A potential energy function isgiven by U = Ax3 + B x2 C x + D. For positive para

Question

A potential energy function isgiven by
U = Ax3 + B x2 C x + D. For positive parametersA, B, C, D this potential has twoequilibrium positions, one stable and one unstable. Find thestable equilibrium position for A= 2.65 J/m3 , B = 1.85 J/m2 , C = 1.9 J/m, and D = 2.8 J . Answer in units of m. b) Unstable? A potential energy function isgiven by
U = Ax3 + B x2 C x + D. For positive parametersA, B, C, D this potential has twoequilibrium positions, one stable and one unstable. Find thestable equilibrium position for A= 2.65 J/m3 , B = 1.85 J/m2 , C = 1.9 J/m, and D = 2.8 J . Answer in units of m. b) Unstable? b) Unstable?

Explanation / Answer

The force on the particle is F = - dU / dx and dU / dx = 0 forequilibrium dU / dx = 3 A x2 + 2 B x - C = 0 Substituting the values given for A, B, C I obtained x =.225 and x = -.400 d2 U / d x2 = 6 A x + 2 B d2 U / d x2 = 7.28 d2 U / d x2 = -2.66 If d2 U / d x2 is positive thenthe tangent to the force curve is increasing when F = 0 (forequilibrium) and the equilibrium is stable So x = .225 is stable and x = -.400 is unstable

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