A particle of mass 0.6 kg moves along x in the potential U(x) = ax⁴ − bx², with a = 12 J m⁻⁴ and b = 5 J m⁻². (a) Derive F(x) and find all equilibria. (b) Classify their stability. (c) Use a first-order force expansion about either stable equilibrium to calculate the small-oscillation angular frequency. No dissipative force acts.

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F = −4ax³ + 2bx. Equilibria x = 0 (unstable), x = ±0.4564 m (stable); small-oscillation ω = 5.774 rad s⁻¹. A conservative one-dimensional force is the negative potential derivative: F(x) = −dU/dx = −4ax³ + 2bx. Equilibria require F = 0, not necessarily U = 0. Factor the force as 2x(b − 2ax²). This gives x = 0 and x = ±√(b/2a) = ±0.4564 m. The curvature is U″ = 12ax² − 2b. At zero it is −2b, a potential maximum and unstable equilibrium. At either nonzero equilibrium it is 4b, a minimum and stable equilibrium. Write x = x₀ + ξ near a minimum. To first order F ≈ −U″(x₀)ξ = −4bξ, so mξ″ = −4bξ and ω = √(4b/m) = 5.774 rad s⁻¹. Both minima have U = −b²/(4a) = -0.5208 J and equal curvature, so they give the same small-oscillation frequency. The approximation requires displacement ξ small enough that nonlinear terms are negligible.

Equilibria in a quartic potential · set 2

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