Questions: Stability of Equilibrium: Stable, Unstable, and Neutral

Question 1

A ball rests at the bottom of a smooth bowl (equilibrium A) and at the top of a smooth hill (equilibrium B). A tiny nudge causes the ball at A to return to its original position, while the ball at B rolls away. Which potential energy interpretation explains this difference?

• A. At A, dV/dq > 0; at B, dV/dq < 0, so forces act in opposite directions
• B. At A, V is at a local minimum; at B, V is at a local maximum
• C. Both are true equilibria, but A has energy dissipation and B does not
• D. Both are equilibria because dV/dq = 0; only the size of the perturbation determines which is stable

Answer: (B)

Both points satisfy dV/dq = 0, confirming both are genuine equilibria — the force condition alone cannot distinguish them. Stability depends on the second derivative: at A, d²V/dq² > 0 (potential minimum), so any displacement increases V and creates a restoring force pointing back. At B, d²V/dq² < 0 (potential maximum), so any displacement decreases V and creates a force pointing further away. The common misconception is that 'equilibrium' implies 'stable' — it does not.

Question 2

As an axially loaded column is compressed toward its buckling load, what happens to the natural frequency of the column's lateral bending mode?

• A. It increases, because the column stores more strain energy and becomes stiffer
• B. It remains constant until the column suddenly buckles at the critical load
• C. It approaches zero, marking the transition from stable to unstable equilibrium
• D. It jumps discontinuously when the buckling load is reached

Answer: (C)

Compressive loading reduces the effective bending stiffness of the column. Natural frequency is proportional to the square root of stiffness, so it decreases continuously toward zero as the load approaches the buckling load. At exactly the buckling load, d²V/dq² = 0 — stiffness vanishes, frequency reaches zero — marking the stability boundary. Buckling is a stability failure, not a material strength failure: it occurs when the potential energy well loses its local minimum, often well below the yield stress.

Question 3

A point where dV/dq = 0 and d²V/dq² = 0 is expected to be a neutral equilibrium.

• T. True
• F. False

Answer: False

When d²V/dq² = 0, the second derivative test is inconclusive. You must examine higher-order derivatives to classify the equilibrium. Neutral equilibrium is one possible outcome (if the potential is genuinely flat in the neighborhood), but the point could also be stable or unstable depending on the sign of the leading higher-order even derivative.

Question 4

For a conservative system, the buckling of a slender column is fundamentally a stability failure rather than a strength failure: the effective stiffness of the bending mode reaches zero before the material yields.

• T. True
• F. False

Answer: True

Buckling occurs when d²V/dq² = 0 — the restoring force against lateral deflection disappears. This can happen at loads far below the material's yield strength, because stability is governed by geometry and bending stiffness, not tensile capacity. A slender steel column may buckle elastically, recovering its shape if the load is removed, precisely because the failure is one of equilibrium stability, not material fracture.

Question 5

A system is in equilibrium at a point where d²V/dq² < 0. What does this tell you about the system's behavior after a small perturbation, and why is the potential energy criterion — not just the force condition dV/dq = 0 — necessary to answer this question?

Model answer: The system is in unstable equilibrium. At d²V/dq² < 0, the potential energy is at a local maximum — any displacement lowers V, generating a force F = −dV/dq directed away from equilibrium, which accelerates the system further from it. The force condition dV/dq = 0 only locates equilibrium points; it cannot distinguish stable from unstable ones. The second derivative encodes the local curvature of V, and it is that curvature — positive (bowl), negative (hill), or zero (flat) — that determines whether perturbations decay, grow, or persist.

Finding dV/dq = 0 answers 'where is equilibrium?' The second derivative answers 'what kind?' A potential maximum (d²V/dq² < 0) produces destabilizing forces; a minimum (d²V/dq² > 0) produces restoring forces. Without examining V beyond the first derivative, stability is unknown — which is why equilibrium analysis alone is insufficient for most engineering problems.