)`

Explanation

To solve the problem, we need to find the maximum speed of a 1 kg particle given its potential energy function and total mechanical energy. The potential energy is given by:

\[ V(x) = \frac{x^4}{4} - \frac{x^2}{2} \]

The total mechanical energy \( E \) is given as \( 2 \, \text{J} \).

### Step-by-Step Solution:

1. **Identify the relationship between potential energy and kinetic energy:**
The total mechanical energy \( E \) is the sum of kinetic energy \( K \) and potential energy \( V \):
\[ E = K + V \]
Therefore, we can express kinetic energy as:
\[ K = E - V \]

2. **Find the critical points of the potential energy:**
To find the points where the potential energy is at a minimum (which corresponds to maximum kinetic energy), we need to find the derivative of the potential energy and set it to zero:
\[ \frac{dV}{dx} = 0 \]
Calculate the derivative:
\[ \frac{dV}{dx} = x^3 - x \]
Setting this equal to zero gives:
\[ x^3 - x = 0 \implies x(x^2 - 1) = 0 \]
This results in:
\[ x = 0, \quad x = 1, \quad x = -1 \]

3. **Determine the minimum potential energy:**
We need to evaluate the potential energy at the critical points \( x = 0, 1, -1 \):
- For \( x = 0 \):
\[ V(0) = \frac{0^4}{4} - \frac{0^2}{2} = 0 \]
- For \( x = 1 \):
\[ V(1) = \frac{1^4}{4} - \frac{1^2}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \]
- For \( x = -1 \):
\[ V(-1) = \frac{(-1)^4}{4} - \frac{(-1)^2}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \]

The minimum potential energy occurs at \( x = 1 \) and \( x = -1 \), where \( V = -\frac{1}{4} \, \text{J} \).

4. **Calculate maximum kinetic energy:**
Using the total mechanical energy:
\[ E = 2 \, \text{J} \]
The maximum kinetic energy occurs when potential energy is at its minimum:
\[ K_{\text{max}} = E - V_{\text{min}} = 2 - \left(-\frac{1}{4}\right) = 2 + \frac{1}{4} = \frac{9}{4} \, \text{J} \]

5. **Relate kinetic energy to speed:**
The kinetic energy is given by:
\[ K = \frac{1}{2} mv^2 \]
Given that \( m = 1 \, \text{kg} \):
\[ \frac{9}{4} = \frac{1}{2} \cdot 1 \cdot v^2 \]
Simplifying gives:
\[ \frac{9}{4} = \frac{1}{2} v^2 \implies v^2 = \frac{9}{4} \cdot 2 = \frac{9}{2} \]

6. **Calculate maximum speed:**
Taking the square root:
\[ v_{\text{max}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}} \, \text{m/s} \]

### Final Answer:
The maximum speed of the particle is:
\[ v_{\text{max}} = \frac{3}{\sqrt{2}} \, \text{m/s} \]