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

`U=5x(x-4)=5x^2-20x`
`F=-(dU)/(dx)=-10x+20`
i.e., force is not contant. KE or speed of the particle will be maximum at the mean position where force becomes zero.
`F=0` or `x=2m`
Acceleration experienced by the particle is
`a=(F)/(m)=(-10x+20)/(0.1)=-(100x-200)`
i.e., particle executes SHM.
As `omega^2=100`
`omega=10`
Hence `T=(2pi)/(omega)=(pi)/(5)s`