# Can you excite coupled pendula at the frequency difference?

Imagine the typical system of coupled oscillators = two pendula joined by a weak spring.

The system is oscillating at the complex motion which arises when you displace one pendulum, say pendulum 1, and then release it. This motion can be viewed as the superposition of two normal modes, the symmetric one (both pendula oscillate in phase, almost as if the spring were not there, at frequency = f1) and the antisymmetric (the pendula oscillate out of phase, compressing and stretching the spring, at f2). In this complex motion the pendula oscillate at the average frequency of the normal modes [(fi + f2)/2] and their amplitude is varying, since energy is being transferred from one to the other, being the rate of change of the amplitude [(f1 – f2)/2]. But the natural frequencies of the system are f1 and f2.

First question: is the frequency of the change of amplitude [(f1 – f2)/2] or (f1 – f2)? In the analogous case of a beat sound wave made of two waves of close frequencies, I am pretty sure that it is [(f1 – f2)/2], because I drew it and only thus did I manage to draw the envelope of the beat wave that marks its amplitude changes. But I am not sure with the coupled oscillators since I often see references to the “frequency difference”. Anyhow let us call this frequency the intermediate frequency or IF.

Second question: imagine that every time that pendulum 1 returns to its origin (the peak of its varying amplitude, the trough for pendulum 2) I give it a gentle pull or push; so I do it at a frequency equal to IF. I understand that this is a “sharp impulse”, mathematically a Dirac function. As it is repetitive, it could be decomposed into the IF and its harmonics through Fourier series. Will I find among them any of the natural frequencies (f1 and f2)? I don’t see it, but on the other hand this procedure looks efficient, it seems that the energy input thus injected should be absorbed by the system...

Third question: what if instead of sharp impulses I apply a sinusoidal driving force equal to the one that set this system in motion at the start?

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Nov 30 '16 at 1:20

# Problem setup

Suppose we have two pendula with the same length $l$, and a string with string constant $k$ that connects them, then the equations of motion are $$\ddot{\theta_1} = - \frac{g}{l} \theta_1 - \frac{kl}{m}(\theta_1 - \theta_2) + F_1(t) \>,\\ \ddot{\theta_2} = - \frac{g}{l} \theta_2 + \frac{kl}{m}(\theta_1 - \theta_2) + F_2(t) \>,$$ where $\theta_1$ and $\theta_2$ are the angular displacements of pendulum 1 and 2, respectively. We have also assumed that the oscillation is small, and that the spring is at is natural length when both pendula are vertical, so the stretch of the spring is just $l(\theta_1 - \theta_2)$. Finally, $F_1(t)$ and $F_2(t)$ are the driving forces that are applied to each pendulum. In the question, you proposed two situations, both have $F_2 = 0$. The first situation has $F_1 = f_0 \operatorname{III}_{2\pi/\omega}(t-t_0)$, where $\operatorname{III}_T(t)$ is the Dirac comb function of period $T$, and the second situation has $F_1 = f_0 \sin(\omega t+ \phi)$. We then want to find out how the system responds to driving forces with different frequencies $\omega$, specifically what happens when $\omega = \omega_1, \omega_2, (\omega_2-\omega_1)/2$.

To simplify the form of the model a little bit, we will rewrite it in normal coordinates. Just as our intuition tells us, the normal modes should correspond to when the pendula are swinging in phase and when they are out of phase by $\pi$. From this we know that the normal coordinates are $q_1 = (\theta_1 + \theta_2)/2$ and $q_2 = (\theta_1 - \theta_2)/2$, corresponding to normal frequencies $\omega_1^2 = g/l$ and $\omega_2^2 = g/l + 2 kl/m$. Thus we can rewrite the system as $$\ddot{q_1} + \omega_1^2 q_1 = \frac{F_1 + F_2}{2} \>,\\ \ddot{q_2} + \omega_2^2 q_2 = \frac{F_1 - F_2}{2} \>.$$ First off, if we set both driving forces to zero, we can simply solve the two decoupled harmonic oscillator equations and get, $$q_1(t) = C_1 \cos\omega_1 t + C_2\sin\omega_1 t \>,\\ q_2(t) = C_3 \cos\omega_2 t + C_4\sin\omega_2 t \>,$$ where $C_1, C_2, C_3, C_4$ can be determined from initial conditions. We will call these solutions $\bar{q}_1(t)$ and $\bar{q}_2(t)$ later to simplify notations.

