Second order linear ODE determine coefficient A vertical spring has the spring constant $32$ N/m. A body with the mass 2 kg is attached to its end and is at equilibrium. At the time $t=0$ a variable force $$F(t)=12\sin(\omega t)$$ is applied in the direction of gravity. Describe the motion of the body for different values of the force's angle frequency $\omega $.$$$$ 
My idea is to denote the springs vertical position with $y(t)$. If we assume that downwards is the positive direction then the acceleration should be described as 
$$y''(t)=6\sin(\omega t) - 16y(t)$$
The homogenous solution is
$$y_h = A_1\sin(4t)+B_1\cos(4t)$$
Assume the particular solution is on the form
$$y_p = A_2\sin(\omega t) $$
Deriving this twice and plugging in to the original equation gives
$$16A_2-A_2\omega^2=6, A_2=\frac{6}{16-\omega^2}$$
and so the general solution would be
$$y=A_1\sin(4t)+B_1\cos(4t)+\frac{6\sin(\omega t)}{16-\omega^2}$$ 
but since $y(0)=0$ we can remove the cosine term and so
$$y=A_1\sin(4t)+\frac{6\sin(\omega t)}{16-\omega^2}$$ 
Now how do I determine $A_1$? And I understand that I need to consider the special case $\omega=4$.

The equation is:
$$y''=6\sin\omega t-16y$$
Let $y=a\sin(\omega t+\theta)$
Now:
$$-a\omega^2\sin(\omega t+\theta)=6\sin\omega t-16a\sin(\omega t+\theta)$$
Or:
$$\underbrace{(16-\omega^2)a}_{\alpha}\sin(\omega t+\theta)=6\sin\omega t$$
Now:
$$(\alpha\cos\theta-6)\sin\omega t+(\alpha\sin\theta)\cos\omega t=0\\
\sin\left(\omega t+\arctan\frac{\alpha\sin\theta}{\alpha\cos\theta-6}\right)=0$$
...
