I have this problem where I am trying to calculate $d(t)$ and $v(t)$ of a mass m on a spring, dropped from a displacement $A$, without using anything else than Hooke's law and energy calculations. Meaning I don't necessarily trust the solution on Wikipedia about $d(t)$ = $A\cos \omega t$, where $\omega$ = $\sqrt\frac{k}{m}$ after all I have read about pendulums being generally treated with the "small angle approximation".
So, here is what I did:
EDIT: In order to facilitate any differentiation tricks I ensure monotony of all functions by restricting this to a single drop of the spring from A to 0.
(i) $F(d)$ = $-kd$ (Hooke's law)
(ii) $K(d)$=$\int-kx dx$ from $x=d$ to $x=A$. This should be equal to $\frac{k}{2}(A^2-d^2)$.
(iii) $K(v)=\frac{1}{2}mv^2$
(iv) I equate $(ii)$ and $(iii)$ and solve for $d$: $d(v)$ = $\sqrt{A^2-\frac{mv^2}{k}}$
(v) I then differentiate this by $v$: $t$ = $d'(v)$ =$\frac{mv}{k\sqrt{A^2-\frac{mv^2}{k}}}$ (am I crazy?)
(vi) and solve for $v$ to obtain $v(t)$: $v(t)$ =$\frac{Akt}{\sqrt{m(m+kt^2)}}$
(vii) I then insert the expression for $v(t)$ in $(iv)$ to obtain $d(t)$: $d(t)$ = $A\sqrt{\frac{m}{m+kt^2}}$
I'm done, but my result disagrees with Wikipedia. It's quite possible that I have made other mistakes in the process, but judging from the answer that I have now approved, it seems I was much too optimistic anyway, doing t=d'(v). t=d'(v) means "Time is the rate of change of displacement with respect to velocity". Are there any situations at all where this makes sense?
EDIT: In the situastion where a particle starts at displacement $d_0$ and has initial velocity $0$ and is worked upon by a single constant force $f=ma$, then $d(t)=d_0+\frac{1}{2}at^2$. $v(t)=d'(t)=at$, so $t=v(t)/a$, which can be written $t(v)=v/a$, since $v(t)$ is injective. Then $d(v)=d_0+\frac{1}{2a}v^2$ and $d'(v)=v/a=t$. So here it works.
EDIT: This question was inspired by a much easier homework assignment, for which calculating v(t) and d(t) was not even necessary. Having solved the assignment, I was still interested in the theoretical questions described here. In light of all this, I have removed the homework label.