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title: "A particle of mass \\(m\\) is constrained to move along the \\(x\\)-axis in a potential energy field given by \\(U(x) = \\dfrac{1}{2}kx^2 + \\dfrac{1}{4}bx^4\\), where \\(k\\) and \\(b\\) are positive constants. The particle is released from rest at \\(x = A\\) and oscillates symmetrically about the equilibrium position at the origin. By applying conservation of mechanical energy, the particle’s maximum speed at \\(x = 0\\) is derived as \\(v_{\\max} = A\\sqrt{\\dfrac{k}{m} + \\dfrac{bA^2}{2m}}\\). Which of the following statements correctly describes the limiting behavior of \\(v_{\\max}\\) as \\(A \\to 0\\) and as \\(A \\to \\infty\\), along with the appropriate physical justification?"
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url: "https://nerd-notes.com/ubq/124609/"
date_modified: "2026-09-28T14:08:48+00:00"
---

# A particle of mass \(m\) is constrained to move along the \(x\)-axis in a potential energy field given by \(U(x) = \dfrac{1}{2}kx^2 + \dfrac{1}{4}bx^4\), where \(k\) and \(b\) are positive constants. The particle is released from rest at \(x = A\) and oscillates symmetrically about the equilibrium position at the origin. By applying conservation of mechanical energy, the particle’s maximum speed at \(x = 0\) is derived as \(v_{\max} = A\sqrt{\dfrac{k}{m} + \dfrac{bA^2}{2m}}\). Which of the following statements correctly describes the limiting behavior of \(v_{\max}\) as \(A \to 0\) and as \(A \to \infty\), along with the appropriate physical justification?

A particle of mass \(m\) is constrained to move along the \(x\)-axis in a potential energy field given by \(U(x) = \dfrac{1}{2}kx^2 + \dfrac{1}{4}bx^4\), where \(k\) and \(b\) are positive constants. The particle is released from rest at \(x = A\) and oscillates symmetrically about the equilibrium position at the origin. By applying conservation of mechanical energy, the particle's maximum speed at \(x = 0\) is derived as \(v_{\max} = A\sqrt{\dfrac{k}{m} + \dfrac{bA^2}{2m}}\). Which of the following statements correctly describes the limiting behavior of \(v_{\max}\) as \(A \to 0\) and as \(A \to \infty\), along with the appropriate physical justification?

![A graph of potential energy U versus position x on bare Cartesian axes. The horizontal axis is labeled x with an origin at 0, and the vertical axis is labeled U. Two symmetric curves open upward from the origin. The first curve is drawn with a solid line and represents the anharmonic potential U(x); it passes through (0,0) and rises steeply on both sides of the origin. The second curve is drawn with a dashed line and represents the reference harmonic potential U_0(x); it also passes through (0,0) and lies strictly below the solid curve for all nonzero values of x. Two vertical dotted lines extend upward from tick marks at -A and A on the horizontal axis to intersect the curves. A legend in the upper-right corner indicates the solid line as U(x) and the dashed line as U_0(x). No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790604528-N7Uhmp.jpg)

- **A.** As \(A \to 0\), \(v_{\max} \propto A\) because the quadratic term dominates the potential energy; as \(A \to \infty\), \(v_{\max} \propto A^2\) because the restoring force is dominated by the quartic term, doing greater work than a linear spring over the same displacement.
- **B.** As \(A \to 0\), \(v_{\max}\) approaches the non-zero constant \(\sqrt{\dfrac{k}{m}}\); as \(A \to \infty\), \(v_{\max} \propto A^2\) because the quartic potential term increases the effective stiffness without bound.
- **C.** As \(A \to 0\), \(v_{\max} \propto A\) because the quadratic term dominates the potential energy; as \(A \to \infty\), \(v_{\max} \propto A\) because the period of oscillation remains independent of amplitude for any symmetric potential well.
- **D.** As \(A \to 0\), \(v_{\max} \propto A^2\) because higher-order terms dominate near the origin; as \(A \to \infty\), \(v_{\max} \propto A^4\) because the kinetic energy is directly proportional to the total mechanical energy.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/124609/*