I am looking for textbooks or papers that provide an analysis for a series of damped springs. I am having a tricky time working out the details on my own. I know that if $F=-k\Delta x$ a series of springs can be described as follows: $$F=-k_{eq}*(\Delta x_1+\Delta x_2+..\Delta x_n), \text{where} \frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}+...\frac{1}{k_n} $$

If instead, we had $F=-kx -c \frac{\partial x}{\partial t}$, and we put $n$ such springs in series could we find an equivalent form of:

$$F=-k_{eq}*(\Delta x_1+\Delta x_2+..\Delta x_n)-c_{eq}(\frac{\partial x_1}{\partial t}+\frac{\partial x_2}{\partial t}+...+\frac{\partial x_n}{\partial t})$$

I couldn't find an analytical closed form solution like for the undamped case, and I was thinking about setting up a numerical model to approximate the equivalent coefficients, but I figure this is an old question and surely there is some good analysis on this set-up already performed. Can anyone help point me in the right direction for relevant reference material :)

  • $\begingroup$ It's hard to understand what you're trying to do. Usually the individual springs don't have significant damping forces associated with them, so when we put springs in series we don't significantly increase the damping force. Instead, damping is mainly associated with the mass – part of the oscillatory system that you seem to be ignoring! The mass usually has a much larger surface area at right angles to the oscillation direction than do the springs. $\endgroup$ Commented Apr 28, 2022 at 22:21
  • $\begingroup$ Hey Phillip thanks for responding! I guess you could think of this as looking at how a force is transmitted through a column of spheres being described by the linear-dashpot model. In this case the springs do have a significant damping because its really a proxy for energy lost during deformation. OR put another way, I AM interested in a system which is a point mass attached to a series of springs with a significant dissaptive term. It might not be common, but that's what Im interested in exploring ! $\endgroup$
    – Chair
    Commented Apr 28, 2022 at 22:42
  • $\begingroup$ I see. I'll think again. $\endgroup$ Commented Apr 29, 2022 at 6:52

1 Answer 1


I don’t have a particular reference to propose. However, in general, you won’t be able to fond an effective $k$ and $c$. The simplest way to see this is going to frequency space. I’ll choose the convention $x=\hat x e^{j\omega t}$. You this revert back to the undamped case with complex coefficients and an $\omega$ dependence. After calculation, you’ll find $\hat F = -K\hat x$ with: $$ K = \frac{1}{\sum \frac{1}{k_i+j\omega c_i}} $$ From this you can see that unless you have a dramatic simplification (like in the case where all the constants are equal), you can’t retrieve an effective damped oscillator. This is simply because if it were the case, the coefficient would be of the form $K=k_{eq}+c_{eq}j\omega$. However, in general, the coefficient is a rational function perhaps of degree $1$, but with a numerator possibly of degree $n$ in its irreducible form.

However, we can get some intuition from the low and high frequency regimes. For $\omega\to0$ you retrieve $K=k_{eq}+c_{eq}j\omega+o(\omega)$ with $$ k_{eq} = \frac{1}{\sum \frac{1}{k_i}} $$ $$ c_{eq} = k_{eq}^2\sum \frac{c_i}{k_i^2} $$

In the opposite extreme, for $\omega\to\infty$ you get $K=c_{eq}j\omega+ k_{eq}+o(1)$ with: $$ c_{eq} = \frac{1}{\sum \frac{1}{c_i}} $$ $$ k_{eq} = c_{eq}^2\sum \frac{k_i}{c_i^2} $$

and in between you typically get a complicated interpolation. This should come to no surprise as complicated analogue filters were made by simply putting impedances in series and parallel. Hope this helps and tell me if you need more details.


After reading @Philip Wood’s comment, I have to admit I did the calculations without worrying too much about the physical relevance. However, this kind of reasoning can be transferred to electronics and becomes a study of real capacitors (with resistivity) in series.

  • $\begingroup$ I think this is exactly what I'm looking for! I'm just working my way through your response. Have to brush up on frequency transforms real-quick and then I'll see if I can make sense of your work. Thank you! $\endgroup$
    – Chair
    Commented Apr 28, 2022 at 22:57

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