# Real Hooke's law for springs

jaumzaum
I was wondering which equation do springs obey better:
$$F=-kx$$
$$F=-k ln(x/x_0)$$

The first is Hooke's law, but the second comes when we consider the relative deformation instead of the absolute deformation. I am asking because I haven't seen any website stating the second equation, I just got it when I integrated the relative strain equation.

Homework Helper
Gold Member
How is this relative deformation defined? Is ##x## zero when there is no deformation? If so, what happens to the logarithm?

• jaumzaum
How is this relative deformation defined? Is ##x## zero when there is no deformation? If so, what happens to the logarithm?

How is this relative deformation defined? I would say ##\Delta x/x##
Is ##x## zero when there is no deformation? I would say so
If so, what happens to the logarithm? It approximates a linear function. But only when x->0, right?

Homework Helper
Gold Member
How is this relative deformation defined? I would say ##\Delta x/x##
I am trying to understand what you mean by this ratio. Say I have a spring of relaxed length 10 cm and I extend it to 11 cm. What are ##x##, ##\Delta x## and ##F##?
Say I compress the same spring from 10 cm to 9 cm. What are ##x##, ##\Delta x## and ##F## now?
Is the magnitude of the force symmetric about the equilibrium position?
Is ##x## zero when there is no deformation? I would say so
If so, what happens to the logarithm? It approximates a linear function. But only when x->0, right?
What about the equiibrium position itself? Have you tried calculating ##\ln\left(\dfrac{0}{0}\right)##? L' Hospital's rule doesn't work and you cannot simplify the fraction by dividing numerator and denominator by zero so that the ratio is equal to one.

• jaumzaum
jaumzaum
I am trying to understand what you mean by this ratio. Say I have a spring of relaxed length 10 cm and I extend it to 11 cm. What are ##x##, ##\Delta x## and ##F##?
Say I compress the same spring from 10 cm to 9 cm. What are ##x##, ##\Delta x## and ##F## now?
Is the magnitude of the force symmetric about the equilibrium position?

What about the equiibrium position itself? Have you tried calculating ##\ln\left(\dfrac{0}{0}\right)##? L' Hospital's rule doesn't work and you cannot simplify the fraction by dividing numerator and denominator by zero so that the ratio is equal to one.
Say I have a spring of relaxed length 10 cm and I extend it to 11 cm.
I would say for Hooke's law: ##F=-k_1(1)=-k_1##
Using the logarithm formula: ##F=-k_2 ln(11/10)##

Say I compress the same spring from 10 cm to 9 cm.
Hooke's law: ##F=k_1##
Logarithm formula: ##F=-k_2 ln(9/10)##

Staff Emeritus
Is x zero when there is no deformation? I would say so

But what does your equation say? And if you and your equation say different things, what is right?

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• jaumzaum
jaumzaum
But what does your equation say? And if you and your equation say different things, whatt is right?

Sorry for the equations, I really meant:

Hooke: ##F=-k_1\Delta x = -k(x-x_0)##
Logarithm: ##F=-k_2 ln(x/x_0)##

Where k1 and k2 are different constants with different units and x is the length of the spring in a given instant and x0 is the initial length of the spring.

When ##\Delta x<<x_0, ln(x/x_0) \approx \Delta x/x_0##
The two equations are similar:
##-k_1 \Delta x = -k_2 \Delta x /x_0 ##
##k_1 =k_2/x_0##

Mentor
The correct equation for a linear spring should be ##F=-k(x-x_0)##, where ##x_0## is the unextended length and x is the extended. For the logarithmic strain spring, the corresponding relationship is ##F=-k'\ln{(x/x_0)}##. But how are the two related? For the linear spring, we can also express the force in terms of the linear strain: $$F=-kx_0\frac{(x-x_0)}{x_0}$$For the logarithmic strain spring, we can write: $$F=-k'\ln{\left(1+\frac{(x-x_0)}{x_0}\right)}$$If we expand the logarithmic strain equation to linear terms of ##(x-x_0)/x_0##, we obtain: $$F\approx -k'\frac{(x-x_0)}{x_0}$$So, taking ##k'=kx_0##, the two approximations provide identical predictions for small strains. So one would choose to work with the one (if either) which best matches the behavior of the actual spring being considered.

• vanhees71, jaumzaum, etotheipi and 1 other person
jaumzaum
The correct equation for a linear spring should be ##F=-k(x-x_0)##, where ##x_0## is the unextended length and x is the extended. For the logarithmic strain spring, the corresponding relationship is ##F=-k'\ln{(x/x_0)}##. But how are the two related? For the linear spring, we can also express the force in terms of the linear strain: $$F=-kx_0\frac{(x-x_0)}{x_0}$$For the logarithmic strain spring, we can write: $$F=-k'\ln{\left(1+\frac{(x-x_0)}{x_0}\right)}$$If we expand the logarithmic strain equation to linear terms of ##(x-x_0)/x_0##, we obtain: $$F\approx -k'\frac{(x-x_0)}{x_0}$$So, taking ##k'=kx_0##, the two approximations provide identical predictions for small strains. So one would choose to work with the one (if either) which best matches the behavior of the actual spring being considered.

Thanks Chestermiller! I was thinking the linear e quation was just an approximation of the logarithmic one. So by what I understood there are more than one type of spring right? Some of them obey better the logarithm equation while others obey better the linear equation? When one say that one spring have a completely elastic behavior, will this spring be a linear or a logarithm strain spring (what law will it obey better ?). Also, in real life springs, what law they tend to obey better?

Mentor
Thanks Chestermiller! I was thinking the linear e quation was just an approximation of the logarithmic one. So by what I understood there are more than one type of spring right? Some of them obey better the logarithm equation while others obey better the linear equation? When one say that one spring have a completely elastic behavior, will this spring be a linear or a logarithm strain spring (what law will it obey better ?). Also, in real life springs, what law they tend to obey better?
There are many strain measures out there, not just these two, and their common characteristic is that they all reduce to the linear strain in the limit of small strains. Fortunately, all materials reduce to linear behavior in the limit of small strains. The other strains are derived from large deformation theory for non-linear elasticity. They satisfy the requirements of isotropy and symmetry in a tensorial 3D framework. But there is no one single strain measure that by itself works for all large strains of real materials (even for uniaxial deformations), and tensorial proper non-linear representations are required. If you want to learn more about this, Google large deformation elasticity of Cauchy-Green deformation tensor.