I am trying to solve a wave equation of motion for the transverse vibrations of a viscoelastic string fixed at each end to get the $Q$ (decay rate) of each harmonic (partial).

The goal is to allow approximate simulation of the vibrating string with a series of damped harmonic oscillators (one for each harmonic) while avoiding the computationally expensive process of finite difference modeling.


An excellent answer for how to do this was given here for the longitudinal waves of the viscoelastic string. However, the equation of motion gets more complex for the transverse waves.

Equation of Motion

I am modeling the string as a series of Kelvin-Voigt damped mass springs like this:

enter image description here

This creates an equation of motion for transverse vibrations (where $w$ is transverse 1D displacement) in the simplest form of:

$ρw_{tt} = (σ + \frac{T}{A})w_{xx}$

Where $T$ is base tension, $A$ is cross sectional area.

In a Kelvin-Voigt damper, you have for stress:

$σ = Eε + \eta ε_t$

And in transverse waves, you can approximate the strain with:

$ε = \frac{1}{2}w_xw_x$

$ε_t = w_xw_{xt}$

This then gives our simplest possible equation of motion as:

$ρw_{tt} = (\frac{E}{2}w_xw_x + η w_xw_{xt} + \frac{T}{A})w_{xx}$

Longitudinal Solution

In the longitudinal case (where $u$ is longitudinal displacement), the $Q$ was solved simply knowing that the equation of motion for each harmonic $n$ as any damped oscillator is:

$u_{tt} +2ζωu_t+ω^2u =0$

It was then possible to solve $Q=\frac{1}{2ζ}$, as the equation of longitudinal motion for each harmonic was described by the following (which matched the above perfectly except different constants):

$ρu_{tt}=−k^2(Eu +ηu_t)$

Transverse Problem

In the transverse case, the equation of motion is much more complicated with now impossible to eliminate derivatives of $x$ (which are not present in the longitudinal version).

Does this mean it is impossible to reasonably describe the transverse vibrations of a viscoelastic string with a series of damped mass spring oscillators?

Ie. Is finite difference modeling the only way to then solve this, given the equation of motion above?

This seems to me to imply the $Q$ of the harmonics change over time in the viscoelastic string based on the total average $ε$ and $ε_t$ in the string at any time. Is this correct?

Is there still any way to solve this in terms of any approximation for the average result over time that doesn't require finite difference modeling?


If one can estimate the total stretch in the "string" at any time (total strain and solve also total rate of strain change), might it be possible to estimate the overall $Q$ of any given harmonic at that point?

I can approximate the total $ε$ on the full "string" at any moment by summing the series of harmonic oscillator outputs to get the amplitude at each time step and measuring the "length" of the resulting "string" created by the outputs over 1/2 of the fundamental's period. I can solve approx total $ε_t$ per sample also from this output.

It seems the $Q$ of each harmonic must be proportional to the total strain and rate of total strain change at each time point, right?

I presume there must be some approximation method (whether this way or another way) that could be applied but I don't have the knowledge or education to know.

  • $\begingroup$ Correct, $w$ is transverse displacement. I updated it to clarify. See equation 5 here: kapitaniak.kdm.p.lodz.pl/papers/2014/Marynowski_Kapitaniak.pdf He also comments "perturbed stress gradient $σ_x$ is small and can be neglected" hence the simplified form. It being non-linear, that means Q is always chaging, right? But one must still be able to approximate the overall $Q$ for a given harmonic at any point in time still I presume. Each harmonic must have a certain overall $Q$ at any point, and I suspect this correlates to the total/average $ε$ & $e_t$ in the string at any time, right? $\endgroup$
    – mike
    Jan 20 at 23:29
  • $\begingroup$ Hard to forecast the evolution of a non-linear system $\endgroup$
    – basics
    Jan 21 at 12:13

1 Answer 1


If you want to think in terms of the weakly nonlinear limit, perhaps it's best to systematically include all the third order terms. I have: $$ \begin{align} \epsilon &= \frac12w_x^2-\frac18w_x^4 \\ \frac{\partial\epsilon}{\partial w_x} &= w_x-\frac12w_x^3 \\ \sigma &= \sigma_0+E\epsilon+\eta\epsilon_t\\ &= \sigma_0+\frac E2 w_x^2+\eta w_xw_{xt}\\ \rho w_{tt} &= \left(\sigma \frac{\partial\epsilon}{\partial w_x}\right)_x \\ &= \left(\sigma_0w_x-\frac {\sigma_0}2w_x^3+\frac E2w_x^3+\eta w_x^2w_{xt}\right)_x \end{align} $$ with $\sigma_0$ a background stress (i.e. $T/A$ in terms of background tension and cress section area), $\rho$ the volume mass density, and $E$ the Young modulus. I'll assume the string has length $L$ and is clamped at its ends (Dirichlet boundary conditions). Neglecting the higher order terms, the modes are: $$ h_n = \sqrt2\sin(k_xx)\\ k_n = \frac{n\pi}L\\ u=\sum_{n=1}^\infty u_nh_n \\ u_n=\int_0^1u(x)h_ndx $$ and the corresponding frequencies are: $$ \omega_n = ck_n \\ c = \sqrt{\frac{\sigma_0}{\rho}} $$ with $c$ the celerity.

