I understand that at resonance, a spring-mass-damper system will only require input energy from a forcing that matches the energy dissipated per cycle by the damper.

Let's say that you have a system with a natural frequency $f_o$, are forcing it at some frequency that is roughly twice that frequency, and I let the system get to a steady state so that it is oscillating as a sinusoid with some amplitude and frequency ($x(t)=X\sin{(\omega t+\phi)}$). I have simulated this exact thing a bunch, and find that the net work done by the harmonic forcing ($\int_0^T F\sin{(\omega t)}\dot{x} dt$) over a period is equal to the net work done by the damper ($\int_0^T c\dot{x}^2 dt$).

Since the system isn't at resonance, I would think that inertial and elastic work should not exactly cancel over a period and the forcing should have to do extra work to compensate for this off-resonance behavior. How would I show this mathematically? Every time I try to do this, I find that the net work of the forcing exactly equals the work dissipated by the damper (which I am aware is just conservation of energy). If I try to integrate the power associated with the mass movement ($\int_0^T m\ddot{x}\dot{x} dt$) or the power associated with the spring ($\int_0^T kx\dot{x} dt$), both of these integrals give 0 over a period. What is happening with energy exchange between the mass kinetic energy and the spring elastic energy when you are at steady state oscillations away from resonance?

  • $\begingroup$ By definition of steady state, since energy stays constant, you must have energy balance between forcing and dissipation. Your math is correct, but your intuition is wrong. $\endgroup$
    – LPZ
    May 1 at 16:05
  • $\begingroup$ Thanks @LPZ! So how does energy balance at steady state jive with the idea that a system oscillating at resonance has the most efficient energy transfer between kinetic and elastic potential energy? It would seem that this transfer efficiency is 100% regardless of whether or not you are at resonance. $\endgroup$
    – Ethan
    May 1 at 19:45
  • $\begingroup$ No problem and welcome to Physics SE! Actually, what is optimal at resonance is not the efficiency since it is always at 100% according to the previous reasoning. It is rather the absolute amount of the flowing energy that peaks at resonance. Even if the forcing amplitude is fixed, the energy injected/dissipated still depends on frequency and is maximal at resonance (by definition). $\endgroup$
    – LPZ
    May 1 at 22:16

1 Answer 1


for a mass spring system you obtain this differential equation:

$$ \ddot x=-\omega^2\,x\tag 1$$

where $~\omega^2=\frac km~$

multiply equation (1) with $~\dot x~$

$$\dot x\ddot x=-\omega^2\,\dot x\,x\\ \frac {d}{dt}\left(\frac {\dot x^2}{2}\right)= -\omega^2\,\frac {dx}{dt}\,x\\ \int{d} \left(\frac {\dot x^2}{2}\right)= \int -\omega^2\, {dx}\,x\\ \underbrace{\frac {\dot x^2}{2}}_{K(t)}= -\underbrace{\omega^2\frac {x^2}{2}}_{P(t)} $$

thus the integration of the Lagrange L(t) from zero to a period T is zero. $$\int _0^T \underbrace{\left[K(t)-P(t)\right]}_{L(t)}\,dt=0$$


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