Consider a helix that is non-rigid in which there are several evenly spaced springs that are parallel to the helical axis. The springs connect subsequent turns in the helix and all have the same spring constant. Assuming there is no damping, what would be the effective spring constant of the net spring system?

Any general ideas would be appreciated. (Such as the system should be regarded as a mixture of parallel/series springs, etc.)Poorly done paint visual of the problem


1 Answer 1


In general, problems like this are best approached by determining the relationship between displacement (the length by which the entire system extends) and potential energy. For a hook attached to the end of the multi-sping, for small displacements it will be the case that (with potential energy $u$, displacement $x$ and spring constant $k$), $$ U = \frac{1}{2}kx^2, k=\frac{2U}{x^2} $$ We are guaranteed this relationship for small displacements because I can choose my $x=0$ origin point to be in a minima of $U(x)$, and then I can take a second order expansion of $U(x)$, throw away the constant because potential is only determined up to a constant anyways, have no linear term because I chose $x=0$ to be a critical point, and be left with the quadratic term above. Now, since potential is additive, all I have to do is figure out how the springs will elongate as I stretch the system, and then I may simply apply Hooke's law individually to each of them.

I can see one case where this is easy: when the big helix extends and contracts while keeping its helical shape. This would happen when the big string is much stiffer than the small springs (so that they don't bend it much), and also when the small springs are directly above one another so that each point on the main helix is pulled equally from above and below. Finally, this would happen in the limit where the small springs were very dense on the main helix. In all of these cases, the main helix would extend as a sping, so that,

$$ K = \frac{2}{X^2}\left(\frac{1}{2}k_0X^2+\sum^{N}_{n=1}\frac{1}{2}k_nx_n^2\right) = k_0+\sum^{N}_{n=1}\frac{x_n^2}{X^2}k_n = k_0+k_1\frac{Nx_1^2}{X^2} $$ Where $K$ is the sping constant of the whole system, $k_0$ is the sping constant of the main helix, $X$ is the total displacement, $x_n$ is the displacement of the $n$th small spring, and $k_n$ is the spring constant of the $n$th small sping. The last step was taken assuming that each spring had the same spring constant, and that the stretching of the helix extended them all equally. The ratio $Nx_1/X$ is a system constant that can be found from the geometry of the helix. With a moments thought, we may realize that $x_1/X$ must be equal to the ratio of the lengths of the main spring and the small springs. So, if $N$ is the number of springs, $L$ is the length of the big spring, $l$ is the lengh of the small springs, and $k_0$ and $k_1$ are the spring constants of the big and small springs respectively, then the combined spring constant $K$ must be,

$$ K = k_0+k_1\frac{Nl}{L}\frac{l}{L} $$ Where it is noteworthy that $Nl$ is the combined length of every small spring.

Outside of that simple case, the main spring may be twisted out of its helical shape - leaving a much more complicated problem.

  • $\begingroup$ Thank you for the thorough response. One thing I am not seeing right now is why the term $\frac{x_1}{X}$ isn't squared in the last equality. Would you mind elaborating on this? $\endgroup$
    – Kylexi
    Oct 27, 2018 at 15:51
  • $\begingroup$ Because I was tired and forgot to type it. It's fixed now. :) $\endgroup$ Oct 27, 2018 at 17:10
  • $\begingroup$ Understandable :) thank you so much for the input. $\endgroup$
    – Kylexi
    Oct 27, 2018 at 19:15

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