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In a damped oscillator, the damping term is represented by a velocity dependent force $b \ \dot{x}$. This makes sense if the damping is due to viscosity of the medium.

Is this modeling correct for the energy dissipation due to heating of the spring? I understand that this heating also comes from friction but I can't visualize if it can also be modeled as velocity dependent.

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  • $\begingroup$ At high Reynolds number $\text{Re}$ the drag force takes on the form $c \dot{x}^2$, which makes the EoM a darn sight harder to solve... $\endgroup$
    – Gert
    Sep 11 '20 at 17:24
  • $\begingroup$ And I don't think the spring heats up. The heat loss is due to viscous dissipation. $\endgroup$
    – Gert
    Sep 11 '20 at 17:29
  • $\begingroup$ Do you mean the heating due to the bending and unbending of the metal of the spring? $\endgroup$
    – nasu
    Sep 11 '20 at 17:59
  • $\begingroup$ @Gert It does heat up, but usually not much unless you exceed the elastic limit of the material. On the other hand, if you take an ordinary wire nail and start to drive it into a block of very hard material, you can make it hot enough to burn your fingers with a few hammer blows. $\endgroup$
    – alephzero
    Sep 11 '20 at 22:02
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The standard way to model the internal energy dissipation (i.e. hysteretic damping, or in viscoelastic materials) for oscillatory motion is to make the stiffness term a complex number. The imaginary part of the number represents the energy dissipation.

The energy dissipation depends on the amplitude of vibration, but unlike viscous damping, if the amplitude is constant the same amount of energy is dissipated in each cycle of the vibration, independent of the frequency. This is consistent with experiments.

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You can do it in an indirect way.

For example, consider a heating model where the energy being dissipated into heat at a given moment in time is proportional to the kinetic energy of the spring at that time, multiplied by some “dissipation constant”. In that way, the amount of energy going into heating the spring becomes implicitly dependent on the velocity of the spring thanks to $KE = \frac{1}{2}mv^2$.

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