# How would I solve a Spring problem,

## Main Question or Discussion Point

...if the Spring does not obey Hooke's Law. For example, F=-kx³ or F=-kx^4, etc. This is just curiosity on my part, but I would still like to know.

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Homework Helper
...if the Spring does not obey Hooke's Law. For example, F=-kx³ or F=-kx^4, etc. This is just curiosity on my part, but I would still like to know.
Possible, of course. What you asked about is non-linear behaviour, which is in general more hard to describe.

Then the spring does not behave elastically. Unless it does but not in accordance with Hooke's law, pretty unlikely considering all the empirical observations from labs all around the world.

There's quite a lot of rheological possibilities, visco-elastic, plastic-elastic etc.. There is maths available to explain some of these things.

Homework Helper
Then the spring does not behave elastically. Unless it does but not in accordance with Hooke's law, pretty unlikely considering all the empirical observations from labs all around the world.
Linearity should not be self-understood when mentioning elasticity. Elasticity is a material property which states that the material recovers its original configuration after deformation. But, the relation between stress and strain may or may not be linear.

Regarding Hooke's law and empirical observations, I'm not so sure that everything obeys Hooke's law so politely. Take the typical example of the stress-strain diagram resulted by a material uniaxial compression test. The diagram is linear to a certain point. Above that point the behaviour is everything else except linear. And the behaviour above that very point can be very important when concluding something about the material.

Linearity should not be self-understood when mentioning elasticity. Elasticity is a material property which states that the material recovers its original configuration after deformation. But, the relation between stress and strain may or may not be linear.

Regarding Hooke's law and empirical observations, I'm not so sure that everything obeys Hooke's law so politely. Take the typical example of the stress-strain diagram resulted by a material uniaxial compression test. The diagram is linear to a certain point. Above that point the behaviour is everything else except linear. And the behaviour above that very point can be very important when concluding something about the material.
That's because Hooke's law describes the linear elastic behaviour of a material, it doesn't account for hysteresis (which would mean your spring isn't perfectly elastic) and it doesn't account for other non-linear behaviour beyond the yield point. Read my earlier post, I already mentioned this with regards to plastic and viscous behaviour, probably should have put brittle in there too. You can only use Hooke's law for elastic behaviour!!

Thus if your spring doesn't conform to Hooke's law it isn't perfectly elastic.

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...if the Spring does not obey Hooke's Law. For example, F=-kx³ or F=-kx^4, etc. This is just curiosity on my part, but I would still like to know.
You could solve problems using conservation of energy equations by finding the work done by the nonlinear force. If the "spring" follows $$F = kx^3$$, the work done for stretching the spring a length x would be

$$W = Fs = \int kx^3 \, dx = \frac{1}{4}kx^4$$

instead of the usual $$W = \frac{1}{2}kx^2$$ when $$F = kx$$.

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Homework Helper
You can only use Hooke's law for elastic behaviour!!
That is correct.

Thus if your spring doesn't conform to Hooke's law it isn't perfectly elastic.
But that is not correct, since, on a conventional stress-strain diagram the 'proportionality point' (i.e. the maximum stress which undergoes Hooke's law) is below the 'elastic point' (i.e. the maximum stress for which the material behaves elastic after removing the stress).

ZapperZ
Staff Emeritus
There's something strange here in this thread, and it isn't the original question. Maybe it is obvious here to everyone but me.

Here is the OP:

Plasma said:
How would I solve a Spring problem,...

...if the Spring does not obey Hooke's Law. For example, F=-kx³ or F=-kx^4, etc. This is just curiosity on my part, but I would still like to know.
Somehow, no one seems to be asking what EXACTLY is this person trying to "solve"!! Yet, I see a series of responses on I don't know what.

This is because, if all we care about is the equation of motion, i.e. solving for "x", then it really doesn't matter if it is F=kx, F=kx^2, F=kx^3, F=kx^n, etc, does it? I mean, just write down the differential equation and SOLVE! It is nothing more than a math problem. Whether it obeys Hooke's Law or not is irrelevant!

But is this what is being asked? Unlike some of you, I am not that certain and I am certainly not going to guess.

Sometime, people, we need to look at the question and figure out if it can be answered. If not, we'd be wasting time barking up the wrong tree.

Zz.

russ_watters
Mentor
...if the Spring does not obey Hooke's Law. For example, F=-kx³ or F=-kx^4, etc. This is just curiosity on my part, but I would still like to know.
ZZ is right, the premise of the thread is flawed and pointless. The question is circular: you would solve it according to whatever law it obeys, which you haven't stated because this is an artificial hypothetical. Ie, if the system (you wouldn't call it a spring because it isn't one if it doesn't behave like a spring) has a cube relationship between force and displacement, you'd solve it according to that.

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you would need a complicated nonlinear mathematical model to describe its behavior