At `t = 0` aparticle of mass `m` start moving from rest due to a force `vecF = F_(0) sin (omega t)hati` :

Options

A. Particle perform `SHM` about its initial position of rest
B. Particle perform `SHM` with initial position as extreme position with angular frequency `omega`
C. At any instant, distance moved by the particle equals its displacement, from the initial position
D. initial velocity of particle increases with time but after time `t = 2 pi// omega` it becomes constant

Answer

C

Explanation

If a particle `S.H.M.` If comes to rest at exterm position but at external position , acceleration of the particle is maximum position .Therefore resultantforce on the particle will also the maximum pasition But force on the given by `F = F_(s)` on as and at `t = 0` the particle was at when it is at rest Therefore this particle cannot `S.H.M.` Therefore (a) and (b) are wrong
In the particle case the force is zero at the initial moment and always act along `x- axis` velocity the force start position `x-` dirfection` Therefore particles start to acceleration along `x-` dirfectionand continutes position up to the instant given by `omega t = pi` after particle moment form because negative hence position start to retard but the impule of the force from ` t = 0` to `t = pi//omega t` is numerically to its impluse from `t = (pi)/(omega)` to `t = (2pi)/(omega)` hence the particle will comes to rest particle contrinuiously increase and and then it decrease particle move because negative hence the particle contitues to move along positive direction of a `x-`axis hebnce its displacement at any instant will be equal to the distance moved by the particle from the initial position
Hence c is correct
After time `t = (2pi)/(omega)` force because position again it mean the particle start to acceleration again along position direction of a `x-` axis .Therefore velocity never becomes correct