# Trouble reconciling conservation of energy on vertical mass-spring

I'm picturing a vertical mass-spring system. This system is at rest at its equilibrium point. If we pull the mass downward, we remove potential energy from it. If we let go, ut will have no kinetic energy and a minimum potential emergy for a moment. The spring will have energy stored at this point.

As the mass goes upward to the equilibrium point, the spring loses all its stored energy, the mass gains maximum kinetic energy, and the mass gains potential energy. I find no trouble here. If we continue, the mass will continue upward to a distance from the equilibrium point approximately equal to the distance we pulled it down initially. Now, the block will have maximum potential energy, as it is highest off the ground, the block will have no potential energy, and the spring will have maximum potential energy stored again.

Comparing the top and bottom of the path, how can the block have the same kinetic energy and the spring have the same potential energy, but the block gains potential emergy as it goes upward? Where is this energy coming from? I'm assuming the spring is massless and we have no friction, air resistance, etc.

• You answered your own question: the "equilibrium point" (i.e. the point where the string is not stretched) can't be exactly in between the top and bottom of the spring's path. Instead, it's shifted. – knzhou Jan 29 '17 at 4:24

You need to add the energies together with some care.
Assume that the system is a spring, spring constant $k$ and a mass $m$ (and the Earth). Assume that the gravitational potential energy (GPE) and the spring potential energy (SPE) are both zero at the level of the bottom of the unextended spring.

The equilibrium position for the spring-mass system occurs when the condition $mg = kB$ is satisfied where $B$ the extension of the spring.

Let the amplitude of motion be $A$.

Now working out the energy of the system $E(x)$ when the kinetic energy is zero.

$E(B-A) = \frac 1 2 k (B-A)^2 - mg (B-A) = \frac 1 2 kA^2 - \frac 12 k B^2$

$E(B+A) = \frac 1 2 k (B+A)^2 - mg (B+A) = \frac 1 2 kA^2 - \frac 12 k B^2$

The energy of the system is the same at both positions.

A similar analysis could be done by assuming that the gravitational potential energy was zero at a different position, for example at $x=B$ where the energy of the system is now $\frac 1 2 kA^2 + \frac 12 k B^2$

The problem is in the statement " the mass will continue upward to a distance from the equilibrium point approximately equal to the distance we pulled it down initially". This isn't true for a vertical spring, precisely because of the gravitational potential energy. The difference between the distance you stretch the spring and the distance it compresses from its original hanging point is equal to mg/k, where m is the mass, g is acceleration due to gravity on Earth's surface, and k is the spring constant.

Notice that for a horizontal spring, the spring will compress exactly as much as it is stretched. Interesting to note that the stiffer the spring, the smaller the difference, but since it goes like (1/k) the difference is never zero. But, in zero gravity, the spring will act like a horizontal spring.