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I understand stiffness as the extent to which an object (e.g. a mass spring) resists deformation from an applied force, or the rigidity of an object. And I understand damping as the energy dissipative properties of an object/system (e.g. a mass spring) under cyclic stress. In the context of a spring/oscillator, damping is what causes a spring to eventually stop oscillating.

But I don't understand the real differences between the two. Couldn't a spring's stiffness explain it's damping? Why not?

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  • $\begingroup$ Can you make it clear how you think that stiffness can be used to explain damping? Damping requires energy loss and if you consider a spring system with no damping but a finite stiffness then you will see that the energy is converted from KE of the mass to PE of the spring in a periodic way. $\endgroup$
    – biryani
    Commented Apr 29, 2016 at 5:14
  • $\begingroup$ When stiffness results in a resistance to an applied force, isnt that energy loss? Forgive my ignorance; My research has nothing to do with physics.. can you explain what you mean by energy being converted from KE of the mass to PE of the spring in a periodic way? $\endgroup$ Commented Apr 29, 2016 at 5:23

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So given a spring with spring constant $k$ can one predict what dissipative force the spring will exert when extended?
The answer is "No" because they depend on different things.
The stiffness depends on the elasticity of the bonds between the atoms/molecules which make up the spring and the damping depends on the permanent distortion of the bonds which are not (simply) related.


The equation of motion of a mass at the end of a spring undergoing damped harmonic motion can be written as

$$m\ddot x + c \dot x + k x=0$$

where $x$ is the displacement at a time, $c$ is a constant relating the frictional force opposing the motion to the velocity $\dot x$ of the mass of the object $m$ and $k$ is the spring constant.

$c \dot x$ is often called the damping term and $kx$ the stiffness term.
As far as I know there is no simple (or even complex) relationship between $c$ and $k$ and indeed the $c$ in this context is to do with the air rather than the spring.
$c$ tells you something about the frictional force which intimately means that it is to do with the reduction of mechanical energy of the system and $k$ tells you something about how the spring resists changes to its length.

In the context of your question the parameter $c^2 - 4mk$ is important because it controls the type of damping which the spring-mass system undergoes and that parameter contains both $c$ and $k$.
So you could say that the spring constant $k$ does influence the type of damping that exists which in turn means that the spring constant also influences the rate at which mechanical energy of the spring-mass system is reduced.

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Spring stiffness is not responsible for energy loss. Consider a spring with stiffness $k$ but no damping. The work done in compressing the spring by a distance $x$ will be stored as the potential energy(PE) of the spring. For a spring compressed by a distance $x$, the PE is given by $\frac12 k x^2$. This energy is not lost and can be used by letting the spring expand (Old mechanical pendulum clocks work on this principle, the energy of a compressed spring drives the clock while the spring slowly expands). This is similar to how you have to do work to move a mass up against gravity, but the work you do is not lost. It is stored as the potential energy of the mass and can be reused.

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  • $\begingroup$ Does a spring with higher stiffness lead to greater damping? How is it possible for a spring with stiffness to have no damping? $\endgroup$ Commented Apr 29, 2016 at 6:14
  • $\begingroup$ Practically there will be both stiffness and damping in a spring. But in theory they are separate phenomena. I am not aware of any relation connecting stiffness and damping of actual springs. It might depend on the material used etc. $\endgroup$
    – biryani
    Commented Apr 29, 2016 at 6:20
  • $\begingroup$ Ok. So its not true (for real life systems) that a very stiff spring given enough force to start oscillating would dampen more quickly than a less stiff spring given the same driving force? $\endgroup$ Commented Apr 29, 2016 at 6:35
  • $\begingroup$ How fast oscillations in an over damped system dies off is given by the damping ratio. It depends on both the stiffness and the damping coefficient. Check (en.wikipedia.org/wiki/Damping#System_behavior) for the complete analysis. $\endgroup$
    – biryani
    Commented Apr 29, 2016 at 6:54

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