The potential energt of a particle of mass 0.1 kg, moving along the x-axis, is given by `U=5x(x-4)J`, where x is in meter. It can be concluded that

Options

A. the particle is acted upon by a constant force
B. the speed of the particle is maximum at `x=2m`
C. the particle executes SHM
D. the period of oscillation of the particle `((pi)/(5))` s

Answer

B, C, D

Explanation

`F=(-du)/(dx)=-[10x-20]=20-10x`
`a=(20-10x)/(m)=100(2-x)`
`a=0` at `x=2` So `V` is maximum at `x=2` This is equation of SHM so particle executes SHM
`omega^2=100impliesomega=10`
`T=(2pi)/(omega)=(pi)/(5)s`