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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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A particle free to move along the x-axis has potential energy given by U(x)=k[1−exp⁡(−x2)]U\left( x \right) = k\left[ {1 - \exp \left( { - {x^2}} \right)} \right] for −∞≤x≤+∞- \infty \le x \le + \infty where k is a positive constant of appropriate dimensions. Then
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Correct option: D
For small displacements from x=0, the motion is simple harmonic
Option D is correct because for small displacements from x=0, the potential energy can be approximated as a quadratic function, leading to simple harmonic motion.

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