# 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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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.
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