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InterviewSolution
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A particle free to move along the (x - axis) hsd potential energy given by `U(x)= k[1 - exp(-x^2)] for -o o le x le + o o`, where (k) is a positive constant of appropriate dimensions. Then.A. for small total displacement from `x = 0`, the motion is simple harmonicB. if its total machanical energy is `k//2`, it has its maximum kinetic energy at the origin.C. for any final norezero value of `s` there is a force directed away from the origin.D. at points away from the origin the particle is in unstable equilibrium. |
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Answer» Correct Answer - A Since `F = (dU)/(dv)= 2kx exp(-x^(2))` `F = 0` (at equilibrium as x = 0) `U` is minimum at `x = 0` and `U` and `U_(min) = 0` U is minimum at `x rarr +- oo and U_(min) = k` The particle would oscillation about `x = 0` for small displacement from the origin and it is in stable equilibrium at the origin |
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