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Here we have Hooke's Law (not in vector form),

$$F=kx,$$

with $k$ being the spring constant and $x$ being the amount displaced from equilibrium (wiki).

My question deals with the fact that it seems like we are missing something here; let me explain. Say you buy two identical springs and they both have the same $k$ and have 10 windings per meter. Now you cut one meter off of one, hence the only difference is one is originally 3 meters and the other is 4 meters (hence their equilibrium locations).

They both have 10 windings per meter when un-stretched/compressed.

Now, logically, the force is also proportional to the amount of (windings per meter - original windings per meter) also even as a young boy I knew that the longer I made the yarn, the further I could stretch it (same force yet different $x$). That being said, both springs are now stretched one meter. According to Hooke's Law the same amount of force was used to do this. But this doesn't resound well with the fact that now one has 7.5 windings per meter and the other has 8 windings per meter.

So can someone please explain where the logic is no longer sound.

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  • $\begingroup$ just a comment - you must have to evaluate the "windings per meter" at equilibrium, not when they are stretched. When they are stretched, the force is different both because of the density of windings, but also the amount of compression. I don't think you want to conflate the two. $\endgroup$
    – levitopher
    Commented Apr 4, 2018 at 16:48

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The following should be instructive. Suppose you have two identical springs with spring constant $k$. If you connect the springs in series the combined spring constant is $k/2$. If you combine them in parallel the new constant is $2k$.

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  • $\begingroup$ Wow, yeah, that cleared up the confusion. Thanks! $\endgroup$
    – Josh
    Commented Apr 4, 2018 at 17:32
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Springs of different geometries (windings, spacing, thickness, length etc.) have different spring constants $k$. The two springs you are describing would have different $k$'s. Hooke's Law would correctly show what your intuition tells you: that they show different spring forces for the same extension.

Were the spring forces equal in spite of different geometries, then they must differ on other parameters (material, hardness, temperature or alike).

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    $\begingroup$ Funny enough, this is basically the same as I just posted; but you seem to take the approach of "different k's" while I interpreted it as "same k" with different geometry. $\endgroup$
    – JMac
    Commented Apr 4, 2018 at 17:11
  • $\begingroup$ @Steeven So if I have a spring with a spring constant $k$ and cut it in half, I have changed the spring constant? I thought the spring constant had more to do with stiffness (which doesn't depend on length). $\endgroup$
    – Josh
    Commented Apr 4, 2018 at 17:17
  • $\begingroup$ @Josh The way to define spring constants is by extending the spring, measuring the force and devide those two numbers. Anything about this spring that causes the force, that being stiffness, geometry, hardness etc, is then automatically included in the spring constant. Stiffness is not the only dependency, no. $\endgroup$
    – Steeven
    Commented Apr 4, 2018 at 18:06
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I believe you're confusing yourself by trying to consider the winding numbers and such, when you already have a spring constant.

The spring constant depends on many factors, such as the diameter of the spring material, the diameter of the helix, the pitch of the helix, the material itself, how the ends are made and connected, etc. It also doesn't necessarily vary linearly with these variables.

You can have two completely different springs with the same k value; completely different length, diameter, material, windings, etc. Two springs of different rest lengths and the same spring constant do not necessarily have much in common besides the spring constant. The two springs will likely have different ranges where Hooke's law is (approximately) accurate as well.

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  • $\begingroup$ I agree, I think this best describes the problem. In order for two springs to have the same k but different windings, something else has to be different too. $\endgroup$
    – Maury Markowitz
    Commented Apr 4, 2018 at 17:10
  • $\begingroup$ @JMac I clarified my question because I didn't think $k$ was dependent on length (so if you had identical springs and cut one in half you would be changing it's spring constant?). $\endgroup$
    – Josh
    Commented Apr 4, 2018 at 17:31
  • $\begingroup$ @Josh Absolutely. You might be interested in springs in series and parallel. If you take two springs with the same k and put them in series (end to end) the resulting combination would actually be far less stiff. It's because you're bending the material uniformly; the bigger length of material, the less you have to move every individual part to get the whole spring to move the same amount. $\endgroup$
    – JMac
    Commented Apr 4, 2018 at 17:37
  • $\begingroup$ @JMac Thanks, I actually saw my2cts response before I saw your reply and completely understand now. For some reason I was all bent out of shape not knowing what was going on! $\endgroup$
    – Josh
    Commented Apr 4, 2018 at 17:43
  • $\begingroup$ @Josh It's just one of those things that can intuitively mislead you when yo don't really dive into it. Thankfully it's usually pretty easy to understand what really happens; you just have to first recognize that you were thinking of it the wrong way (like you've done here). It's the kind of thing you don't really think of until it comes up. $\endgroup$
    – JMac
    Commented Apr 4, 2018 at 17:46
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Here is a picture of your two springs with the same stretching force $F$ applied to both springs.

enter image description here

Each metre of spring is extended by $x$ so the longer spring has a total extension of $4x$ and a spring constant of $\frac{F}{4x}$ and the shorter spring has a total extension of $3x$ and a spring constant of $\frac{F}{3x}$.

The spring constant of the shorter spring is larger than for the longer spring.

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