An ODE for a driven, undamped, oscillator might be written as $$ m\frac{d^2y}{dt^2}+ky=\sin{(\sqrt{k/m})}t $$ If the initial conditions are $y(0)=0$ and $y'(0)=0$ and the values of parameters are $m=1$ and $k=1$, the solution is $$ y=\frac{1}{2}(\sin t-t\cos t). $$ That is, we expect an increasing amplitude over time, a resonance.

But an oscillating system has numerous resonance frequencies, $n\omega$ where $n=1,2,3,...$ for $\omega=\sqrt{k/m}$ . But the solution for $n>1$ does not imply an increasing amplitude, e.g $n=2$ gives us $$ y=\frac{-2}{3}\sin{t}(\cos(t)-1) $$ I can derive the resonance frequencies (the frequency which gives us a standing wave) for a wave on string, based of the wave speed and the string's length, and if we apply a driving force in any resonance frequency we expect an increasing amplitude in the fundamental frequency as well as in higher modes. Is the above ODE not applicable in such cases?


2 Answers 2


An oscillating system may have numerous resonances, but not that system! That ODE represents a driven harmonic oscillator with exactly one resonant frequency and, indeed, if you apply an out-of-resonance driving force you get a bounded solution.

Systems with more than one resonant frequency are represented by higher-order differential equations or by partial differential equations (as in the case of continuous systems), but note also that, in general, the resonant frequencies need not be harmonics of a fundamental.


The full solution is $y(t) = \dfrac{\sin (\omega t) -\omega t \cos (\omega t)}{2 m \omega ^2}$ where $\omega^2 = \dfrac km$.
It may not look dimensionally correct but one must remember that there is a constant $1$ which multiplies the right-hand side of the differential equation and that $1$ has the dimensions of a force.

Let $\omega = 3$ and $m=1$ the graph of $y$ against $t$ looks like this.

enter image description here

As there is no damping term and the forcing frequency is equal to the natural frequency of oscillation of the system (the resonance condition) energy is pumped into the spring-mass system from the driver with no energy being lost by the system so the amplitude of motion of the mass increases without bound.

Note that there is only one natural frequency of oscillation with one mass and one spring.
As the number of masses and springs in a system increases so does the number of natural frequencies of oscillation of the system and hence the number of resonance frequencies.

So with two masses and three springs,

enter image description here

there are two resonant frequencies as explained here.

With a stretch string which can be thought of as very many masses (molecules) connected together with very many strings (bonds between molecules) there are very many natural frequencies of oscillation and hence resonant frequencies.

  • $\begingroup$ Thanks for the correction of the solution of the differential equation. I updated my post with parameter values for $m$ and $k$. I think I'm a little confused over the fact that a spring attached to a vibrator may have $2, 3, 4,...$ nodes (e.g standing waves with different wavelength). But this is maybe to compare with your string example above. $\endgroup$
    – magistern
    Nov 5, 2022 at 13:03
  • $\begingroup$ Why has this answer been given -1? I would like to know as then I can try and improve my answer. $\endgroup$
    – Farcher
    Nov 6, 2022 at 8:24
  • $\begingroup$ In my opinion your explanation and the linked information is excellent! @Farcher $\endgroup$
    – magistern
    Nov 6, 2022 at 13:30
  • $\begingroup$ The answer is mostly factually correct (though I don't understand the initial statement that the OP's differential equation solution is wrong -- with unit coefficients, it appears to match the OP's solution), but the core answer ("The ODE here is for a single-mode system, and multiple resonances require multiple modes") feels lost, especially with the initial statement that the ODE solution is wrong. $\endgroup$
    – RLH
    Nov 6, 2022 at 17:21
  • 1
    $\begingroup$ @magistern - Apologies for making the comment that your solution to the differential equation was incorrect due to the fact that I missed your statement that some of the parameters you set to be unity. I have now corrected my answer. $\endgroup$
    – Farcher
    Nov 6, 2022 at 19:42

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