Question

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 1

(i) `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.
(ii) The mean position of the particle is X=0 or x-2=0, which gives x=2m
(iii) At x=2 m (mean position)
`U_("min")=20+(2-2)^(2)=20J`
`therefore` Maximum kinetic energy of the particle is, `K_("max")=E-U_("min")=36 -20=16 J`