# Why does the force of a spring = -kx

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1. Mar 8, 2015

### Essence

I know this is an idealization, but I'm wondering if there is a reason this idealization works so well. Is it related to the coil of the spring or the bonds between each atom in the spring?

I have considered using
$F = \frac{kq_1q_2Nsin\left(\theta \right)}{r^2}$

Where:

$\frac{kq_1q_2}{r^2}$ would just be the force between two atoms in the spring

N would be the number of atoms

$\sin \left(\theta \right)$ would be the portion of the force vector parallel to the length of the spring

Unfortunately these variable seemingly can't be written as a function of each other so my attempt ends there (flop). Any ideas?

** Well actually that seems to suggest(ish) that $F = - k/x^2$

Last edited: Mar 8, 2015
2. Mar 8, 2015

### Staff: Mentor

The equation for the spring can be developed using stress-strain analysis. This would involve using the 3D version of Hooke's stress strain law. The primary kinematics of the deformation involve rotation of the wire cross sections relative to one another, and the kinematically associated axial displacements.

Chet

3. Mar 8, 2015

### Staff: Mentor

If you have a potential energy which is quadratic and has a minimum at x=0 then the force is -kx. Since all minima look approximately quadratic locally, you get -kx being a good approximation of the force locally also.

4. Mar 8, 2015

### Essence

Thank-you. Now I know where to look :). Much appreciated.

5. Mar 8, 2015

### Essence

Ok. But there could be a better approximation if the ideal function really wasn't quadratic (you'd just take the derivative of the real function). But I understand this line of thought. I'm guessing you dumbed it down for me so I would get a rough idea, which I appreciate. I'm guessing that there is a derivation that implies that the potential energy function will start to look like a quadratic more so than something else in the stress strain analysis, but it's probably complicated. Since a proof for that may span some text I can look that up.

Thanks by the way.

6. Mar 8, 2015

### Staff: Mentor

There is, but the stress-strain analysis essentially assumes F=-kx anyway, so it seems a little circular to me to use it to prove F=-kx. All you would be doing is proving that the thing you assumed at a microscopic level leads to a similar law at the macroscopic level.

Are you familiar with a Taylor series expansion? You can take any smooth function and expand in a Taylor series. Since we are talking about a potential energy the constant term can arbitrarily be set to 0, and since we are looking around a minimum, the coefficient of the x term is 0, so the first non-zero term is x^2 and the associated error is on the order of x^3. If we need a better approximation then we could take an x^3 term also for which the error would be on the order of x^4. We can build a force which is more and more complicated to get the accuracy we need, but it will always start out with the first approximation being quadratic and coming to Hooke's law.

Real springs don't follow Hookes law exactly, you would have to put in some of these corrections if you wanted to analyze the deviation from ideal.

7. Mar 8, 2015

### Essence

Oh forgot about Taylor. I now know where this comes from. Thank-you.

D