Question

A system with effective mass (m) has a potential energy given by (U(x) = U'-2(x/x')²+(x/x')⁴) where (U') and (x') are positive constants and (U(0)) = the point where force on the particle is 0. Classify stable and unstable and find the value of (U(x)) at these equilibrium points.

a) Stable at (U(x) = U'), Unstable at (U(x) = -U')
b) Stable at (U(x) = -U'), Unstable at (U(x) = U')
c) Stable at (U(x) = 2U'), Unstable at (U(x) = -2U')
d) Stable at (U(x) = -2U'), Unstable at (U(x) = 2U')

Answer 1

To classify the equilibrium points as stable or unstable, we need to analyze the second derivative of the potential energy function U(x). Based on the information provided, we cannot determine the correct classification or values of U(x) at the equilibrium points.

Step-by-step explanation:

To classify the equilibrium points as stable or unstable, we need to analyze the second derivative of the potential energy function U(x).

If the second derivative is positive, the point is stable, and if it is negative, the point is unstable. Let's find the second derivative:

1. Calculate the first derivative of U(x) with respect to x: U'(x) = -4(x/x') + 4(x/x')^3
2. Calculate the second derivative by differentiating U'(x) with respect to x: U''(x) = -4/x' + 12(x/x')^2

From the second derivative, we can see that it depends on the value of x' (the positive constant).

Therefore, based on the given options, none of them can be confirmed as the correct classification of stable and unstable equilibrium points.

Additionally, we cannot determine the exact values of U(x) at the equilibrium points without knowing the values of U' and x'.