Oscillatory motion with a suspended mass

[Jukai]

Homework Statement

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.

Homework Equations

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..

 

[anathema]

It seems like you are on the right track with your solution. To account for the mg factor, you can simply subtract it from both sides of the equation. This will not affect the oscillatory motion, as the gravitational force is constant and does not change with time. So your equation would become ma = -kx - mg, and then you can proceed with your integration as before.

As for your question about the mass setting a new equilibrium position without engaging in oscillatory motion, it ultimately depends on the initial conditions of the system. If the mass is released with a certain amount of initial velocity, it will continue to oscillate around the new equilibrium position. However, if it is released with no initial velocity, it will simply reach the new equilibrium position and remain stationary.