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I was solving a practice Physics GRE and there was a question about springs connected in series and parallel. I was too lazy to derive the way the spring constants add in each case. But I knew how capacitances and resistances add when they are connected in series/parallel. So I reasoned that spring constants should behave as capcitances because both springs and capacitors store energy.

This line reasoning did give me the correct answer for how spring constants add, but I was just curious if this analogy makes sense, and if it does, how far one can take it. That is, knowing just that two things store energy, what all can you say will be similar for the two things.

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4 Answers 4

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Electrical analogies of mechanical elements such as springs, masses, and dash pots provide the answer. The "deep" connection is simply that the differential equations have the same form.

In electric circuit theory, the across variable is voltage while the through variable is current.

The analogous quantities in mechanics are force and velocity. Note that in both cases, the product of the across and through variables has the unit of power.

(An aside, sometimes it is convenient to use force and velocity as the across and through variables respectively while other times, it is more convenient to switch those roles.)

Now, assuming velocity is the through variable, velocity and electric current are analogous. Thus, displacement and electric charge are analogous.

For a spring, we have $f = kd \rightarrow d = \frac{1}{k}f$ while for a capacitor we have $Q = CV$.

For a mass, we have $f = ma = m\dot v $ while for an inductor we have $V = L \dot I$

Finally, for a dashpot, we have $f = Bv$ while for a resistor we have $V = RI$.

So, we have

$\frac{1}{k} \rightarrow C$

$m \rightarrow L$

$B \rightarrow R$

For a nice summary with examples, see this.

UPDATE: In another answer, RubenV questions the answer given above. His reasoning requires an update.

Alfred Centauri's answer is not correct. The analogy he mentions is true, but it is irrelevant as it does not tell you anything about components in series or in parallel.

In fact, it is relevant and it does tell you everything about components in series or in parallel. Let's review:

When two circuit elements are in parallel, the voltage across each is identical.

When two circuit elements are in series, the current through each is identical.

This is fundamental and must be kept in mind when moving to the mechanical analogy.

In the mechanical analogy where a spring is the mechanical analog of a capacitor:

force is the analog of voltage

velocity is the analog of current.

Keeping this in mind, consider two springs connected in mechanical parallel and note that the velocity (rate of change of displacement) for each spring is identical.

But recall, in this analogy, velocity is the analog of current. Thus, the equivalent electrical analogy is two capacitors in series (identical current).

In series, capacitance combines as so:

$\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$

With the spring analogy, $C \rightarrow \frac{1}{k}$ , this becomes:

$k_{eq} = k_1 + k_2$

The key point to take away from this is that mechanical parallel is, in this analogy, circuit series since, in mechanical parallel, the velocity (current) is the same, not the force (voltage).

For example, consider dash pots (resistors). Two dash pots in "parallel" combine like two resistors in series, i.e., the resistance to motion of two dash pots in "parallel" is greater then each individually.

Now, if the roles of the analogous variables are swapped, if force is like current and velocity is like voltage, then mechanical parallel is like circuit parallel. However, in this analogy, mass is like capacitance.

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You cannot draw this conclusion from a naive "they both store energy" argument (though you can use it as a mnemonic if you find it helps you).

Capacitances are usually measured in farads which corresponds to coulombs squared per joule while spring constants are usually measured in newtons per metre equivalent to joules per metre squared.

If you did your calculations with joules in a consistent position in the unit, for example using using reciprocal capacitance (elastance?), I think you would find the parallel and series composition calculations were reversed (and reciprocal capacitance would behave more like resistance and inductance).

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I intended to post this as a comment to Alfred Centauri's above post, but I do not have enough reputation for this yet, so I will state it here:

Alfred Centauri's answer is not correct. The analogy he mentions is true, but it is irrelevant as it does not tell you anything about components in series or in parallel. To prove my point: as he illustrates, the analogy connects C with 1/k, however as the OP noted, both C and k have the same addition rules, whereas the C <-> 1/k link would suggest that they should be opposite.

The RLC-circuit vs. spring analogy is based on both differential equations having the same structure, however the DE determines the evolution of the system for fixed constants (R, L, C, or gamma, m, k) and there is no reason to suspect that the analogy also works for how you add up many components into one effective component, and indeed, the analogy breaks down in those cases.

