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

Explanation

(a) `F=-(dU)/(dx)=-2(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` or `x - 2 = 0, which gives `x = 2m`
( c) Maximum kinetic energy of the particle is,
`K_(max)=E - U_(min) = 36 - 20=16J`.