Question

A particle of unit mass is moving in a one-dimensional potential \( V(x) = x^2 - x^4 \). The minimum mechanical energy (in the same units as \( V(x) \)) above which the motion of the particle cannot be bounded for any given initial condition is:

Hint:

For potential problems, always find the critical points by differentiating the potential and analyzing the behavior of the particle at these points.

Answer 1

Correct Answer: 0.24

Step 1: Understanding the potential function.
The potential is given by \( V(x) = x^2 - x^4 \). To find the minimum energy at which the particle can be bound, we need to analyze the potential's local minima and maxima. Step 2: Determine the critical points.
To find the critical points, take the derivative of the potential \( V'(x) = 2x - 4x^3 \). Setting \( V'(x) = 0 \), we get the equation \( x(2 - 4x^2) = 0 \), which gives the critical points at \( x = 0 \) and \( x = \pm \frac{1}{\sqrt{2}} \). Step 3: Minimum energy.
The energy required to escape from the potential well is the energy at the local minima, which is between 0.24 and 0.26 units.