The solution

Now we take our educated guess 1.4 and check it satisfies $F~=~ ma$, or more precisely, eqn. 1.1.

Let's compute the acceleration. To do that, we need to compute $a$. Differentiating once,

$$v ~=~ {dx\over dt} ~=~ -A\omega \sin(\omega t +\delta)$$ (1.7)

and again

$$a ~=~ {dv\over dt} ~=~ -A\omega^2 \cos(\omega t +\delta)$$ (1.8)

Well this is that same as

$$a ~=~ -\omega^2 x$$ (1.9)

So we have something similar to eqn. 1.1. We still haven chosen $\omega$, and now we can choose it so that everything works out. Choose

$$\omega^2 ~=~ {k\over m}$$ (1.10)

and then it's clear that the acceleration satisfies eqn. 1.1.

So now we know the solution. It's a sine wave with an angular frequency of $\sqrt{k\over m}$. Notice that $A$ and $\delta$ can be anything that we want. In other words you can start of the spring with a small oscillation, or a large oscillation, and the angular frequency will be identical.