A mass of 200 grams stretches a spring 49/80 meters.
I've found the spring constant k.
m= 200g = .2kg
L= 49/80 = .6125
I've found w, T, and f for the free undamped motion.
The general solution, I've found to be
u(t) = c1cos(4t)+c2sin(4t)
So now the question is: Suppose that the mass initially starts at rest in its equilibrium position. Solve for the equation of motion.
So far so good. At time t=0 the spring is in the equilibrium position, so u(0) = 0. From that it must be that the unknown constant c1 = 0, and your equation is now:
Now take the derivative:
At time t = 0 you say that the mass is at rest, so u'(0) = 0. Which means c2 = 0.
So the equation of motion is: u(t) = 0.
In other words it doesn't move at all. If a spring/mass is in the equilibrium position and is not moving, there's nothing that's going to make it start moving. Hence u(t) = 0. Usually for these type of problems they give you an initial deflection, or velocity, or acceleration, but in this case it's simply at rest.
OK, I see what the problem is. Looking at the attachment it appears that this problem is a continuation of part (c), not a problem on its own. So there is an exterior force of 6 sin(wt) acting on the mass. The basic equation governing all this is:
Using the coefficients you already worked out yields:
You have been told that the equation of motion is . Can you take it from here?