Spring force that grows not proportional to displacement

Do springs that change their proportionality 'constant' as a function of their winding number per length? I want a spring that obeys $F(x)=kx^2$ instead of $F(x)=kx$.

To my intuition if I decrease the winding/length ratio linearly from one end to the other end of the spring, I should get something like $F(x)=kx^2.$ Is this true?

• Small side note, you want your forces to be $F(x)=-kx$ or $F(x)=-kx^2$, otherwise you don't have a spring, you have a force that drives whatever experiences the force off to infinity. – Aaron Stevens Jul 23 '18 at 14:12
• @AaronStevens His force law works if $x<0$ :-) – garyp Jul 23 '18 at 20:10
• @garyp Yes this is true for just the squared one I suppose, but I don't think this is what the goal is. – Aaron Stevens Jul 23 '18 at 20:28
• @AaronStevens ... and your force law does not work for $x<0$. I'm just pointing out pedantically, and with some attempt at being light-hearted, that your math expression is not quite right, which is what you were saying about his. But I agree totally that pedantry about math expressions does not help the OP – garyp Jul 24 '18 at 2:38
• @garyp Ah yes. Silly oversight on my part. Thanks for pointing it out. – Aaron Stevens Jul 24 '18 at 9:23

the way this is done in practice is to use a spring with variable winding pitch, in which upon compression the windings that are on the close-wound end progressively compress into solidity and stop deflecting. this reduces the active length of the spring, which makes it stiffer.

This is a common trick to stiffen the suspension response of a pickup truck, where you want stiff springs under a heavy load and softer springs in the unloaded state.

It looks like a spring with a variable pitch won't work as you expect, i.e., it'll still be linear.

To prove it, we can, for example, take a compression spring and treat it as a number of short springs connected back to back.

When this spring is compressed, all short springs will experience the same compression force and will shrink by $\Delta x_i$, which will be proportional to that force.

Since the total shrinkage of the spring, $\Delta x$, is a sum of all individual shrinkages, it will also be proportional to the applied force.

Obviously, this will be the case even if spring constants of individual short springs are not the same. Therefore, a spring with a variable pitch may have a different overall spring constant, but it will remain linear, i.e. $F=-k'x$ will hold.