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The potential energy of a particle oscillating on x-axis is given as U = 20 + (x – 2)2

Here U is in joules and x in metres. Total mechanical energy of the particle is 36 J.

(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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By assuming x – 2 = X, we have F = – 2X

Since, F  – X

The motion of the particle is simple harmonic

(b) The mean position of the particle is X = 0 or x – 2 = 0, which gives x = 2 m

(c) Maximum kinetic energy of the particle is,

Kmax = E – Umin = 36 – 20 = 16 J.

Note : Umin is 20 J at mean position or at x = 2 m.

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