In other words, to understand why capacitors add the way they do, one must go back all the way to their definitions.

For a spring: F = kx, where x is the stretching from equilibrium, and F is the restoring force. For a capacitor: CV = Q, where Q is the amount of charge on one side, and V is the restoring potential. One can again note the C <-> 1/k link (i.e. 1/k and C are the factors in front of the restoring force, if you will). The reason that 1/k and C do not add in the same way, however, is because of the physically different behaviour of the restoring forces:

Consider two springs in parallel: a certain stretch x will now cause much more restoring force, both springs independently trying to undo the stretching; this translates itself into $k_{net} = k_1 + k_2$. However, if one considers two capacitors in parallel, then putting a certain amount of charge Q on them will lead to a smaller voltage difference than for the one-capacitor case, since the Q is now diluted over two capacitors (*); taking into account that C is the factor in front of the potential, this is summarized in $C_{net} = C_1 + C_2$ (C is like a measure of how easily one puts charge on, whereas k is a measure of how difficult it is to extend the spring; the reason they add the same way is because adding more springs in parallel makes stretching harder, whereas adding capacitors makes adding charge easier).

The conclusion is as follows: indeed, C and 1/k are analogous, but it is rather despite this connection than thanks to it that they add in similar ways, the reason for this being in their definitions and the opposite behaviour of the restoring forces in terms of which these constants are defined.

(*) Another reason is that parallel voltage differences are not cumulative, whereas parallel forces are.

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I believe it is the case that you've reasoned incorrectly here. When two capacitors are in parallel, the voltage across each is identical. When two capacitors are connected in series, the current through each is identical. This is fundamental. Now, when two springs are connected side by side, the velocity is the same, not the force. So, in the analogy where velocity is like current, the two springs are not connected in "parallel", i.e., their "currents" are the same, not their "voltage". In series, capacitances don't add. The answer I gave is correct. – Alfred Centauri Nov 5 '12 at 13:16
In series, $C_{eq} = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2}}$. Substituting $\frac{1}{k}$ for C, we get $\frac{1}{k_{eq}} = \frac{1}{k_1 + k_2}$ or $k_{eq} = k_1 + k_2 $ as desired. – Alfred Centauri Nov 5 '12 at 13:20
When I said "The analogy he mentions is true, but it is irrelevant" I was referencing to your $$\frac{1}{k} \to C$$ analogy you listed at the end of your post (in what seemed to be the conclusion), and I still stand by my statement (based on your post back then). That being said, the aforementioned analogy is relevant if you also include the analogy "series <-> parallel (and vice versa)" which you indeed discussed in the update to your post (so I have no more problems with your modified answer, so thank you for expanding!). – Maxwell Bunyan Dec 7 '12 at 10:46

Stiffnesses add when the springs are connected "in parallel" (side by side): $F = F_1 + F_2$; $x_1 = x_2 = x$; $k_x = k_1 x + k_2 x$; $k_x = k_1 + k_2$.

Compliances add when the springs are connected "in series" (end to end): $F_1 = F_2 = F$; $x = x_1 + x_2$; $s_ F = s_1 F + s_2 F$; $s_ = s_1 + s_2$.

This is what I mean by "the math changes"-- the physics is obviously the same, but the form of the equations is different. I am calling into question your ability to jump from "they store energy" to "they add in parallel (and add as inverses in series)" by pointing out that with the use of compliances rather than stiffnesses, the opposite equations (representing of course the same physics) are obtained.

As for inductors-- why are capacitors a better analogy than inductors?Hm. How does your reasoning stand up to the following objections?

We are used to working with the stiffness or rate of springs k, defined by $F = k x$. But it is no less reasonable to work with the compliance $s = 1 / k$, defined by $x = s F$. This does not change the fundamental fact that springs store energy, nor that they can be viewed as the electrical analogues of capacitors. But the math changes!

Alternatively, consider inductors, which also store energy, and are typically characterized by the inductance L defined by $V=L. di/dt$. Do inductances add like capacitances or like resistances?

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protected by Qmechanic Mar 18 '13 at 13:21

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