# Sinusoidal driving

This system is actually not too difficult to solve when the driving force is a sine wave or a delta function, but why labour ourselves when Mathematica can help? In the second situation (sinusoidal driving), assuming $\omega\neq \omega_1$ or $\omega_2$, we have $$q_1(t) = \bar{q}_1(t) - \frac{f_0}{2(\omega^2 - \omega_1^2)} \sin(\omega t+\phi) \>,\\ q_2(t) = \bar{q}_2(t) - \frac{f_0}{2(\omega^2 - \omega_2^2)} \sin(\omega t+\phi) \>,$$ This is the free oscillator solution with an extra sine wave at the frequency of the driving force added to each mode. Notice that all of these terms are bounded, so the amplitude is also bounded. However, we can also see that when $\omega\rightarrow \omega_1$ or $\omega_2$, the term resulting from driving blows up to infinity, which is hallmark for resonance. Now if we solve the equation with $F_1 = f_0 \sin(\omega_1 t+\phi)$, the solution to $q_2$ does not change, but $q_1$ becomes $$q_1(t) = \bar{q}_1(t) + \frac{f_0}{8\omega_1^2}\sin(\omega_1 t + \phi) - \frac{f_0}{4\omega_1} t \cos(\omega_1 t + \phi) \>.$$ The terms resulting from driving changed. Notice especially that the last term contains a factor of $t$, which means the amplitude of the oscillation grows without bound. The $\omega=\omega_2$ case is very similar, where the solution to $q_2$ is just the above result with $\omega_1$ swapped out for $\omega_2$. Now what happens when $\omega=(\omega_2 - \omega_1)/2$? Well, it is exactly the same as the off resonance case above, with $\omega$ substituted for $(\omega_2-\omega_1)/2$. Since no singularity occurs at this value, it would not cause any numerical difficulty. So the short answer to your question in the title is, no.

# Delta function and Dirac comb driving

Moving on to the delta function case. Before we march on ahead into the rather complicated Dirac comb case, let's first look at the result of delta function pulse at $t_0$, $F_1 = f_0 \delta(t-t0)$. The results are \begin{gather} q_1(t) = \bar{q}_1(t) + \frac{f_0}{2\omega_1} H(t-t_0) \sin \omega_1(t-t_0) \>,\\ q_2(t) = \bar{q}_2(t) + \frac{f_0}{2\omega_2} H(t-t_0) \sin \omega_2(t-t_0) \>, \end{gather} where $H(t)$ is the Heaviside step function. Effectively, a delta pulse injects some energy into the system at the time when it's applied, and changes the amplitude a little, but does not introduce new frequencies.

For the full on Dirac comb function, we have, \begin{gather} \begin{split} q_1(t) & = \bar{q}_1(t) - \cos \omega_1 t\int_1^t \frac{f_0}{2\omega_1} \operatorname{III}_{2\pi/\omega}(t'-t_0)\sin\omega_1 t' d t' \\ & \quad + \sin \omega_1 t\int_1^t \frac{f_0}{2\omega_1} \operatorname{III}_{2\pi/\omega}(t'-t_0)\cos\omega_1 t' d t' \>, \end{split} \\ \begin{split} q_2(t) & = \bar{q}_2(t) - \cos \omega_2 t\int_1^t \frac{f_0}{2\omega_2} \operatorname{III}_{2\pi/\omega}(t'-t_0)\sin\omega_2 t' d t' \\ & \quad + \sin \omega_2 t\int_1^t \frac{f_0}{2\omega_2} \operatorname{III}_{2\pi/\omega}(t'-t_0)\cos\omega_2 t' d t' \>. \end{split} \end{gather} This looks a bit daunting, but remember that the Dirac comb is just the sum of a bunch of delta functions, so the integrand here only takes nonzero values at discrete points, i.e., when $t' = t_0 + \frac{2\pi}{\omega} k$. We essentially just need to count how many of these points are between $1$ and $t$. Thus we have the result \begin{gather} q_1(t) = \bar{q}_1(t) +\frac{f_0}{2\omega_1} \sum_{k\in \mathcal K(t)} H\left(t-\left(t_0 + \frac{2\pi k}{\omega}\right)\right) \sin \omega_1 \left(t-\left(t_0 + \frac{2\pi k}{\omega}\right)\right) \>,\\ q_2(t) = \bar{q}_2(t) +\frac{f_0}{2\omega_2} \sum_{k\in \mathcal K(t)} H\left(t-\left(t_0 + \frac{2\pi k}{\omega}\right)\right) \sin \omega_2 \left(t-\left(t_0 + \frac{2\pi k}{\omega}\right)\right) \>, \end{gather} where $\mathcal K(t)$ is the set of values for $k$ such that $t_0 + \frac{2\pi}{\omega} k$ lies between $1$ and $t$.

Let's interpret this solution. Formally, this looks just like summing the extra terms from the delta pulse solutions with $t_0$ taking different values. This means that every time $2\pi/\omega$ units of time passes, the delta pulse hits, and adds a sine wave to the original oscillation, always at the normal frequency $\omega_i$, and always with the same amplitude $f/2\omega_i$ ($i = 1, 2$).

What's important is realising that each of these added oscillation has a phase shift $2\pi\frac{\omega_i}{\omega}$ from the previous one. If $\omega_i/\omega$ is not an integer or rational number, given enough time, the phase shift will cause the added oscillation to average each other out. If $\omega_i/\omega$ is an integer or rational number, then after a certain number of cycles, the phase shift will go back to a value it has previously taken. (Correction: even when it's a rational number, the phase shifted waves will still add up to zero, so for there to be resonance, $\omega$ must be an integer multiple of $\omega_i$, which confirms the argument in the question body.) This means that the added oscillation will have a net effect on the system, and eventually increases the energy of the system. (This should remind us of pushing someone on a swing. If you push at a frequency that doesn't match the natural frequency of the swing, you won't get it to work.) The effect is strongest when the ratio $\omega_i/\omega$ is $1$, because in this case every period the delta pulse kicks in at the same time, and the compounded effect is the greatest.

