Why doesn't amplitude increase when the frequency of external periodic force increases above the natural frequency of the vibrating object?

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    $\begingroup$ Try pushing a kid on a swing more often than the natural frequency. $\endgroup$ – Pieter Feb 4 '17 at 15:42
  • $\begingroup$ The math behind a driven damped oscillator is e.g. reviewed here: physics.stackexchange.com/q/228279/2451 $\endgroup$ – Qmechanic Feb 4 '17 at 16:04

In one sense this is a consequence of the meaning of resonance. Resonance is identified by a maximum in amplitude. If amplitude increased above (or below) the resonant frequency then we have not found a maximum in amplitude.

Perhaps what you are really asking is, why is there a maximum in amplitude when driving frequency $f$ equals the natural frequency $f_0$? This was answered by What is the qualitative cause for a driven oscillator to have a max. amplitude during resonance?

At the simplest level this is because at the natural frequency $(f=f_0)$ the driving force is always in phase with the natural motion of the system. The driving force is then always adding energy to the system, which will increase indefinitely unless there is some form of damping (eg friction) which removes energy from the system at a faster rate as amplitude increases.

At all other frequencies the driving force is sometimes in phase and sometimes out of phase with the natural motion of the system. Sometimes it adds energy, sometimes it takes energy away.

However, this assumes that both the natural motion and the driving force are both sinusoidal. If the driving force is applied as an intermittent impulse or "kick" then sub-harmonic frequencies of $f_0/n$, where $n$ is an integer, will also result in resonance. For example, if the system oscillates once every second, giving it a "kick" in the right direction once every 2 or 3 or 4 etc seconds will also increase the amplitude of the system, provided that the energy imparted with each kick is not outweighed by the loss of energy in between. See Non-resonant but efficient frequencies.

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A simple physical explanation follows directly from Newton's second law, $F = ma$.

Suppose you keep the amplitude $F_0$ of the external force $F = F_0 \cos \omega t$ constant, and increase the frequency $\omega$. The amount of time that the force is acting in one direction is inversely proportional to the frequency - it is a quarter of the vibration period, or $(1/4)(2 \pi / \omega) = \pi / ( 2\omega)$.

So as the frequency increases, the structure doesn't have enough time to move far, before the force reverses and accelerates it in the opposite direction.

If $\omega$ is much higher than the resonant frequency, the stiffness of the structure is not very important, because it never gets time to move far, and $Kx$ is always small compared with $F$.

There are situations where the amplitude of $F$ is not constant, but proportional to $\omega^2$. One example is the force acting on a rotating shaft which is not perfectly balanced - the radial force on a small element of material $dm$ of the rotating structure is $r \omega^2\, dm$.

In that situation, the amplitude of vibration is approximately constant for frequencies well above the resonant frequency. In fact the shaft is trying to rotate about its mass center, not its geometrical center, and the vibration amplitude is simply the distance between those two points.

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