Undergrad Hooke's Law: Changes for Large Displacements

[Lukeblackhill]

Morning,

I've come across this statement in Berkeley Physics Course, Vol.1 - Cp. 5 (pg.149):

"For sufficiently small displacements such a force may be produced by a stretched or compressed spring. For large elastic displacements we must add terms in higher powers of x to Eq. (,5.19): Fx = - Cx."

I tried to find on internet explanations about how this change in Hooke's law happens when large displacements are involved, but I couldn't find a convincing explanation. Could anyone here help me?

Cheers,
Luke.

 

[sophiecentaur]

I couldn't find a convincing explanation.

Here's an arm waving one. It is easier to appreciate if you are dealing with a straight wire, rather than a coiled spring but the principle can be extended (no pun intended).

The Young Modulus of a material is fundamental to the material and is given by Stress / Strain where stress is the force per unit area and strain is the proportional change in length. It is constant. When the displacement is significant, assuming that the volume of the wire doesn't change, the cross sectional area will reduce so the stress will increase. Hence the 'Force' that's used in Hooke's law does not result equal steps in Stress. Or, the other way round, although the stress may be proportional to the strain, the Force is not proportional to the extension.
The wire in your average helical (or leaf) spring doesn't change its thickness appreciably as it is extended over its normal operating range so Hooke works quite well enough.

 

[Dr.D]

Hooke's law is a linear relation between force and deformation. What the Berkeley folks are saying is simply that, for large enough deformations, the linear relation fails and a nonlinear relation (represented by a power series in the deformation) is required. This is essentially what Sophie said above.

 

[Dale]

There is some function F=g(x) that represents the force at an arbitrary displacement from equilibrium. So you can do a Taylor expansion about the equilibrium point. That will give

$g(x)=g(0)+g’(0)x+g’’(0)x^2+g’’’(0)x^3+...$

At the equilibrium point g(0)=0, by definition, so Hooke’s law is the first order expansion

$g(x)=g’(0)x+O(x^2)$

If you need higher accuracy then you just add more terms. There is actually no physics in the statement, just a little math