A simple harmonic oscillator of angular frequency 2 rad/s is acted upon by an external force F = (sin t) N. If the oscillator is at rest in its equilibrium position at t=0, its position at later times is proportional to:
A. $cost-\dfrac{1}{2}sin2t$
B. $sint-\dfrac{1}{2}sin2t$
C. $sint+\dfrac{1}{2}sin2t$
D. $sint+\dfrac{1}{2}cos2t$

Hint: We will first take the general equation of the force having its two parts of the solution; a particular solution and the specific solution. Then we will simplify the two solutions and will insert the given value of the angular frequency in the equations at taking the time, t = 0. Thus, we will get the final result.

Complete step by step solution:
From the equation of motion, we have
$F(t)=kx+m\ddot{x}$ ………. (i)
$\dfrac{F(t)}{m}={w_0}^2 x + \ddot{x}$ ………. (ii)
In equation (ii), its general equation has a sum of two parts. Let the first part solution be x=P(t), which satisfies the given equation and is called the particular solution. And the second part solution be x=s(t) which also satisfies the equation (ii) is the specific solution.
$\dfrac{{d^2}P(t)}{dt^2}+{w_0}^2 P(t)=\dfrac{F(t)}{m}$
Now, we will try a solution of type $P(t)={A_1}sinwt$ where the frequency and the forcing frequency are same and is equal to 1, and thus ${w_0}=2\; rad/s$
$\therefore \dfrac{sint}{m}={2^2}{A_1}sin(t)-{A_1}sin(t)$
$\implies {A_1}=\dfrac{\dfrac{1}{m}}{4-1}=\dfrac{1}{3m}$
And, we can also write the specific solution by
$\dfrac{d^2 S(t)}{dt^2}+{w_0}^2S(t)=0$
For which, we have the solution given in the form of SHM S(t), such that
$S(t)={A_2}sin({w_0}t-\phi)={A_2}sin(2t-\phi)$, where the given initial conditions determine ${A_2}$ and $\phi$.
Thus, we can write the given solution by
$x(t)=P(t)+S(t)=\dfrac{1}{3m}sin(t)+{A_2}sin(2t-\phi)$
According to question, t = 0, therefore x(t) = 0 and $\dfrac{dx}{dt_{t=0}}=0$
Substituting this, we get
$0=0+{A_2}sin(0-\phi)$
$\implies \phi = 2k\pi$, having k as an integer
$\therefore \dfrac{dx}{dt}=\dfrac{1}{3m}cos(t)+({A_2}\times 2\times cos(2t-2k\pi)$, which is time t = 0.
$\implies {A_2}=\dfrac{-1}{6m}$
Now, let us substitute the value of $A_2$ in the equation of x(t).
Thus, $x(t)=\dfrac{1}{3m}sin(t)-\dfrac{1}{6m}sin(2t-2k\pi)$
Putting the value of k = 0, we get
$x(t)=\dfrac{1}{3m}(sin(t)-\dfrac{1}{2}sin(2t))$
Hence, option b is the correct answer.

Note: The general equation of force will be used and its vector equation will be simplified to obtain the simple harmonic motion equation of the oscillator. Then, as the oscillator is in equilibrium at t = 0. We get the equation of position of the oscillator.