A mass m is suspended on a vertical spring. The mass is released from the equilibrium position of the spring without the mass. Find the position of the mass as a function of time, while neglecting friction.
ma=-kx + mg
The Attempt at a Solution
I set the downward motion as positive, thus explaining my - and + choices in the equation. I posed x=Acos(wt) + Bsin(wt) and integrated the ma=-kx + mg equation twice to get a function of x(t). My main problem is that I don't know what to do with the mg factor. When I integrate I get a gt²/2 factor and that's obviously not oscillatory motion.
If the mass is released from the equilibrium position of the spring without the mass, will the mass simply set a new equilibrium position on the spring without engaging an oscillatory motion? I realize this might be a very simple problem, but I haven't done oscillatory motion in a really long time..