I am somewhat confused about the amplitude of forced vibrations at non-resonance driving frequencies.

If I was to assume that there was no / negligible damping present, then at resonance, the amplitude of the harmonic oscillator would continue to grow with each oscillation?

If this is the case, then is this also true for driving forces that are not at resonance, or would the oscillator reach a constant amplitude?



2 Answers 2


Yes, for the harmonic oscillator, the amplitude diverges with time at resonance, as already discussed in Undamped Resonance of a Classical Harmonic Oscillator.

In general, what happens away from resonance depends both on the system (e.g., harmonic oscillator or anharmonic?), on the forcing (very high or low frequency, non-periodic?), and possibly on the starting point of the system. For example, the movement can be chaotic for the cubic (anharmonic) potential

$$ U(x) = \alpha x + \beta x^3, $$

which is the well-known Duffing oscillator.

Generally, the amplitude away from resonance will not diverge for small but nonzero damping, but will also not be necessarily constant; it'll grow during some oscillations, and decrease with others. When the forcing injects energy (i.e, in the instants its work is positive - when the force is in the same direction as the oscillation speed), then the amplitude increases, while decreasing otherwise (negative work done by forcing, which happens when it acts against the momentary movement).

The sinusoidally forced harmonic oscillator will, after a transient behavior, oscillate at the forcing frequency at constant amplitude. If the forcing is periodic, but not sinusoidal, or the oscillator not harmonic, then the steady-state amplitude is constant, but the movement is described by an infinite sum of sines (at multiples of the forcing frequency, i.e., a Fourier series). For example, Keith Fratus notes gives, for an arbitrarily forced harmonic oscillator, the following solution:

$$ x(t) = \sum_{n=1}^\infty \frac{\sqrt{a_n^2+b_n^2}}{\sqrt{(\omega^2-n^2\lambda^2)^2+4\beta^2n^2\lambda^2}}\cos{(n\lambda t - \delta_n-\phi_n)}, $$

where $\lambda$ is the angular frequency of the forcing, $\omega$ the oscillator frequency, and $\beta$ the damping constant.

Related discussions are A conceptual doubt regarding Forced Oscillations and Resonance and Physical reason behind having greater amplitude when driving frequency$ < $ natural frequency than that when driving frequency $>$ natural frequency.

  • $\begingroup$ For "small but nonzero" damping, the undamped motion always dies to zero eventually, and the only motion is at the forcing frequency, not the system's resonant frequency. For zero damping, the motion is the sum of two components at the two frequencies. This is fundamentally different. If the forcing frequency changes with time, that complicates the solution a bit, but the general principle is still true - if the damping is non-zero, the resonant frequency component is too always small to matter, of you look at the motion over a long enough time scale. $\endgroup$
    – alephzero
    Sep 13, 2017 at 3:06
  • $\begingroup$ @alephzero, I included remarks on the periodically forced harmonic oscillator, though the answer is much more general than that. In particular, you're mistaken about the the nonperiodic forcing, because then the oscillator typically won't have a steady state, i.e., will be in a perpetual "transient" behavior that doesn't have a constant amplitude. $\endgroup$
    – stafusa
    Sep 13, 2017 at 6:22

You are correct that for resonant frequencies, the amplitude will grow with every oscillation (until something breaks, runs out of space, or cannot apply force to it).

For forces not at resonance, the amplitude will not really "reach a constant", nor will it grow infinitely.

The amplitude will grow and shrink, creating a second periodic oscillation with a longer period (see beat in acoustics).

To visualize what is happening, consider the case where the frequency of the driving force is near the resonant frequency, and the driving force is also a sinusoidal wave. When in phase, the two forces act together to form positive interference, graphically it's the equivalent of adding the two waves together and getting a bigger wave (see superposition for more information on the subject).

If this is allowed to keep happening, the wave will keep growing forever. Thankfully, if the frequencies aren't exactly the same, it won't happen.

Instead, because the frequencies are different, the driving force will become out of phase with the oscillator over time (the closer the frequency to resonant, the slower this process). After some time, it will form destructive interference and the amplitude will greatly decrease.

The cycle then repeats, with the phase going back towards resonance. This will form the "beat" pattern (as demonstrated graphically in the link earlier).

  • $\begingroup$ You have the right idea, but you can't really have "destructive interference" between two different things, like a force and a displacement. A better way to describe it is to say that the relative phase between the force and displacement is continuously changing. If the force is doing positive work (= force $\times$ distance) on the system at one phase angle, it will be doing negative work later, when the phase has shifted by 180 degrees, and over a complete "360 degree cycle" of phase angles the positive and negative work will cancel out to zero. $\endgroup$
    – alephzero
    Sep 13, 2017 at 3:01

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