It's known first order dynamical systems had one energy storage (example C, in RC circuits) these systems act as a filter but don't resonate, on the other hand a second order system had two energy-storage elements (C and L, in RLC circuits) and can exhibits resonance effect.

In a vibrating string there is resonance effect, usually analyzed from wave equation, but as a dynamical system there seem to be just one element for energy storage (the string inertia), how can this be thought as a linear dynamic system, i.e.: where is the second energy storage equivalent that allows a 'resonance behaviour as a second order system'?

  • 2
    $\begingroup$ If the only relevant dynamical ingredient in a string were its inertia, then if you pushed on it, it would move uniformly at constant velocity. $\endgroup$ Mar 6, 2019 at 19:59

3 Answers 3


In a vibrating string, the inertance arises from the mass per unit length of the string. The compliance arises from the string's elasticity, or "stretchiness". The energy storage in the inertance (because of its velocity) goes to zero at the moment of peak displacement of the string, at which point the energy storage in the stretch of the string goes through a maximum. Since these two energy storage mechanisms are out of phase, the system will resonate.

  • $\begingroup$ Straight to the point, those are the two elusive dynamic elements that I was looking for, I still would like to see them appearing in wave equation, as clear as I can see them in the linear second order system, but closer now. By now, both answer deserve the bounty. $\endgroup$
    – HDE
    Mar 8, 2019 at 0:04
  • $\begingroup$ @HDE But for a standing wave, one does not really need the wave equation to describe the oscillation. Energy is going back and forth between kinetic energy and potential elastic energy, with the same phase for all points on the string. Dynamically, it acts just like a mass-spring system. $\endgroup$
    – user137289
    Mar 9, 2019 at 8:06

The first system that you mentioned is the LC circuit, which I will take to be the series circuit. Let us review this, as we will compare it to the string equation later. The governing differential equation is $$ L\ddot{I} + \frac{1}{C}I = 0 \ . $$

$$ \ddot{I} = - \frac{1}{LC}I \ . $$

I have used the compact notation for time derivatives, but the equation is the same as on Wikipedia except with zero resistance, so that we can have perfect harmonic solutions. This current $I$ is the current at some instance of time, everywhere in the circuit.

The general solutions to this differential equation are exponentials; $$ I = A_1 e^{iwt} + A_2 e^{-iwt} \ . $$ The constants have to be determined by some boundary conditions, that is, initial conditions. The frequency $\omega$ is $1/\sqrt{LC}$. The reason for oscillations is that the second derivative is proportional to the function itself.

The wave equation, in one dimension, reads $$ \ddot{u}=c^2 \frac{\partial^2 u}{\partial x^2} \ , $$ Well we already see a problem, these differential equations are different. We can massage this into a form that would look better, but first lets play with its solutions. The general solutions to this equation are $$ u(x,t) = f(x+ct)+g(x-ct) \ , $$ where $f,g$ are arbitrary wave forms that propagate in opposite directions. But we want a resonant solution, so lets fix the string at $0$ and $1$, and hence the functions are sinousoids of some frequencies, $sin(2\pi n(x-ct))$ and $sin(2\pi n(x+ct))$. Now we can go ahead and shove this back into the string differential equation, and only evaluate the spatial derivatives: $$ \ddot{u} = -(2\pi n)^2 c^2 u \ . $$ Well is that not much better? I am sure at this point you can see the similarity to the original equation.


The attentive reader may have already seen this coming, but let us spell it out. Take the original string DE and Fourier transform it from $x$ space to $k$ space, I will leave it as an exercise to show that the answer is $$ \ddot{u} = (2\pi i k)^2 c^2 u \ . $$ For some fixed frequency, we get back precisely the LC circuit differential equation.

Well great, we are done. It is time to just interpret the result. We can identify $1/LC$ with $(2\pi c k)^2$, by observation.

But really I do not think it is moral to do more than state that the differential equations have the same form. You could check out the derivation of the wave equation for the case of a string, in which the second time derivative comes from Newtons law $F=ma$, then you can identify this with the contribution from the inductor to the current equation. I do not know if it is morally acceptable to start interpreting "voltages" in the string equations and so on, I would say the safest option is to observe that the differential equations admit harmonic solutions, despite the fact the physical systems are different. If you add resistance terms, they come with one time derivative into both equations, and so again the dissipation solutions would be the same.

  • $\begingroup$ This is a great math answer, I agree oscillations come from second derivative proportional to the function itself, besides this @niels nielsen also hits the nail on the two "energy storage elements" that I was looking for, 1-inertia due to mass and 2-potential energy due to string elasticity. Although that 2nd element doesn't appear in this answer, I think both deserve the bounty. Finally and bold enough, Wikipedia also have wave equation derived from Hooke's Law (elasticity). $\endgroup$
    – HDE
    Mar 7, 2019 at 23:50

Equations of motion for transverse oscillations

we are working in the $(x\,,y)$ plane, the string have its endpoints fixed at $(0,0)$, and $(a,0)$. In a transverse oscillation the x-coordinate of any point on the string dose not change in time. the transverse displacement at any point is given by the y-coordinate.

The EOM's

$$\frac{\partial^2\,y(x,t)}{\partial x^2}-\frac{\mu_0}{T_0}\,\frac{\partial^2\,y(x,t)}{\partial t^2}=0\tag 1$$

Where: $T_0$ is the string tension force, and $\mu_0$ is string mass per unit length. the total string mass is then $M=\mu_0\,a$

Boundary condition and initial condition

Dirichlet Boundary condition


Neumann Boundary condition

$$\frac{\partial y}{\partial x}(t,x=0)=\frac{\partial y}{\partial x}(t,x=a)=0$$


Ansatz :

$$y(t,x)=y(x)\,\sin(\omega\,t+\phi)\tag 2$$

where $\omega$ is the angular frequency and $\phi$ is the phase.

equation (2) in (1) we get:

$$\frac{d^2 y(x)}{dx^2}+\omega^2\frac{\mu_0}{T_0}\,y(x)=0\tag 3$$

The differential equation (3) is solved in term of trigonometric functions, and with the dirichlet boundary condition we get:


plugging $y_n(x)$ in (3) we find the frequencies

$$\omega_n=\sqrt{\frac{T_0}{\mu_0}}\frac{n\,\pi}{a}=\sqrt{\frac{T_0}{\mu_0}\,\frac{1}{a^2}}\,{n\,\pi}=\sqrt{\frac{T_0}{a\,M}}\,{n\,\pi}=\sqrt{\frac{k}{M}}\,{n\,\pi}\quad,n=1\,,2\ldots$$ .

so we have the same "energy source" as for a rigid body oscillation with mass $M$ and stiffness $k$


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