JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Critical Thinking

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    A particle free to move along the x-axis has potential energy given by \[U(x)=k[1-\exp {{(-x)}^{2}}]\] for \[-\infty \le x\le +\infty \], where k is a positive constant of appropriate dimensions. Then                   [IIT-JEE 1999; UPSEAT 2003]

    A)             At point away from the origin, the particle is in unstable equilibrium

    B)             For any finite non-zero value of x, there is a force directed away from the origin

    C)             If its total mechanical energy is k/2, it has its minimum kinetic energy at the origin

    D)             For small displacements from x = 0, the motion is simple harmonic

    Correct Answer: D

    Solution :

                    Potential energy of the particle \[U=k(1-{{e}^{-{{x}^{2}}}})\]             Force on particle\[F=\frac{-dU}{dx}=-k[-{{e}^{-{{x}^{2}}}}\times (-2x)]\]             F\[=\,-2kx{{e}^{-{{x}^{2}}}}\]\[=-2kx\left[ 1-{{x}^{2}}+\frac{{{x}^{4}}}{2\,!}-...... \right]\]             For small displacement \[F=-2kx\]             Þ \[F(x)\propto -x\] i.e. motion is simple harmonic motion.


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