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Here is a description of the motion of two springs in series. The premise is that the force on the two springs is the same.

This is derived from the following reasoning: when I pull the mass with a certain force $F$ at some point I reach the equilibrium position.

So every piece is stationary and in particular the point between the springs is stationary: we conclude that the forces applied in that point (the two elastic forces) sum up to zero. We derive that the elastic forces are equal and they're equal to $F$.

The problem is: this is valid in the equilibrium position. I can't understand why this conclusion is extended for every position in order to derive the equation of motion:

enter image description here

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Because the springs are considered massless. So if there were a force difference between the ends, you would get infinite acceleration. And this is valid not only for the ends, but for any two points, if there is no mass between them.

The issue is somehow explained in this post:
Is the tension in both ends the same (on a massed string)?

And indeed, if the springs are massive (easier: if you have one massive spring), you get another value for the period. That's the same issue - why should we take the tension inside one string to be equal?
So, to sum it up: if the part between two points is massless, then the force on both sides (i.e. the strain in the end-points) has to be exactly equal. Thus, the strain inside the spring is equal everywhere. Only this fact allows you to talk about the elastic force of the spring. Otherwise is would change througout the spring.

The reasonig is the same as for a static situation; equal forces are implied by a finite mass and zero acceleration -- as well as by zero mass and finite acceleration.

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  • $\begingroup$ so the important point here is that massless points (such as the conjunction point between the springs) can't accelerate, isn't it? and this is valid for that point (but also for any two points without mass in between), not just for the start condition (springs at rest) but throughout all the motion.. am I correct? $\endgroup$ – Surfer on the fall Apr 24 '16 at 16:26
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    $\begingroup$ yes, approximately :) "can't accelerate" is not the right phrase - the massless parts can accelerate very well, and they will always do so as soon as there is any non-zero force; the argument needs not the points, but the part in between two points - if this is massless, then the force on both sides (i.e. the strain in the end-points) has to be exactly equal. Thus, the strain inside the spring is equal everywhere. Only this fact allows you to talk about the elastic force of the spring. Otherwise is would change througout the spring. $\endgroup$ – Ilja Apr 24 '16 at 16:46
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    $\begingroup$ Yes, this force is a somehow antiintuitive concept (I think you would have found it in the links explained a bit). I mean the magnitude of the force. The direction is... in both directions :) think of the tension in a string/spring as the force you would need to hold it together if there were a cut. You see, you need to pull both pieces in opposite directions then. It's just like pressure in water, by the way - which direction does it have? There is only a direction if you consider some surface - then you need a certain force to keep the water from pushing through that surface. $\endgroup$ – Ilja Apr 24 '16 at 19:55
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    $\begingroup$ yes, they can, and they will. They will do anything to keep the forces equal! But if the forces were not equal, there would be nothing (no mass, no friction, ...) to stop the whole spring from moving instantaneously to a position where the forces are equal. Newtons law is not only valid for rigid bodies; it holds for any system, that the net force times the total mass must be the acceleration of the center of mass. It's not that infinite acceleration is "bad" as you wrote above, it just will instantaneously abolish its cause. The same argument as for why electric field inside metals is zero. $\endgroup$ – Ilja Apr 24 '16 at 20:57
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    $\begingroup$ oh, we somehow ended the converstation, but are you convinced at last? :) system mostly move to equllibrium, because that's where the force is pointing to. This is then called a stable equillibrium, because small deviations dont matter, they are returned. An unstable equillibrium (pencil standing on its point) are more rare, and in this case a stretched spring will move to it's equillibrium position. Because deforming a spring from the equally stretched state creates a reaction opposing the deformation. $\endgroup$ – Ilja Apr 28 '16 at 21:30
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I think you are asking why this argument remains valid when the mass is oscillating, and does not only apply when it is static.

I think the answer to this is that the forces in the springs depend only on their extensions ($F=kx$), not how quickly the extension is changing ($F=kx+b\dot x+c\ddot x$). So at any given extension $x$ the tension is the same regardless of the motion.

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