At t  = 0, a particle of mass m starts moving from rest due to a force F→ = F0sin(ωt)i^

If a particle performs S.H.M., it comes to rest at extreme position but at extreme position, acceleration of the particle is maximum possible. Therefore, resultant force on the particle will also be maximum possible. But force on the particle is given by F= Fo sin ωt and at t = 0, the particle was at rest, it means resultant force on the particle is zero when it is at rest. Therefore, this particle cannot perform S.H.M. Therefore (a) and (b) are wrong.In this particular case the force is zero at the initial moment and always act along x-axis. Initially the force starts to increase from zero initial value along positive x-direction. Therefore, particle starts to accelerate along x-direction and continues to accelerate till the force remains positive. It remains positive up to the instant given by ωt = π. After that moment, force becomes negative. Hence, particle starts to retard. But the impulse of the force from t = 0 to t = πω is numerically equal to its impulse from t = πω   to t = 2πω. Hence the particle will come to rest at It means from t = 0 to t = πω velocity of the particle continuously 0) increases and then it decreases from t = πω to t = 2πω. It means the velocity of the particle never becomes negative. Hence, the particle continues to move along positive direction of x-axis. Hence, 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 = 2πω, force becomes positive again. It means the particle starts to accelerate again along positive direction of x-axis. Therefore, velocity never becomes constant.