The potential energy of a particle oscillating along x-axis is given as `U=20+(x-2)^(2)` Here, `U` is in joules and `x` in meters. Total mechanical energy of the particle is `36J`. (a) State whether the motion of the particle is simple harmonic or not. (b) Find the mean position. (c) Find the maximum kinetic energy of the particle.
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(a) `F = -(dU)/(dx) = - 2x(x-2)` By assuming `x - 2 = X`, we have `F = -2x` Since, `F prop - X` The motion of the particle is simple harmonic (b) The mean position of the particle is `X = 0 rArr x -2 = 0`, which gives `x = 2 m` (c) Maximum kinetic energy of the particle is, `K_("max") = E - U_("min") = 36 - 20 = 16J`
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