The nonlinear term adds interactions between them. You can look at how the harmonics correspondingly deform.

Say you weakly excite the harmonic $h_m$ with amplitude $A$. At leading order, it is an undamped harmonic oscillator: $$ w^{(0)} = A\cos(\omega_mt)h_m(x) $$ At next order, you only have self interactions since the other modes are zero at leading order. Using: $$ \langle ((h_m')^3)',h_m\rangle = -\frac32k_m^4 $$ the full nonlinear equation involving only self interactions is: $$ \rho\ddot w_m = -\sigma_0k_m^2w_m-\frac32k_m^4\left(\frac{E-\sigma_0}2w_m^3+\eta w_m^2\dot w_m\right) $$ so it is a mix of a Duffing and Van der Pol oscillator. Generally, the damping is in the nonlinear term, so comparing it to a damped harmonic oscillator is not obvious. Indeed, instead of an exponential decay in amplitude, you'd rather expect a power law decay. A rough justification of this is that for small amplitudes, the $w_m^3$ terms are negligible, so $w_m$ oscillates at frequency $\omega_n$. The long term decay, when averaging over the period $2\pi/\omega_n$, is solely governed by the damping term $w_n^2\dot w_n$ which by dimensional analysis gives: $$ \langle w_m\rangle \sim \frac1{\sqrt t} $$ which you can check numerically. Therefore, there is no quality factor to define as the decay is not exponential. The scaling argument is pretty robust, and you'd expect it to be correct in general. This includes accounting for interactions between modes and for the fully nonlinear equation and not the truncated ones. Furthermore, the coupling with longitudinal modes should not change anything either since they should decay exponentially, so comparatively much faster.

In order to identify the transverse modes as damped oscillators, you could amend the model by saying that the string lies in a viscous region. This would add a viscous force: $$ \rho w_{tt}+\lambda w_t = \left(\sigma \frac{\partial\epsilon}{\partial w_x}\right)_x $$ which would then give you a $Q$ factor. However just from the assumption of viscoelastic string, I don't see how to get transverse modes to be damped harmonic oscillators.

Note that a bit like in your previous question, the strength of the nonlinearity is non uniform for all modes as it grows as $n^4$. Therefore, the weak nonlinear assumption breaks down.

Hope this helps.

  • $\begingroup$ Again, incredibly helpful @LPZ. More than words can say. You have my immense gratitude. My only formal physics education was a first year physics course. So I haven't known even where to begin. It's been a puzzle for years. You explaining all this is amazing. I am learning through it. You might also be shocked by how many published articles try to decompose a vibrating string into exponentially decaying oscillators. It doesn't seem to work in my experience but I couldn't understand why. Just a few small points: $\endgroup$
    – mike
    Jan 21 at 21:38
  • $\begingroup$ (1) I believe $T$ must divided by $A$ (cross sectional area) in stress equations as stress is by definition $\frac{N}{m^2}$ so I believe your equation should be $σ = \frac{T}{A} + Eε + ηε_t$. I don't want to mess up something in your equations though as I don't know exactly how that transmits through the rest. I fear to touch it myself. $\endgroup$
    – mike
    Jan 21 at 21:39
  • $\begingroup$ (3) I will actually want very much to try to simulate your oscillator equation and see how it performs. I am now quite eager. I can follow your derivation (perhaps as a child learns to walk), but it is incredible to me and I don't want to mess it up. You said you made things dimensionless by setting $L=1$ and $ρ=T$. Is it in any way possible for you to update the equations to keep those terms? It would be incredibly helpful as I will use $L$, $ρ$, $T$, and $A$ (along with $E$ and $η$) to parameterize. Thanks again so much either way for everything. It has meant a lot. $\endgroup$
    – mike
    Jan 21 at 21:40
  • $\begingroup$ (2) As mentioned in the other thread, as per kapitaniak.kdm.p.lodz.pl/papers/2014/Marynowski_Kapitaniak.pdf in $ρw_{tt} = (σ + \frac{T}{A})w_{xx}$, $ρ$ is mass density ($kg/m^3$) not linear density ($kg/m$). (They use $ρ_l$ for linear density when applicable.) Connected to point 1, is your $ρ$ then certainly mass density or linear density? Thanks for any clarification. $\endgroup$
    – mike
    Jan 21 at 21:51
  • $\begingroup$ Thank you for the updates! You have been very generous and this has all made my day. I am eager to test this new oscillator. :) I will need to add an external force $F$ (to excite it). I presume this would then be added as: $ρ w_{tt} = -σ_0k^2w - \frac{3}{2}k^4(\frac{E-σ_0}{2}w^3 + ηw^2w_t) + \frac{F}{AL}$ (ie. force divided by cross sectional area times string length) just based on matching units, correct? $\endgroup$
    – mike
    Jan 22 at 4:24

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