This more or less corresponds to the resonance we see in the sinusoidal driving case. We still have strong resonance at $\omega_1$ or $\omega_2$, but also at many more other frequencies. If we set the driving frequency to $(\omega_2 - \omega_1)/2$, then frequency ratio $2\omega_i/(\omega_1 - \omega_2)$ is only a rational number when $\omega_1$ and $\omega_2$ are rational multiples of each other. Therefore rather different from the sinusoidal case, the Dirac comb driving can excite the system at the frequency difference, given that the ratio between natural frequencies of the system is rational.

# Resonance

After all of this calculation, let's review the concept of resonance. According to Wikipedia,

In physics, resonance is a phenomenon in which a vibrating system or external force drives another system to oscillate with greater amplitude at a specific preferential frequency. Frequencies at which the response amplitude is a relative maximum are known as the system's resonant frequencies or resonance frequencies. At resonant frequencies, small periodic driving forces have the ability to produce large amplitude oscillations.

In a typical driven damped oscillator, the resonance frequency fits the above definition as the frequency of the driving force that produces the maximum amplitude, but this definition does not quite apply to undamped oscillators. For a damped oscillator, the amplitude response is $$\alpha(\omega) \sim \left(\frac{\omega_0^2}{(\omega_0^2 - \omega^2)^2 + \gamma^2\omega^2}\right)^{1/2} \>.$$ Here $\omega_0$ is the natural frequency of the system, $\omega$ is the driving frequency, $\gamma$ is the damping factor. This expression achieves maximum at $\omega = \sqrt{\omega_0^2 - \gamma^2/2}$. If we were to directly set $\gamma$ to zero, however we see that the amplitude response blows up to infinity at $\omega_0$, and indeed this is what we observed in the coupled system as well. What happens then? Well as it turns out, since there is no damping, the system does not dissipate energy, so at the resonant frequency the amplitude just keeps growing. This is confirmed by the above calculation.

Your attempt at illustrating resonance by adding waves of various frequencies does not make sense. A driving force does not dictate directly how an oscillator oscillates. We cannot determine the wave form of the resulting oscillation simply by adding the waveform of the driving force and the free oscillation. We must solve the equation of motion to get the resulting waveform.

# Aside 1: Green's functions

Normally, when we are interested in the response of a system to some outside force, we would invoke Green's function. In fact, put rather simplistically, the Green's function is found by calculating the response of the system when driven by a delta function pulse, so we did in fact encounter it here. It is rather more difficult to calculate the Green's function for a coupled system, since the differential equation becomes 4th order, so I did not follow that approach. (In truth this might still be how Mathematica solved the equations.)

# Aside 2: Quantum mechanics

The way you phrased the question in the title immediately begs the comparison to a two level quantum system. A well studied problem is quantum mechanics is what happens when a two level system is perturbed by a potential that is periodic in time. Moreover, we want to know how the system behaves when the frequency of the driving is near the difference of the energy of the two levels. Comparing your question about a driving force at $\omega_1-\omega_2$ to the quantum mechanical problem might be interesting, but I won't attempt it here, since it should be topic for a different question.

• Can you elaborate on your conclusion for the driving force at $\omega=(\omega_1 - \omega_2)/2$ in the Dirac comb case? Is it simply the same as for the sinusoidal driving case? Is it the answer that $\omega=(\omega_1 - \omega_2)/2$ would initially inject energy into the system but the same would be soon kicked out? – Sierra Nov 30 '16 at 7:54
• @Sierra I added some explanation to the Dirac comb driving at frequency difference. It actually is a "resonant" frequency given some conditions on the natural frequency of the system. – Elliot Yu Nov 30 '16 at 15:16
• @ Elliot Yu: It is an excellent answer. It will take me time to assimilate all the math, so I may be asking you questions from time to time. To start with the basics, so for example with $\omega_1=18$ and $\omega_2=19.8$, could the system be excited with a driving force oscillating at frequency $\omega=0.9$? – Sierra Nov 30 '16 at 21:21
• @Sierra: It sort of depends on your definition of "excite". If you mean will it change the motion and increase the energy/amplitude, then yes, it will. But any old frequency can do that. If you mean will it cause the resonant behaviour I cited in the section titled "Resonace", then no. The amplitude will not increase without bound when driven at $\omega = 0.9$. – Elliot Yu Nov 30 '16 at 23:17
• In case anyone is interested, we clarified that my above numerical example is ok if referred, as it was my intention, to Dirac comb case, because 18 and 19.8 are rational multiples of each other. So sharp impulses (Dirac comb) repeated at frequency 0.9 would cause resonance = increase of amplitude without bound. – Sierra Dec 1 '16 at 0:02