# Energy stored in series springs vs energy stored in parallel springs

I'm trying to understand why the energy stored in a set of series springs is different from the energy stored in parallel springs.

We know that the elastic potential energy stored in a spring system is as follows: $E=\frac{1}{2}k(\Delta l)^2$.

So imagine we have two identical springs each with a spring constant ($k$) of 85 Nm-1

In one system, they are in parallel, supporting a load of 15 N. In another, they are in series, also supporting 15 N.

So the combined spring constant in the parallel system is equal to $2k$, which is 170 Nm-1.

The combined spring constant in the series system is equal to $\frac{k}{2}$, which is 42.5 Nm-1.

Using the energy equation above, the energy stored in the springs is different for both systems, since $k$ is different and so is $\Delta l$.

I understand it from a mathematical point of view, but in terms of energy transfer, I don't understand why the elastic potential energy varies. Some energy is of course lost as thermal energy, but why don't both systems lose the same amount of energy or even gain the same amount of energy in the first place?

• My "common sense" tells me, (with out going trough all the lengthy question an answers,) the force working on both systems is the same, but the displacement isn't, as it's harder to press against parallel springs. W = F*S (very simplified but good enough), therefore work is not the same, so the energy stored isn't. You don't have to think about the spring physics at all. Just imagine what is the difference in potential energy of the weight you put on it. Jan 14, 2017 at 16:29
• Thank you everyone for your answers. It now makes a lot more sense. So the forced applied may be the same, but that does not necessarily account for the work done. So more energy is stored in the series system because the springs are stretched further. I guess the series springs can be thought of as one longer spring, in which case it makes sense that it would have to store more energy to carry the load. Jan 15, 2017 at 11:24

We know that the elastic potential energy stored in a spring system is as follows: $E=\frac 12k\Delta l$.

You are missing a power of 2 here: $E=\frac 12k(\Delta l)^2$

Using the energy equation above, the energy stored in the springs is different for both systems, since $k$ is different and so is $\Delta l$.

• If $k$ was different, then yes: the stored energy must also be different.
• If $\Delta l$ was different, then yes: the stored energy must also be different.

If both are different, then you don't know. You can't conclude that the energy is different then. $k$ is halved, so what if $\Delta l$ is $\sqrt 2$ times as big? Then the same energy is stored.

• Parallel springs help each other in carrying the weight. They can together in total exert double the force. Therefore you can consider this two-spring system as one equivalent spring with double the stiffness (since spring force $F=kx$ doubles with double stiffness $k$ if you compress it just as much): $$k_{parallel}=2k$$ Since they each carry half of the load, they are each only compressed half as much as what they would have been alone: $$\Delta l_{parallel}=\frac12\Delta l$$

• Series springs have no help from each other. If you place a spring on a rigid table or on another spring makes no difference, since you look at the final situation (after compression is done). Therefore the top spring carries the whole load, and the second carries the whole load as well (plus the top spring's weight, but that's usually assumed mass-less). Each is compressed $\Delta l$, so in total the equivalent compression is double: $$\Delta l_{series}=2\Delta l$$That is the same as having an equivalent only half as stiff spring: $$k_{series}=\frac12 k$$

Conclusion:

• $E_{one\;spring}=\frac12k(\Delta l)^2$
• $E_{parallel}=\frac12(2k)(½\Delta l)^2=\frac12\cdot2\cdot\frac14k(\Delta l)^2=\frac12E_{one\;spring}$
• $E_{series}=\frac12(½k)(2\Delta l)^2=\frac12\cdot\frac12\cdot 4k(\Delta l)^2=2E_{one\;spring}$

Yes, the energies stored are indeed different. Because the displacements are different - you store energy in a spring by compressing/extending it, so different compression/extension means different energy stored.

