A particle of mass `m` is under the influence of a force F which varies with the displacement x according to the relation `F=-kx +F_(0)` in which k and `F_(0)` are constants. The particle when disturbed will oscillate

Options

A. about x=0, with `w ne sqrt(k//m)`
B. about x=0, with `omega=sqrt(k//m)`
C. about `x=F_(0)//k` with `omega=sqrt(k//m)`
D. about `x=F_(0)//k` with `omega ne sqrt(k//m)`

Answer

C

Explanation

Restoring force is zero at mean position
`F=-kx+F_(0)implies 0=-kx+F_(0)implies x=(F_(0))/(k)`
i.e. the particle will oscillate about `x=(F_(0))/(k)`
`implies F_(0)=kx implies ma =kx implies a=(k)/(m)n therefore omega=sqrt(k/m)`