So, we have explored several aspects of the behaviour of an elementary oscillator. You know how it oscillates when we leave it free to move, after an initial condition. You know how a short loading, an impulse loading, acts on the oscillator. Now let us see what happens when there is an oscillating loading. This is a very common situation - in a washing machine, the rotating drum is an oscillating source of vibration at the frequency of rotation of the engine. You certainly experienced this. It also happens in acoustics when you sing - you load the acoustical modes of the room with an oscillating pressure. Even the tide brings a periodic loading on the free surface of bays, as we shall see later. Let us see what a periodic loading does on an oscillator. Here is our oscillator equation. Now, f(t) = f_0 sin(Omega t). Here f_0 is the amplitude of the forcing. And Omega is the frequency oscillation of the forcing. As you remember, these are both dimensionless quantities. Omega is the frequency of the loading with respect to the frequency of the oscillator. This means that at Omega = 1 the frequency of the loading is that of the oscillator. Actually, we can simplify even more by using a normalized amplitude q = q/f_0, so that we just have (q)ddot + q = sin(Omega t). What is the solution to this equation? It is actually quite simple to obtain, using our Laplace transform. In the Laplace domain we have L[(q)ddot+ q] = L[sin( Omega t)]. The quantity of the left hand side is something we know and have used before s^2 Q - s q_0 - (q)dot_0 + Q. On the right hand side, we have the Laplace transform of sin(Omega t). This is simply Omega/(Omega^2 + s^2). Finally, I know the Laplace transform of Q as a function of s, as we did before, but now with Omega inside. You remember that the last two terms correspond to the effect of the initial conditions. Let us disregard them at this stage So we just have Q(s) = Omega/[(1+s^2)(Omega^2 + s^2)]. Can we now go back in the time domain, to have the response of the oscillator? We certainly can, because the solution in the Laplace domain is not so complicated. Let us leave aside for the moment the case of Omega=1, when the forcing is exactly at the frequency of the oscillator. I can also write Q as a difference of two parts, right here. This is much better, because you recognize here the Laplace of sin(t) and the Laplace of sin(Omega t). We can now go back into the time domain, and here is the response of our oscillator! Well. The result is a bit surprising. We have an oscillating force at a frequency Omega, and the response is a mix of a motion at the same frequency Omega and of a motion at the original frequency of the oscillator. This second part is just the effect of the zero initial condition I have prescribed. We shall see later that it eventually vanishes. The main part is more interesting. What we find is that the oscillator moves at the same frequency as the forcing, Omega. But the amplitude of the motion depends on Omega. What does it look like? Here it is. It goes from one to infinity at Omega = 1. Then it changes sign and goes back to zero. That is not easy to understand, in terms of physics. I prefer to write this as q = A sin(Omega t - Phi), where A is a positive amplitude and Phi is the phase. What we see is an amplitude that increases as we get close to the case where the forcing is at Omega = 1, and then gets back to zero with a switch in the phase Phi. The switch in phase just corresponds to the change in sign of the amplitude of response. What is that peak? It I called the resonance peak, and we shall see later what happens there. Can we say anything about these evolutions? Yes, certainly. What happens at Omega close to zero ? This is when my force oscillates very slowly. Intuitively I can say that in that case I am just doing successive static loadings of my oscillator. This means that the inertial term in my equation is just negligible versus the stiffness term? The response is just sin(Omega t). Here is what the motion looks like, on a pendulum. No inertial effect. It is just the stiffness of the oscillator that opposes the loading. This is the regime of the stiffness response of the oscillator Let us see what happens at the other limit, of high forcing frequency. In that limit, it is the stiffness term that becomes negligible, in comparison with the inertial term. The external force is just opposed by inertia. This means that the movement is out of phase with the loading, to have an opposed inertia. This also means that the amplitude of the motion is smaller with increasing frequency, because the force is opposed by omega^2 times the displacement. This is the regime of inertial response. We know what happens at low frequencies, the stiffness response, and at high frequencies, the inertial response. But what happens in between? What happens near Omega = 1? And particularly at Omega = 1? At Omega = 1, we have a problem. If q oscillates at the same frequency, then q = sin(t) and (q)ddot = -sin(t). The stiffness and the inertial terms are just opposed and compensate. This means that nothing can oppose the forcing. This is why the amplitude goes to infinity. What happens in practice? If we go back to the Laplace transform for Omega equals one, we have Q = 1/(1 + s^2)^2. The inverse Laplace of this is a bit more complicated and here it is - 1/2[t cos(t)] + sin(t)/2. Now we have something new here! The motion is an oscillation with an amplitude that increases linearly with time, and eventually goes to infinity. Why does it increase? Because nothing opposes motion, and each cycle transfers additional energy to the oscillator. To summarize, here is how an oscillator behaves under harmonic loading, depending on where the frequency of the loading stands in comparison with that of the oscillator. At low frequency, stiffness response. At high frequency, inertial response. At the oscillator frequency, resonant response. Actually, all we found here on what happens on a simple oscillator has a wide range of application. Why? Because the oscillator equation gives you the response of any mode in vibrations! Remember, it is the elementary brick of all our modal dynamics! For instance, when you think of resonance you think of a solid system that vibrates, or sound, in a flute. Here is a simple and rather surprising example. There is bay, in Canada, called the Bay of Fundy. It is an elongated body of water. It has a sloshing mode with a long period of about 12 hours. This sloshing mode is excited by the tides, of course And what is the period of a tide? it is about 12 hours, too. We are near or at resonance and this is why the Bay of Fundy has some of the highest tides in the world, as you can see here, up to 15 or 20 meters. Just because of resonance! We have now come to the conclusion of this series of sequences on the elementary oscillator. This is rather simple material, but it is, as you know, the elementary brick of all vibrations. We shall re-use all this a lot in the next week, on the dynamics of many systems. Next, we are going to give you some additional material on damping and on random loading, and other things!