• When you use the word 'compression' to represent $\Delta l$, does that also represent 'stretching' the spring? Jan 15, 2017 at 11:39
• @kashveyron Yes it does. I say "compression" because that is the case in your example. I have now made a small edit to the last paragraph to make it clear that it counts for both compression and extension. Jan 15, 2017 at 11:57
• Oh I see, so it does not matter in which direction the spring is manipulated, as long as there is some displacement. Thank you for clarifying that. Jan 16, 2017 at 6:59
• Exactly. Since all the formulae work for both cases, also this whole discussion does. Jan 16, 2017 at 10:37

I will try to sanswer the question in as much detail as possible. The energy stored in a spring is NOT $\frac{k\Delta l}{2}$. The spring is modeled as a source of simple harmonic motion. To do that we need equations of the form, $\frac{\mathrm{d}^2x}{\mathrm{d}t^2}=-\lambda x$. So, the easiest way to do it is. $$\vec{F}=-k\vec{x}$$ Now, the potential energy in case of a conservative force is used to define the force in that aspect. $$\vec{F}=-\nabla U\\ \Rightarrow \int\mathrm{d}U=\int\vec{F}.\mathrm{d}\vec{x}=\int_{0}^{\Delta l}kx\mathrm{d}x=\frac{k(\Delta l)^2}{2}\\$$ The difference between the series system and the parallel system is the displacements in each spring. Suppose you have a system of two springs $k_1$ and $k_2$ in series. Then the displacement of each spring to a Force $F$ will be $\frac{F}{k_1}$ and $\frac{F}{k_2}$ respectively and the total displacement which is the sum of the two displacements is modeled as $F/K_{eff}$. That is where you get your version of $k_{eff}$ from in a series system. $$\frac{F}{k_1}+\frac{F}{k_2}=\frac{F}{k_{eff}}$$ Now, for a parallel system the displacement that each spring moves through to balance the force IS the SAME. So, the equation is as shown below. $$k_1x+k_2x=k_{eff}x$$. Now, since the $K_{eff_p}$ and $k_{eff_s}$ (parallel and series spring constants for two springs) are different. The displacement needed to balance similar forces would be different. $$F=k_{eff_p}x_p=k_{eff_s}x_s\\ \text{iff}\quad k_{eff_p}\neq k_{eff_s}\\ \Rightarrow x_p \neq x_s$$. This difference in displacements causes a difference in the gravitational potential energies of the bodies when suspended. (Remember, a body at height $h$ has a gravitational potential energy wrt the earth equal to approximately $U=mgh$). With different values of $h$ because $x_p\neq x_s$, the gravitational potential energy difference reflects in the difference of your sprinbg potential energies due to energy conservation.

The math has been explained very well, but you said that you already understood the math, and it was just the apparent conceptual contradiction about energy transfer that bothered you.

The problem is very subtle and interesting. If you put your "gut reaction" into words, it would probably say that since the 15N force supported is the same for both spring arrangements, they do the same amount of work.   But at the same time, your physics-trained mind knows better.   It knows that work, or energy transfer, only occurs when force is exerted through a distance.

When you hold a book motionless at arm's length out in front of you, no work is done on the book.   Yet after a minute, your body feels like it's doing a lot of work, that energy is being transferred, and it is--in your muscle cells.   So your body is not wrong.   But no energy is being transferred to the book.

These "common sense" things that our bodies and senses tell us are about the world around us are reinforced thousands of times by everyday experiences before we study physics.   They are hard to get rid of.   Aristotle's ideas that heavy things fall faster than light things, and that the force of the hand is still on a rock after you throw it, were believed by many for about 2000 years before Galileo stepped in.

There is no reason why the elastic energy stored in the springs should be the same for series and parallel arrangements. The load $F=mg$ is the same, but the extensions and elastic energy are not, because the combinations have different effective spring constants.

The energy stored in each case is $\frac12 k'x^2=\frac{F^2}{2k'}$ where $k'$ is the effective spring constant. The parallel combination of springs is stiffer and has a higher effective spring constant $k'=2k$ than the series combination which is $k'=\frac12k$. The series combination stores 4x as much energy as the parallel combination. ($k$ is the spring constant for each individual spring.)