Finding equilibrium position of spring-mass system (Apologies for the elementary difficulty I'm having.) If we have two identical blocks connected by a spring and hung by a string, as in the figure below 

what is the displacement of the spring from equilibrium? 

I'm getting confused because equating energy stored in the spring with grav. pot. energy 
  $$mg\Delta x = \frac12 k (\Delta x)^2 \Rightarrow \Delta x = \frac{2mg}{k}$$
  gives a different answer from Hooke's Law
  $$mg = k\Delta x \Rightarrow \Delta x = \frac{mg}{k}.$$
  I know the latter is correct, but I'm not sure why. 

 A: Static equilibrium is achieved when the net force is zero and the system is not moving.  
The position of static equilibrium relative to that when the spring was unstretched is your second answer because both of those conditions are satisfied.  
You first method which equates energy does assume that the mass falls a certain distance and the reduction in gravitational potential energy is equal to the increase in elastic potential energy stored in the spring with the mass not moving ie having no kinetic energy.
However at maximum extension of the spring there is an upward force due to the spring on the mass so that is not the equilibrium position but it is one of the two positions where the mass is at a maximum excursion from the static equilibrium position.
Under the action of the upward force due to the spring the mass will start moving upwards, reach its original position where the spring is not extended and repeat the motion - the mass will oscillate about the static equilibrium position with an amplitude of $\frac{mg}{k}$.
A: If gravitational potential energy change is in the approximation form,
$$\Delta U = mg\Delta x,$$
($x$ has a coordinate direction which is positive opposite to but parallel to the gravitational field), the zero reference is arbitrary. The zero reference for the spring is not arbitrary.
This means that simply equating the $\Delta U$ for the spring and $\Delta U$ for gravitational force is not a valid approach to this problem. On the other hand one might use conservation of mechanical energy in the special case of no external force doing work:
$$K+\Delta U_{spring}+\Delta U_{g} = E$$
$$E_2-E_1=W_{ext}=0,$$
where the subscripts $1$ and $2$ designate two different locations along the path of motion.
It seems that what you have done is to set the zero points for both potential energies at the same vertical location. That's okay, but you have to remember that along the path, the kinetic energy will not necessarily be zero.  If you release the mass at the unstretched spring position with no kinetic energy, the mass will move until the kinetic energy is again zero, but that is not at equilibrium position.  In fact, the equilibrium position is half of that total distance. That's what your answer is telling you.
When the mass reaches the equilibrium position is has kinetic energy (it's moving), but the net force is zero so there is no acceleration to stop it at that position.
The equilibrium position can be found by finding the position of minimum total potential energy, or in this case, finding the maximum kinetic energy position.
