Getting 2 different answers when finding spring constant $k$ when gravity is involved

Question

OK, so I've been wracking my brain for the past hour trying to figure out how to calculate k in a problem like this:

A mass of 10 kg is attached to a spring hanging from the ceiling. It is released, allowed to oscillate, and comes to rest at a new equilibrium point 10 meters below the spring's natural length. What is the value of the spring constant k for this spring?

There's two approaches that are giving me different answers:

1: We can use forces: at equilibrium, the force of gravity will equal the spring force, so mg = kx. This gives a value of (10)(9.8) = k (10) or k = 9.8.

2: We can use potential energy. Before the mass is released, it has gravitational potential energy; at the new equilibrium, it has LESS gravitational PE, but more elastic PE, since it is now stretched. The elastic PE must have been converted from gravitational PE, so dPE (elastic) = dPE (gravitational). Since the change in height is the same as the change in stretch, the h in mgh = the x for spring stretch. So:

dPE (elastic) = dPE (gravitational), h = x

.5kx^2 = mgx

.5kx = mg

k = 2mg/x

Plugging in, we get k = 2(10)(9.8)/10 or k = 19.6, which is twice as much as the k found through the other method.

I must be missing something here, why am I getting two different values for k depending on which approach I use?

Answer 1

After you release the mass it accelerate downwards and at the static equilibrium position it has lost $mgx$ of gravitational potential energy whilst at the same time gaining $\frac12 kx^2$ elastic potential energy and $\frac 12mv^2$ kinetic energy.
So the mass overshoots the static equilibrium position and finally stops when the extension is $2x$.
At this position it has lost $mg(2x)$ gravitational potential energy whilst at the same time gaining $\frac12 k(2x)^2$ elastic potential energy.
Equating these two energies gives the same value of the spring constant as that found using the static extension $k=\frac{mg}{x}$.

The mass then oscillates about the static equilibrium position with reduced amplitude eventually stopping at the static equilibrium the spring-mass system having lost half its mechanical energy due to the resistive forces acting on the system.

Answer 2

Energy isn't conserved in this experiment -- you release the spring, and the spring oscillates, loses energy while settling down, and ends in a new equilibrium after having lost energy. you calculated the turnaround point of the block if the spring was frictionless, and correctly determined that this is further down than the new equilibrium.