A 0.500 kg particle moves in one dimension with potential U(x) = ax⁴ − bx², where a = 20.0 J m⁻⁴ and b = 4.00 J m⁻². Determine all equilibrium positions and classify their stability. Find the small-oscillation angular frequency about either stable equilibrium, the barrier energy relative to a minimum, and the turning points for total energy E = −0.100 J. Explain why a particle with E = −0.100 J cannot cross from one well to the other.

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Equilibria: x = 0 unstable and x = ±0.316 m stable. Small-oscillation ω = 5.66 rad s⁻¹; barrier above either minimum = 0.200 J. For E = −0.100 J, turning points in x² are 0.0293 and 0.1707 m², so the two wells remain disconnected. Force is F = −dU/dx = −(4ax³−2bx). Equilibrium requires 2x(2ax²−b) = 0, giving x = 0 and x = ±√(b/2a) = ±0.316 m. Stability follows from U'' = 12ax²−2b. At x = 0, U'' = −8.00 N m⁻¹, a local maximum; at either nonzero point U'' = 4b = 16.0 N m⁻¹, a minimum. Near a minimum, U is approximately quadratic with effective spring constant U''. Hence ω = √(U''/m) = √(16.0/0.500) = 5.66 rad s⁻¹. The minimum potential is −b²/(4a) = −0.200 J while U(0) = 0, so the central barrier is 0.200 J above either minimum. Set 20x⁴−4x² = −0.100 and let y = x²: 20y²−4y+0.100 = 0 gives y = 0.0293 or 0.1707 m². Since E < U(0), the central region is classically forbidden and trajectories stay in one well.

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