Recover force and classify equilibrium from an energy landscape

Potential belongs to an interaction system

A position-dependent potential represents a conservative interaction in the selected configuration coordinate. Its zero is conventional: adding a constant shifts every energy value but does not change force or energy differences. It is not an additional force on top of the conservative interaction already represented.

Differentiate with the correct sign

For U=Ax⁴−Bx², force is −dU/dx=−4Ax³+2Bx. Setting force zero gives x=0 and x=±sqrt(B/(2A)). The coefficients are positive in this packet. A horizontal tangent of U identifies a candidate equilibrium, but does not classify its response to a displacement.

Test nearby displacements

Curvature is U double-prime=12Ax²−2B. At the center it is −2B, so a small positive displacement gives force further right: unstable. At either outer stationary point curvature is 4B, positive, and small displacements produce restoring forces: stable. Their potentials are −B²/(4A), below the chosen zero at the center.

Recognize the limits of the second derivative test

Zero curvature alone is inconclusive. U=x⁴ has both zero slope and zero curvature at zero, yet its force −4x³ restores displacements from either side. A constant potential over an interval gives zero force throughout and neutral equilibrium there. Stable equilibrium does not imply that a conservative object eventually settles; without dissipation it can keep oscillating.

Reconstruct an energy difference from force

Integrating the derived force from x 1 to x 2 gives minus the potential difference. This supplies an independent sign check. The familiar ideal spring U=kx²/2 follows by integrating restoring force −kx, and a nonlinear spring includes additional terms rather than merely replacing k with an endpoint stiffness. Changing the energy reference cannot repair an incorrect derivative.

Worked original case

A=1,B=4 gives stable coordinates ±sqrt(2), potential −4 J and curvature 16 J/m²; x=0 has curvature −8 J/m² and is unstable.

Investigation

Compare B=2 and B=6 at A=1. Derive all roots, classify with nearby forces, and independently integrate force between the center and a minimum. Explain how adding 100 J to U changes the plot but not motion.

Public numerical transfer practice

Try a new practice case

For U=2x⁴−4x² J in SI units, what is the force at x=1 m in newtons?

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Original teaching, values and calculations, checked 30 October 2026 against the official calculus-based Mechanics framework and OpenStax work and kinetic energy. The numerical packets are supplied or generated, not measured apparatus records. No physical launch, machinery, loading or field intervention is requested. Public feedback and virtual captures do not establish independent mastery or supervised laboratory participation.