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.

The potential energy of a particle oscillating along x-axis is given a

1.	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.
Answer» a. `F = - (du)/(dx) = - 2 (x - 2)` By assuming `x - 2 = X, we have F = - 2X` Since, `F = - X` The motion position of the partical is simple harmonic. b. The mean position of the partical is `X = 0 or x - 2 = 0`, which gives `x = 2m` c. Maximum kinetic energy of the partical is `K_(max) = E = U_(max)` `= 36 - 20 = 16 J`.

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