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.

Explanation

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`.