Non-resonant but efficient frequencies

Question

I understand that if the frequency of a driving force coincides with the natural frequency of an oscillator (say a pendulum), the rate at which energy is transferred to the same is maximized. However, there may be other frequencies that are not so efficient but do transfer energy to the system, i.e. the latter absorbs it.

The usual graphs plot intensity or amplitude versus frequency and they have the look of a steep triangle. So they point at the frequencies surrounding the natural one as the most effective. But what about frequencies that are sub-multiples of the natural frequency ($f_o$), like $f_o/2$, $f_o/3$ and so on. For example, when swinging I take impulse not every time I reach a peak but every other time. It seems to me that a force operating at this frequency should attain as much amplitude increase as the natural one, in every shot, though of course it shoots at lower rate.

Thus I understand that this actor does not shine up in a graph plotting against intensity (which is power/surface, so it has a time-dependence, ok?) but I don't see why these frequencies are not merited in a graph plotting purely against "amplitude", like this one:

A reason may be damping (the effect is wiped out before it can consolidate), but what if the oscillator were ideally free of damping?

Anyhow, leaving aside the graphs, can it be said that those are, after the natural frequency itself, the most effective frequencies in terms of increasing amplitude?

Answer 1

This is a subtle issue! Your intuition is correct (a driving at $f_0/2$ should be very effective) even though the graph seems to contradict this. The reason is that the graph displays the response to a sinusoidal driving force. If you indeed drove the mass sinusoidally at frequency $f_0/2$, it would indeed be ineffective -- you'd be holding onto the mass and trying to make it go half as fast as it wants to go.

However, you're proposing forces that look like sharp impulses, where the impulses come at a frequency $f$. Such a force actually has an infinite number of harmonics, at frequencies $f$, $2f$, $3f$, and so on, so it's, in a sense, equivalent to driving with infinitely many sinusoids at once.

To see why intuitively, consider hearing somebody clapping a few times a second. Since the frequency of the clapping is quite low, you might think it would sound low in pitch, but it actually sounds quite high -- because of the many harmonics created at every individual clap.

It is these harmonics that the mass is listening to when you drive at a frequency $f_0/n$. The mass is most sensitive to the $n^\text{th}$ harmonic, because it has the resonant frequency $n(f_0/n) = f_0$. A more detailed analysis (i.e. taking the Fourier transform of a repeated short square pulse) shows that driving with impulses at frequency $f_0/n$ is almost exactly as effective as driving with impulses with frequency $f_0$, as long as the damping is low and the impulses are sufficiently short.

Answer 2

First I'll try to explain why the amplitude vs frequency diagram only has one maximum, then I'll go back to why this seems to contradict your intuition.

Let's take the simplest forced oscillator formula, with no damping (this won't affect our conclusion), for instance that of a spring undergoing a force $F$ :

\begin{equation} x''(t) + \omega_0^2 x(t) = F(t) \end{equation}

Let's assume that $F = F_0 e^{i\omega t}$. Now, let's try a solution of the form $x(t) = Ae^{i\omega t}$. By injecting these expressions in the equation we find (after diving out both sides by $e^{i\omega t}$):

\begin{equation} -\omega^2 A +\omega_0^2 A = F_0 \end{equation}

And finally : \begin{equation} A = \dfrac{F_0}{\omega_0^2-\omega^2} \end{equation}

It is clear that the amplitude $A(\omega)$ has only one maximum at $\omega_0$ and that's it. There is nothing special with the pulsations $\omega_n = \omega_0 / n$ where $n$ is any integer. Of course, there exist resonant devices where these modes are resonant as well - this is the case of some cavities for instance, where resonant modes are those such that reflected waves interfere constructively, which can happen when it takes them $n$ periods to travel from one side of the cavity to another. But there's no such effect in a linear spring or pendulum.

Now, this applies to a force $F$ of the form $e^{i\omega t}$, which is very specific: this means a sinusoidal force of pulsation $\omega$.

In your experiment, however, $F$ doesn't look like a sinusoid. It is closer to a sharp impulse, which is known as a Dirac function, noted $\delta$ : $F(t) = \delta(t)$. Now the previous analysis is still valuable because any reasonable function $F$ has what we call a Fourier transform $\tilde{F}$ such that : \begin{equation} F(t) = \int \tilde{F}(\omega) e^{i\omega t} d\omega \end{equation} This means that a function $F$ can be represented as a sum of sinusoids. This is very helpful, because the motion equation takes a very simple form when written in Fourier space : \begin{equation} \tilde{x}(\omega) = \dfrac{\tilde{F}(\omega)}{\omega_0^2 - \omega^2} \end{equation}

Now, when $F(t) \simeq \delta (t)$ (i.e. it is very high valued around $t=0$ when the impulse is exerted and 0 otherwise), the Fourier transform is a constant function. This means a very short impulse contains every frequency, including the resonant frequency. So in the end, since $\tilde{F}(\omega) = cst = F_0$ : \begin{equation} x(t) \sim F_0 \int \dfrac{e^{i\omega t}}{\omega_0^2 - \omega^2} d\omega \end{equation} It is very clear that the frequencies around the resonant frequency dominate, and we have approximately: \begin{equation} x(t) \propto e^{i\omega_0 t} \end{equation}

So in the end, because an impulse "contains" every frequency, it also contains the resonant frequency, whose effect dominates.

Answer 3

The details they don't put in physics textbooks, but that you likely learn from engineering experience is that resonance depends not only on internal structure of the system but how energy flows in and out. A resonant system tends to trap energy and that energy may or may not necessarily be admissible at the resonant frequency of the system. It depends on the internal structure.

Although the more common means of energy transfer for a swing is a cycle by cycle push in one direction, it's also possible to have a person in front also deliver a push such that the rate of energy input to the system is doubled. Each input, the same amplitude but 180 degrees out of phase with one another. You show only the amplitude frequency response of your swing (pendulum) system but there is also a phase component and this illustrates how the 180 degree out of phase push works. The phase plot below shows that approaching from lower frequency the near zero degree phase signal is admissible, and approaching from high frequency, 180 degrees. The change in phase across resonance is very sharp when the system is high Q (very little damping, very low energy loss).

For practical purposes the swing system depends on the swing's bearings and drag forces at some point to reach an energy flow that is equal and opposite to the rate of energy input from the push. Otherwise the swinger will eventually loop over the top! In principle for linear systems zero damping means all the energy entering the system stays there, and the resonant peak approaches infinite amplitude. But for practical, real systems there are nonlinearities that limit trapping of energy. Energy has a tendency to find a way out sometimes breaking the system (like the Tacoma Narrows Bridge collapse).

For the swing system the structure is such that energy rate and phase (in the case of two people pushing) must be specific but that's not necessarily true for all resonant systems. Consider the singing rod that's often used in physics demos on resonance. The energy in this case is supplied by slip-stick friction between rosin coated fingers and the surface of the rod; essentially broad band colored noise vibrations entering the rod's surface. In this case the internal structure of the rod filters out and concentrates energy from the noise input at the natural frequency of the rod. The rod admits and traps only a narrow band of the input excitation. The remaining band of frequencies are mostly dissipated as heat at the rod's surface.