Solving Spring Force Equation: 2X

[sophiecentaur]

The equation F=kx describes the force for all values of acceleration. But only when acceleration is zero.

That doesn't make sense; I guess you didn't really mean it. The spring doesn't 'know' it's accelerating. It can only respond to the forces on it. That force will be equal to ma, where m is the mass on the end and a is the acceleration. Zero acceleration will produce zero force, whatever the stiffness of the spring.

 

[jbriggs444]

That doesn't make sense; I guess you didn't really mean it. The spring doesn't 'know' it's accelerating. It can only respond to the forces on it.

Possibly we are not talking about the same situation. I am envisioning an anchored spring with one end held stationary. The force on the free end will not match f=kx. Neither will it match f=ma.

 

[sophiecentaur]

Possibly we are not talking about the same situation. I am envisioning an anchored spring with one end held stationary. The force on the free end will not match f=kx. Neither will it match f=ma.

I am picturing a spring with a mass on one end and a force applied to the other. (Isn't that the model in this thread?) But, whether the other end is fixed or not, the spring will always obey F=kx (where x is the extension)
Actually, anchoring one end of a massless spring is only the same as having a very large mass on it; the acceleration will be zero(ish).
One problem with this whole thread is that we are really dealing with a dynamic situation; some comments acknowledge it and others don't.

 

[jbriggs444]

I am picturing a spring with a mass on one end and a force applied to the other. (Isn't that the model in this thread?) But, whether the other end is fixed or not, the spring will always obey F=kx (where x is the extension)
Actually, anchoring one end of a massless spring is only the same as having a very large mass on it; the acceleration will be zero(ish).
One problem with this whole thread is that we are really dealing with a dynamic situation; some comments acknowledge it and others don't.

My spring has mass, so indeed, we are not talking about the same situation. I'll defer to the OP to decide what interpretation he or she has in mind.

 

[sophiecentaur]

My spring has mass, so indeed, we are not talking about the same situation. I'll defer to the OP to decide what interpretation he or she has in mind.

That's good. So neither of us is a total idiot.
It only goes to show, yet again, how a question that's well defined (diagrams whenever possible) makes answering a lot easier.
I am surprised that, with so much apparent computer literacy around, people can't bring themselves to do more diagrams and, even then, they do them with dreadful Painting packages with a mouse - very hard to make sense of. A picture can be worth a thousand words. ""

 

[Nidum]

I am surprised that, with so much apparent computer literacy around, people can't bring themselves to do more diagrams and, even then, they do them with dreadful Painting packages with a mouse - very hard to make sense of. A picture can be worth a thousand words.

Very true . Good diagrams not only aid understanding of a problem - they often give indication of how to solve a problem or in some cases actually give the solution .

Long forgotten methods like polygons of forces , relative velocity triangles and graphical integration also still have considerable merit in solving practical problems .

 

[Chandra Prayaga]

I am quoting the problem stated by UMath1: "Initially the spring is not stretched or compressed so x=0. If I were to compress this spring I must then apply a nonzero force. By Newton's 3rd Law the spring must also apply a nonzero force, even though kx=0. So my speculation is that the force of the spring is only equal to kx when it is not accelerating." I believe sophiecentaur already answered that question: "You cannot instantly apply a 10N force to the spring unless you displace it instantaneously by10k. You have to start with an infinitesimal reaction force". Now I am only amplifying sophiecentaur's remarks: If you want to apply a non-zero force on the spring, that force starts from zero and gradually increases to 10 N or whatever. The word "gradually" does not mean "slowly". The rise from zero to 10 N could happen in a few microseconds, but it is still gradual. During the entire time, the spring keeps getting more and more stretched (or compressed), and exerts an opposite force equal to what you are applying, and to kx.

 

[UMath1]

Jbriggs444 has the situation I am describing, the spring does have mass and is anchored. All I am saying is I think that the force the spring applies cannot be equal to kx when it is accelerating or experiencing a net force. This is my thinking. If you were to compress the spring an infintesimal amount, dx, you would still have to apply a nonzero force. Sure, the force does not have to be constant, it can gradually increase. But you cannot possibly have any compression when the force is zero. That means that at some time interval the force must be nonzero while x=0. This means that the force of the spring is only kx when it is not accelerating.

 

[sophiecentaur]

It's a pity you didn't make that clear with a diagram, then.

 

[jbriggs444]

Jbriggs444 has the situation I am describing, the spring does have mass and is anchored. All I am saying is I think that the force the spring applies cannot be equal to kx when it is accelerating or experiencing a net force. This is my thinking. If you were to compress the spring an infintesimal amount, dx, you would still have to apply a nonzero force. Sure, the force does not have to be constant, it can gradually increase. But you cannot possibly have any compression when the force is zero. That means that at some time interval the force must be nonzero while x=0. This means that the force of the spring is only kx when it is not accelerating.

A complete analysis of this situation would require us to model the spring as a sequence of little point masses and ideal springs linked together. This is the basic notion of "finite element analysis". If one were to improve the model by making the springs tinier and the masses smaller, the limit yields a differential equation that can describe the behavior of the spring under any particular driving pattern that you might choose. In complete generality that equation would be difficult or impossible to solve analytically. However, with some simplifying assumptions, one of the generic solutions that it can yield is a wave equation -- real springs with mass support waves.

It is very much simpler to assume ideal massless springs. This applies both for practical purposes and teaching purposes. Nobody wants to solve an intractable differential equation when designing a mechanism. And nobody wants to have to teach a first year physics class about finite element analysis, differential equations and numerical methods in order to have them work a homework exercise involving springs.

All that aside, what I have written previously stands. If you have a static spring with mass then f=kx is still valid. But in order to disturb its center of mass, you need an additional increment of f=ma above and beyond this equilibrium force in order to succeed in disturbing the spring's center of mass.

 

[sophiecentaur]

But you cannot possibly have any compression when the force is zero. That means that at some time interval the force must be nonzero while x=0.

As has already been said, you seem to be looking for some sort of paradox and there isn't one here. If you want to question the logic behind this then you would have to look at every relationship in Physics between two variables. If the function that describes the relationship is a continuous one then what happens at the origin shouldn't give you any cause for concern. We're talking cause and effect and an infinitessimal change in the cause will produce an infinitessimal change in the effect. Volts and current. Force and Acceleration, Heat and Temperature rise all accept that you can move away from the origin of your graph without violating anything. The next step in an effect is due to the previous step plus a perturbation due to a cause.
x = x₀ +vt
and many many more like that.