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?
 A: 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. 
A: 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.   
