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I have a system where two particles $x_1$ and $x_2$ in one dimension are connected by a spring and a dash in parallel. This is analogous to the Kelvin-Voigt model for viscoelastic materials. The two particles are also driven by two forces $F_1$ and $F_2$, as in the diagram below.

enter image description here

I'm having difficulties deriving the equations of motion for the two masses: the only strategy I can think of is to start from force balance equations, even though this is probably not sufficient. For a spring with a rest length of $l_0$, the force balance equations of both particles will read (if there is no motion there is no damping force!):

$$ F_1=k(x_1-x_2-l_0) \qquad \qquad F_2=-k(x_1-x_2-l_0) $$

Question: In the Kelvin-Voigt model, the relations are written in terms of stress, strain and (for the viscosity part) strain rate. This is appropriate for a continuum medium, but here I'd like to find the equations of motion for the two degrees of freedom of the two masses. How can these "continuum" Kelvin-Voigt relations be transformed into the equations of motion for a discrete set of masses? Which are the equations of motion for the two masses?

Note: If the masses are connected by a spring and dash in series, we have the Maxwell model, see this diagram. The same general question is also relevant to this case.

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  • $\begingroup$ Isn't this a physics forum? its more difficult to post here a question than getting a paper into nature physics. $\endgroup$
    – jarhead
    Jul 22, 2023 at 17:57
  • $\begingroup$ Honestly, experiences like this makes me happy that AI is taking over such that forums like this will be redundant along with all the "reviewers" $\endgroup$
    – jarhead
    Jul 22, 2023 at 17:58
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    $\begingroup$ there is a continuous stream of questions being posted. I think it's a pity that it has been closed and I voted to reopen but you also have to ask yourself if there is something wrong in the way you post questions here (e.g. the duplicate thing is not seen well, and also asking homework-like questions). My suggestion is: remove the duplicate and insert the diagram of the model here. With this edit I am confident that it will be reopened. $\endgroup$
    – Quillo
    Jul 22, 2023 at 18:19
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    $\begingroup$ I've edited the question as you suggested. where do I ask to reopen it? $\endgroup$
    – jarhead
    Jul 22, 2023 at 19:33
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    $\begingroup$ I also deleted the duplicate $\endgroup$
    – jarhead
    Jul 22, 2023 at 19:34

1 Answer 1

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The Kelvin-Voigt (KV) is the most simple viscoelastic model (it is the prototype of rubbery materials). For the "continuum" version of the model, the total stress will be the sum of the stress in each component. Similarly, in the "discrete" version, the total internal force will be the sum of the force imparted by the spring and by the dash.

The KV relations are $\sigma_{el} = E \epsilon $ for the elastic stress, $\sigma_{visc} = \eta \dot{\epsilon} $ for the viscous stress, where $\epsilon$ is the strain. Since the components are in parallel, $\sigma_{el}+\sigma_{visc}$ is the total internal stress. Force balance tells us that, for applied stress $\sigma_{ext}$, the total stress is zero: $\sigma_{el}+\sigma_{visc}+\sigma_{ext}=0$. The problem is... where is the inertia? There is nothing like $\ddot{\epsilon}$, meaning that the continuum model is valid in the overdamped regime. No worries, we can add something like this by hand (for sure our system of masses has inertia!).

Discrete model - You can see $\epsilon$ as the relative distance between the two masses, $x_1-x_2$ (not that this is an analogy, beware of this "problem" related to finite strains), and $\sigma_{ext}$ as an applied force (in the continuum model $\sigma_{ext}$ is, dimensionally speaking, a force per unit area, while $\epsilon$ is dimensionless). In practice, the KV model $\sigma_{el}+\sigma_{visc}+\sigma_{ext}=0$ translates in $F_{i}+f_i=0$ for both masses ($i=1,2$), where $f_i$ is the total internal mutual force. To go beyond overdamped motion, we have to go beyond force balance: $F_{i}+f_i\neq 0$. Consider the following system:

$$ m_1 \ddot{x}_1 = F_1 + f \qquad m_2 \ddot{x}_2 = F_2 - f $$ where $f$ is the (mutual) internal force between the two masses due to the spring and the dash. The system is such that the total momentum of the system changes only due to the external forces (internal forces can not change the motion of the centre of mass!):

$$ \frac{d}{dt}(m_1 \dot{x}_1+m_2 \dot{x}_2) = F_1+F_2 $$

Now we are almost done: just remember the analogy between force $f$ and internal stresses ($f \leftrightarrow \sigma_{el}+\sigma_{visc}$, The total internal stress in the KV model is the sum of the elastic one and viscous one!). Therefore, $f= - k (x_1-x_2-l_0) - \eta(\dot{x}_1-\dot{x}_2)$, so that:

$$ m_1 \ddot{x}_1 = F_1 - k (x_1-x_2-l_0) - \eta(\dot{x}_1-\dot{x}_2) \\ m_2 \ddot{x}_2 = F_2 - k (x_2-x_1+l_0) - \eta(\dot{x}_2-\dot{x}_1) $$

Overdamped regime - The final system for the 2 masses reduces to $F_1+f=F_2-f=0$ in the overdamped regime, in analogy with the original continuum KV model $\sigma_{ext}+\sigma_{el}+\sigma_{visc}=0$. For $F_1=F_2=0$, any displacement $|x_1-x_2|\neq l_0$ will result in an exponential relaxation, exactly like the continuous KV model.

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