# Springs in parallel store less energy than springs in series ... but where does the excess energy go?

Two springs in parallel with a weight of 15 N will stretch a certain amount, while the same two springs arranged in series will stretch more under the same weight and so store more energy. I'm fine with this part including the calculations.

However, I started to wonder, since you have the same number of springs, and the same weight, the two systems must have the same energy overall. But if less energy is stored in the parallel arrangement, where does the difference in energy go? I suspect it is gravitational potential energy in the weight, but I can't find anywhere that explicitly says this.

Any help would be much appreciated!

• So you need to be careful is stating the situation. "Spring in parallel [...] than in series when subject to the same force". Consider the situation when they are stretched through the same distance to see why the addendum is important. Commented Jun 18, 2018 at 14:59
• Consider if you are hanging the weight from the springs. In the parallel case the weight does not descend as far, therefore it's movement does not release as much energy. Commented Jun 18, 2018 at 15:08

No, there is no reason why the stored energy should be the same in both cases. The only requirement is that the total force supplied by each system of springs is the same, because the two systems are supporting the same load, and each is in equilibrium.

The difference in stored energy does not 'go' anywhere because these are two separate cases. You are not converting one case into the other.

If you did start with the load supported by the springs in series and converted this to the load supported by the springs in parallel, the difference in elastic energy would not in fact equal the difference in gravitational PE of the load!

The tension in each spring would have to be reduced from $W$ to $\frac12 W$ where $W$ is the weight of the load, and the extension of each reduced from $x$ to $\frac12 x$. The elastic energy stored in a spring with tension $W$ and extension $x$ is $\frac12 W x$. The total elastic energy stored in the two springs is initially (ie in series) $2\times \frac12 Wx=Wx$ and finally (ie in parallel) $2\times \frac12 (\frac12 W)(\frac12 x)=\frac14 Wx$. The decrease in elastic PE is $\frac34 Wx$.

After the change from series to parallel, the springs initially supply a total upward force of $2W$ while the weight supplies a downward force of $W$. An additional downward external force is required, starting at $W$ and decreasing to $0$, to prevent the system from gaining kinetic energy as the tension is reduced. The work done against this external force equals the KE which the load would have acquired when it reached the equilibrium position had it been released. In that case it would have overshot the equilibrium position and oscillated indefinitely.

The work done on the load (average force x distance) by the two springs is $W\times (\frac12 x)=\frac12 Wx$, which is the increase in its gravitational PE. The work done against the external force is $(\frac12 W)(\frac12 x)=\frac14 Wx$. The total work done by the two springs is $\frac34 Wx$.

To summarise : the two springs lost total elastic energy $\frac34 Wx$ of which $\frac12 Wx$ increased the gravitational PE of the load while $\frac14 Wx$ was done against an external force.

But if less energy is stored in the parallel arrangement, where does the difference in energy go? I suspect it is gravitational potential energy in the weight, but I can't find anywhere that explicitly says this.

Let's work it out explicitly.

For two springs in parallel, the displacements (from equilibrium) are identical and so

$$F_{||} = -(k_1 + k_2)\Delta z_{||}$$

For two springs in series, the restoring forces are identical and so

$$F_\Sigma = -\frac{1}{\frac{1}{k_1}+\frac{1}{k_2}}\Delta z_\Sigma$$

But, in either case, the total restoring force is the same for a given hanging mass:

$$F_{||}= F_\Sigma$$

Thus

$$\frac{\Delta z_\Sigma}{\Delta z_{||} } = \frac{(k_1+k_2)^2}{k_1k_2}$$

For example, let $k_1=k_2$ and see that $\Delta z_\Sigma = 4\Delta z_{||}$

But the change in gravitational potential energy of the mass is $W = mg\Delta h$ and so, explicitly

$$W_\Sigma = 4W_{||}$$

It's not true that with the same number of springs and the same weight, the two systems must have the same energy overall. Counterexample: if you hang the weight on one spring, you can attach as many (ideal, weightless) springs as you like to the bottom of the weight and it should still have the same energy.

The two weights in the systems move by different amounts - the difference corresponds to the potential energy stored in the springs.