---
title: "A particle of mass \\(m\\) is constrained to move along the \\(x\\)-axis under the influence of a conservative force. The potential energy \\(U(x)\\) of the particle as a function of position \\(x\\) is shown in the graph, with symmetric local minima of value \\(-U_0\\) located at \\(x = -x_0\\) and \\(x = +x_0\\), and a local maximum of value \\(0\\) at \\(x = 0\\). Which of the following statements correctly describes the motion of or the force on the particle?"
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url: "https://nerd-notes.com/ubq/120642/"
date_modified: "2026-08-23T04:42:28+00:00"
---

# A particle of mass \(m\) is constrained to move along the \(x\)-axis under the influence of a conservative force. The potential energy \(U(x)\) of the particle as a function of position \(x\) is shown in the graph, with symmetric local minima of value \(-U_0\) located at \(x = -x_0\) and \(x = +x_0\), and a local maximum of value \(0\) at \(x = 0\). Which of the following statements correctly describes the motion of or the force on the particle?

A particle of mass \(m\) is constrained to move along the \(x\)-axis under the influence of a conservative force. The potential energy \(U(x)\) of the particle as a function of position \(x\) is shown in the graph, with symmetric local minima of value \(-U_0\) located at \(x = -x_0\) and \(x = +x_0\), and a local maximum of value \(0\) at \(x = 0\). Which of the following statements correctly describes the motion of or the force on the particle?

![A grayscale Cartesian coordinate graph showing potential energy U(x) on the vertical axis versus position x on the horizontal axis. The horizontal axis has tick marks labeled -x_0, 0, and x_0. The vertical axis has tick marks labeled -U_0 and 0 at the origin. A smooth, solid curve forms a symmetric double-well potential (W-shape). The curve has a local maximum at the origin (0, 0), curves downward to smooth local minima at (-x_0, -U_0) and (x_0, -U_0), and rises steeply upward for |x| > x_0. Dashed vertical construction lines drop from the horizontal axis ticks at -x_0 and x_0 to each respective minimum, and a horizontal dashed construction line connects both minima across to -U_0 on the vertical axis. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787460147-aAeoc5.jpg)

- **A.** At \(x = \dfrac{x_0}{2}\), the net force on the particle is directed in the negative \(x\)-direction because the potential energy is decreasing with increasing \(x\).
- **B.** The position \(x = 0\) is a stable equilibrium point because the potential energy curve has a horizontal tangent where the net force is zero.
- **C.** If the particle has total mechanical energy \(E = -\dfrac{U_0}{2}\) and is initially at \(x = x_0\), it is trapped in the region \(x > 0\) and oscillates between two turning points where \(U(x) = -\dfrac{U_0}{2}\).
- **D.** For a particle oscillating with total mechanical energy \(E > 0\), the particle achieves its maximum speed at \(x = 0\) because the net force acting on it is zero at that position.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120642/*