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I do not remember when, but a while a ago I was told that for a spring with stiffness $k$ and a mass $m$, the equation $-kx = ma$ never holds. (assume a horizontal spring)

So if I was given a problem like this:

You pull a mass $m$ attached to a spring with stiffness $k$ to the right of the equilibrium point $x = 5\text{ m}$ of the spring and you let it go, what is the acceleration of the mass?

Then is this question meaningless (or any other that is similar)? I forgot what the real question is like, but I have seem such a method was used to tackle these problems? As such an attempt to write $-kx = ma$?

Could someone please jog my memory? I realize I am asking you guys to 'read my mind', but please bear with me.

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Hi lum, and welcome to Physics Stack Exchange! I removed the image from your post because the one you had there showed a completely unrelated problem. If you posted the wrong URL by accident, then feel free to track down the right one and edit it in. – David Zaslavsky♦ Feb 21 '12 at 1:21
I'll point out that many equations in physics are meant to provide models that provide highly accurate descriptions of what experiments show, although strictly speaking they "never hold" exactly, since there are always additional complications that weren't addressed in the model. For most purposes, using the $-kx = ma$ equation would work just fine to answer the question you've posed. – kleingordon Feb 21 '12 at 7:53
Comment to the question(v2): Could someone please equip this question with a more relevant title? – Qmechanic♦ Feb 21 '12 at 17:04

1 Answer

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The spring "with mass" has two possible complications: it's affected by gravity, and it may have a non-negligible kinetic energy (inertia).

You say the spring is positioned horizontally - hence the gravity effect is irrelevant.

As long as you consider only static problems - the kinetic energy is irrelevant as well, and your spring behaves exactly as "massless". The same holds also for the problems with constant velocities (because what matters is not the energy, but its change).

However there will be inevitable effects in problems with accelerations. Again, if you're talking about constant accelerations (such as sliding blocks and etc.) - then the spring would probably deform non-uniformly, but remain "static" in its shape. This may be solved relatively simply. However if accelerations are not constants - you'll get a non-trivial dynamics.

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There's also the complication that if a given string is stretched too far (beyond what is sometimes called a yield point), it will become distorted and no longer obey the Hooke's law relation. – kleingordon Feb 21 '12 at 7:56
@kleingordon: of course. Real-world spring have a so-called "elastic" range – valdo Feb 21 '12 at 21:50

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