What does "w" mean? And where did you get "-25.81"?

The way I would do this problem is, using "force= mass times acceleration", if the x m from its equilibrium position then the force is -200x = 0.3 x'' (negative since the force is opposite x- if x is downward, the force is upward and vice-versa), where x'' is the second derivative of position- the acceleration. That is the "differential equation" x''+ (200/0.3)x= x''+ (266.67)x= 0 which has the general solution \(\displaystyle x= Acos(\sqrt{266.67}t)+ Bsin(\sqrt{266.67}t)= A cos(25.81t)+ Bsin(25.81t)\). We can then calculate that \(\displaystyle v= dx/dt= -25.81A sin(25.81t)+ 25.81B cos(26.81t)\).

We are told that the original displacement is 10mm= 1 cm= 0.01 cm (**not** 0.1) with initial velocity 0 (since it is "released" from that position). That is, we have x(0)= A cos(0)+ B sin(0)= A= -.01 and v(0)= -25.81A sin(0)+ 25.81B cos(0)= 25.81B= 0.

So position x at time t is given by x(t)= -.01 cos(25.81 t). The velocity, as above, is the derivative, 0.2581 sin(25.81 t), and the acceleration is the derivative of that, \(\displaystyle (25.81^2)(0.01) cos(25.81t)= 6.66 cos(25.81 t)\).

So the acceleration is \(\displaystyle \frac{6.66}{-0.01}= -666= -(25.81)^2\) times x. The graph of that is a straight line, passing through the origin with slope \(\displaystyle -(25.81)^2= -666\).

I don't know how much of that makes sense because you don't say what you **do** understand and what you **can** do. I suspect that you were given some formulas but you don't say **what**. You will get a lot better (and simpler) answers next time if you give us all that information to start with.