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
k= 3.2
w^2= 16
I've found w, T, and f for the free undamped motion.
w= 4
T= 1.57
f= .637
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:
$\displaystyle u(t) = C_2 \sin(\omega t)$.
Now take the derivative:
$\displaystyle u'(t) =\omega C_2 \cos(\omega t) $
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.
Please write the problem out exactly as stated. As I noted the problem as you wrote it is a bit strange - I've never seen a problem like this where there is no initial displacement and no initial velocity. But I wonder - the problem says the mass starts at its "equilibrium position," but perhaps what they meant to say is that it starts at its "unstretched position?"
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:
$\displaystyle \sum F = ma$
so,
$\displaystyle 6 sin(2 t) - kx = m\ddot x$
Rearrange:
$\displaystyle \frac {6 \sin(2t)} m = \ddot x + \frac k m x$
Using the coefficients you already worked out yields:
$\displaystyle \frac {6 \sin(2t)}{0.2 kg} = 12A \cos(2t) + 12B \sin(2 t)$
from which
$\displaystyle A = 0, \ B=\frac {30}{12} = 2.5$
You have been told that the equation of motion is $\displaystyle x(t) = A \cos(2t) + B \sin(2t)$. Can you take it from here?