Vertical oscillator with a punctual mass Ok, this is apparently a simple problem.
Consider a mass bound to a vertical oscillator of constant k, at thr equilibrium position, and initial height H.
When letting it move by its own weight, one has the balance $\sum F=0$, and so the elastic force equals gravity force (equilibrium at the lowest part ($v=0$)): $F=P\Rightarrow kH=mg\Rightarrow H=\frac{mg}{k}$.
Ok, now solve it through energies.
One has initial energy $mgH$, and final energy elastic one $\frac{1}{2}kH^2$. And so equating those we reach $H=\frac{2mg}{k}$.
Why those results are different? Thanks
 A: When you apply the equilibrium equation, you have indeed located the position of equilibrium, which is $H = mg/k$ below the release point, given the system had no initial elastic energy. When you release the system, after a while the mass will finally stop at the equilibrium position. How long this while is depends on the damping of system. High damping means the system equilibrates quickly.
When you use conservation of energy, you need to be especially careful. First of all, I'm going to use $x$ instead of $H$ here since they are two different things being measured here. So, initially, by setting the GPE datum at $x$ below the initial height, the system's initial energy is $mgx$. Then, at the lowest point, all the GPE has indeed become elastic potential energy, and we get $x = 2mg/k = 2H$. Now, it is important to note that this corresponding point is the lowest point of vibration, and is not the same as the equilibrium point. This is because, in this model, the velocity at the equilibrium point is not zero.
This may seem counterintuitive that the velocity is not zero at the equilibrium point, even though after a while the system should eventually come to rest at the equilibrium point. The thing is, for a simplified model, we often assume the damping to be zero so we don't have to keep track of losses of mechanical energy to heat. As a result, the modelled system will vibrate forever and ever with constant amplitude of vibration. Real systems don't do this, but this simplification allows us to make use of energy conservation.
The kinetic energy is actually maximised at the equilibrium point, and the regions of no kinetic energy correspond to the highest and lowest points of vibration. The equilibrium point is actually the centre of the vibrational motion, and it should lie right between the highest and lowest points of vibration. The highest point will be the point of release, and the lowest point lies a distance of $x=2H$ below the release point (this is what we calculated with conservation of energy). So, the equilibrium point should lie between these, i.e. $H$ below the release. This matches what we calculated using equilibrium!
So, the key thing to watch out for is that, when using conservation of energy and ignoring damping, don't assume the velocity at the equilibrium point is zero.
Here's a graph to show how the mass' position varies with time, for cases of damping and no damping. (Might replace with nicer looking graphs when I'm free again)

A: When you start with an unstretched spring and extend it by $H$ the energy stored in the spring is $\frac 12 kH^2$ and the work done by the gravitational field is $mgH = kH^2$ as the static equilibrium condition is $mg-kH=0$.
To understand where the "lost" energy ($\frac 12 kH^2$) has gone go back to holding the mass at the end of the unstretched spring and then let go of the mass.
The mass will accelerate downwards and at the static equilibrium position the gravitational field has done work equal to $kH^2$ on the spring mass system, the spring has elastic potential energy $\frac 12 kH^2$ and the mass has kinetic energy $\frac 12 kH^2$.
The mass overshoots the static equilibrium position and finally stops when the extension of the spring is $2H$.
At this position the work done by the gravitational field is $mg2H = 2kH^2$ and the spring elastic potential energy is also $ 2kH^2$.
The mass then moves upwards and executes harmonic motion about the static equilibrium position.
With friction this is damped harmonic motion and the mass finishes up stationary at the equilibrium position.
Going back to you holding the mass and making the mass stop at the static equilibrium position the "lost" energy is equal to the work done on you whilst you were stopping the mass at the static equilibrium position.
A: I think the error is your assumption that the equilibrium position (where $kx=mg$, ie the net force on the mass is zero) is at the lowest point where $v=0$.  In fact if the mass drops a total distance $H$ then the amplitude of oscillation is $\frac H2$, and the equilibrium position is $\frac H2$ below the point of release or above the lowest point.  $v$ is not zero at the equilibrium point, only at the highest and lowest points.  So $\frac {kH}2=mg$ hence $\frac 12 kH^2=mgH$.  The $PE$ lost from highest to lowest point is $mgH$, the energy stored in the spring at the lowest point is $\frac 12 kH^2=mgH$, which is exactly the same.
