The potential energy of a $$1\,kg$$ particle free move along the x-axis is given by $$$V(x) = \left(\frac{x^4}{4} - \frac{x^2}{2}\right)J$$$ The total mechanical energy of the particle $$2\,J$$. Then, the maximum speed (in $$m/s$$) is

Solution & Explanation

1. Relate Kinetic Energy to Total Mechanical Energy

The total mechanical energy ($$E$$) of a particle is the sum of its kinetic energy ($$K$$) and its potential energy ($$V(x)$$):

$$E = K + V(x)$$

We are given the total mechanical energy $$E = 2 \,\, \text{J}$$ and the potential energy function:

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

Substituting these values into the energy conservation equation allows us to express kinetic energy as a function of position:

$$K = E - V(x)$$

$$K = 2 - \left( \frac{x^4}{4} - \frac{x^2}{2} \right) \quad \text{--- (Eq. 1)}$$

2. Find the Position ($$x$$) for Maximum Kinetic Energy

The speed (and consequently the kinetic energy) of the particle reaches its maximum value when the potential energy ($$V(x)$$) is at a local minimum value. To find this minimum, we differentiate $$V(x)$$ with respect to $$x$$ and set it equal to zero:

$$\frac{dV}{dx} = \frac{d}{dx}\left( \frac{x^4}{4} - \frac{x^2}{2} \right) = 0$$

$$x^3 - x = 0$$

$$x(x^2 - 1) = 0 \implies x = 0 \quad \text{or} \quad x = \pm 1$$

To verify which point corresponds to a minimum potential energy well, we take the second derivative:

$$\frac{d^2V}{dx^2} = 3x^2 - 1$$

• At $$x = 0$$: $$\frac{d^2V}{dx^2} = -1 < 0$$ (Local Maximum Potential)
• At $$x = \pm 1$$: $$\frac{d^2V}{dx^2} = 3(1) - 1 = 2 > 0$$ (Local Minimum Potential)

Thus, the minimum potential energy occurs at $$x = \pm 1$$. Let us compute this minimum potential energy value ($$V_{\text{min}}$$):

$$V_{\text{min}} = \frac{(\pm 1)^4}{4} - \frac{(\pm 1)^2}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \,\, \text{J}$$

3. Calculate the Maximum Speed ($$v_{\text{max}}$$)

Substitute the value of $$V_{\text{min}}$$ back into Equation 1 to find the maximum possible kinetic energy ($$K_{\text{max}}$$):

$$K_{\text{max}} = E - V_{\text{min}} = 2 - \left(-\frac{1}{4}\right) = 2 + \frac{1}{4} = \frac{9}{4} \,\, \text{J}$$

The formula for kinetic energy in terms of mass ($$m = 1 \,\, \text{kg}$$) and maximum speed ($$v_{\text{max}}$$) is:

$$K_{\text{max}} = \frac{1}{2} \cdot m \cdot v_{\text{max}}^2$$

$$\frac{9}{4} = \frac{1}{2} \cdot (1) \cdot v_{\text{max}}^2$$

Multiply both sides by 2 to isolate the velocity variable:

$$v_{\text{max}}^2 = \frac{9}{2}$$

$$v_{\text{max}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}} \,\, \text{m/s}$$

Concept Check: Because total energy stays conserved, the particle exchanges potential and kinetic energy back and forth as it oscillates. It travels fastest precisely as it passes through the bottom valleys ($$x = \pm 1$$) of the potential graph where it converts maximum possible potential configurations into pure translational kinetic velocity.

Correct Option Key: Option B ($$3/\sqrt{2}$$)