# Modeling linear spring deformation in time

Suppose I had a spring (at equilibrium) and applied a certain force $F$, causing it to undergo elastic deformation. I know that by applying this specific force, hooke's law tells me that the spring will contract by a total distance $\Delta x=F/k$, where $k$ is hooke's constant.

I want to be able to model this deformation in time. Assuming the force is constant in time, I want to be able to predict the length of the spring at any given time $t$. The standard differential equations are given as:

$$mx'' + \mu x' + kx = F$$

However, I'm having difficulty adusting this equation to the situation I have described above. Certainly, if the spring was at equilibrium at first, there is an acceleration induced. Hence, I believe the $x'$ term should remain. However, if the force is applied constantly, the spring should deform continuously to its steady state position without oscillation (if I'm wrong about this, please correct me!). Hence, I also think there is an argument against including the acceleration term. Also, my spring system doesn't really have a dampening agent, so I'm not sure if it's feasible to include the velocity term either. However, I predict that the length of the spring to decay to its resting steady state length exponentially; indicating that the velocity term might be important to keep.

I'm just not sure which assumptions are correct for my system. Any help with this would be greatly appreciated! :)

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## 1 Answer

Suppose the constant force is gravity. Experience should tell you that if you release a mass attached to a spring is oscillates up and down, so a constant force does not mean there won't be any oscillation.

The $x'$ term is the damping term. If you assume there is no damping the $x'$ term can be omitted, and in that case you just get the equation for a simple harmonic oscillator. The only way you're going to get the length of the spring to decay to its resting steady state length exponentially is if there is damping, and in particular if the system is critically damped or greater.

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I see... So, all three terms would be necessary after all. But if there is no damping mechanism attached to the spring, how would I even begin to assign a value to the damping coefficient? –  Paul Nov 19 '12 at 14:01
If there is no damping mechanism on the spring it will just oscillate forever. In a real spring there will be damping from air resistance and from energy dissipation as the spring flexes (I'd guess air resistance will dominate). To get anything like critical damping you will need to add some form of mechanical damping to the spring. –  John Rennie Nov 19 '12 at 14:32
Can I think of it as though the eternal force that is being applied can also act as its own damper? –  Paul Nov 20 '12 at 16:33
I've lost track of what you're trying to do. If you want your differential equation to describe damped behaviour you need to include a damping term, and we usually take that to be linear in velocity. Note however that in many cases e.g. air resistance, the damping goes as $v^2$ and life gets more complicated. –  John Rennie Nov 20 '12 at 16:45