A simple harmonic oscillator of angular frequency 2 rad/s is acted upon by an external force F=(sint)N. If the oscillator is at rest in its equilibrium position at t=0, its position at later times is proportional to :

Writing equation of motion
mẍ + kx = F(t).. (i)
ẍ + ω²x = F(t)/m.. (ii)
The general solution of equation (ii) consists of a sum of two parts, The first part is solution let's say x = P(t) which satisfies equation (ii), is called as a particular solution. The second part is the solution let's say x = S(t) which satisfies equation (ii) with F(t) = 0, is called a specific solution.
d²P(t)/dt² + ω²P(t) = F(t)/m
We try a solution of type P(t) = A₁sinωt whose frequency is same as of forcing frequency which is equal to 1, and ω = 2rad/s
-A₁sin(t) + 2²A₁sin(t) = sint/m
A₁ = 1/3m
And specific solution is given by
d²S(t)/dt² + ω²S(t) = 0
For which the solution is given as of SHM
S(t) = A₂sin(ωt - φ) = A₂sin(2t - φ)
where A₂ and φ are determined by initial conditions
The general solution is given as
x(t) = P(t) + S(t) = (1/3m)sin(t) + A₂sin(2t - φ)
Given x(t) = 0 at t = 0 and dx/dt = 0
0 = 0 + A₂sin(0 - φ) ⇒ φ = 2kπ where k is an integer
dx/dt = (1/3m)cos(t) + A₂ × 2 × cos(2t - 2kπ)
at t = 0 velocity is zero ⇒ (1/3m) + A₂ × 2 = 0
A₂ = -1/6m
substituting value of A₂ in general solution of x(t)
x(t) = (1/3m)sin(t) - (1/6m)sin(2t - 2kπ)
taking k = 0 ⇒ x(t) = (1/3m)(sin(t) - (1/2)sin(2t))
Hence, correct answer is option B.