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title: "A particle of mass \\(m\\) is constrained to move along the \\(x\\)-axis in a region where its potential energy is given by \\(U(x) = U_0 \\tan^2(\\alpha x)\\), where \\(U_0\\) and \\(\\alpha\\) are positive constants and \\(-\\dfrac{\\pi}{2\\alpha} < x < \\dfrac{\\pi}{2\\alpha}\\). Which of the following expressions correctly gives the acceleration \\(a(x)\\) of the particle as a function of position \\(x\\)?"
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url: "https://nerd-notes.com/ubq/117566/"
date_modified: "2026-08-04T07:49:08+00:00"
---

# A particle of mass \(m\) is constrained to move along the \(x\)-axis in a region where its potential energy is given by \(U(x) = U_0 \tan^2(\alpha x)\), where \(U_0\) and \(\alpha\) are positive constants and \(-\dfrac{\pi}{2\alpha} < x < \dfrac{\pi}{2\alpha}\). Which of the following expressions correctly gives the acceleration \(a(x)\) of the particle as a function of position \(x\)?

A particle of mass \(m\) is constrained to move along the \(x\)-axis in a region where its potential energy is given by \(U(x) = U_0 \tan^2(\alpha x)\), where \(U_0\) and \(\alpha\) are positive constants and \(-\dfrac{\pi}{2\alpha} < x < \dfrac{\pi}{2\alpha}\). Which of the following expressions correctly gives the acceleration \(a(x)\) of the particle as a function of position \(x\)?

- **A.** \(a(x) = \dfrac{2\alpha U_0 \sin(\alpha x)}{m\cos^2(\alpha x)}\)
- **B.** \(a(x) = -\dfrac{\alpha U_0 \sin(\alpha x)}{m\cos^3(\alpha x)}\)
- **C.** \(a(x) = \dfrac{2\alpha U_0 \sin(\alpha x)}{m\cos^3(\alpha x)}\)
- **D.** \(a(x) = -\dfrac{2\alpha U_0 \sin(\alpha x)}{m\cos^3(\alpha x)}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/117566/*