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. `(3)/sqrt(2))`
B. `sqrt(2)`
C. `(1)/(sqrt(2))`
D. `2`

Explanation

To find the maximum speed of a 1 kg particle moving along the x-axis with a given potential energy function, we can follow these steps:

### Step 1: Understand the relationship between total mechanical energy, kinetic energy, and potential energy.
The total mechanical energy (E) of the system is given by the sum of kinetic energy (KE) and potential energy (PE):
\[ E = KE + PE \]

### Step 2: Set up the equation using the given total mechanical energy.
We know that the total mechanical energy is \( E = 2 \, \text{J} \). Therefore, we can write:
\[ KE + PE = 2 \]

### Step 3: Write the expression for potential energy.
The potential energy is given by:
\[ V(x) = \frac{x^4}{4} - \frac{x^2}{2} \]

### Step 4: Find the minimum potential energy.
To find the minimum potential energy, we need to differentiate \( V(x) \) with respect to \( x \) and set the derivative to zero:
\[ \frac{dV}{dx} = x^3 - x = 0 \]
Factoring gives:
\[ x(x^2 - 1) = 0 \]
This gives us the critical points:
\[ x = 0, \, x = 1, \, x = -1 \]

### Step 5: Evaluate potential energy at critical points.
Now, we will evaluate \( V(x) \) at \( x = 1 \) and \( x = -1 \):
\[ V(1) = \frac{1^4}{4} - \frac{1^2}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \]
\[ V(-1) = \frac{(-1)^4}{4} - \frac{(-1)^2}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \]

### Step 6: Substitute the minimum potential energy back into the energy equation.
Now we can substitute the minimum potential energy back into the total energy equation:
\[ KE + (-\frac{1}{4}) = 2 \]
So,
\[ KE = 2 + \frac{1}{4} = \frac{8}{4} + \frac{1}{4} = \frac{9}{4} \]

### Step 7: Relate kinetic energy to speed.
The kinetic energy is given by:
\[ KE = \frac{1}{2} m v_{\text{max}}^2 \]
Since the mass \( m = 1 \, \text{kg} \):
\[ \frac{1}{2} v_{\text{max}}^2 = \frac{9}{4} \]
Multiplying both sides by 2 gives:
\[ v_{\text{max}}^2 = \frac{9}{2} \]

### Step 8: Solve for maximum speed.
Taking the square root of both sides:
\[ v_{\text{max}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}} \, \text{m/s} \]

Thus, the maximum speed of the particle is:
\[ \boxed{\frac{3}{\sqrt{2}} \, \text{m/s}} \]

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