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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. at points 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 originC. if its total mehanical energy is `k//2`, it has its minimum kinetic energy at the originD. for small displacements from `x = 0`, the motion is simple harmonic. |
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Answer» Correct Answer - D Let us plot the graph of the mathematical equation `U(x) = K[1- e^(-x^(2))]` `:. F = -(dU)/(dx) =-2kxe^(-x^(2))` It is clear that the potential energy is minimum at `x =0`. Therefore, `x = 0` is the state of stable equilibrium. Now if we displace the particle from `x = 0`, then for small displacements the particle tends to regain the position `x = 0` with a force `F = 2kx//e^(x^(2))` for `x` to be small `F prop x`. |
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