Hello! Last week we have found that any problem of dynamics could be much simplified by using modal superposition. How was it simplified? Well, instead of having to solve a full problem in space and time, we had only a series of simple problems in time of individual modal dynamics. We went from 4 dimensions (3 in space and 1 in time) to an infinite series of problems of dimension 1 (in time). This means that you can write the displacement of the system as the sum of contributions by each and every mode of the system. And if you do that the problem to solve takes always the same form - a series of independent differential equations. And what is remarkable is that this form is common to all problems in linear dynamics: cables, surface waves in fluids, acoustics. Of course, the modes of your system are going to be different in mode shapes, in frequencies if you consider a cable or acoustics. But the equations that govern their contributions always have the same form. This week we are going to take advantage of this to solve any problem of dynamics. And you will be able to use that approach in a large variety of problems. But for this we need to solve all these new equations, the modal equations. Let us have a closer look at the equation that governs the contribution of each mode, what we noted the q_N(t). This equation is always m_N (q_N)ddot + k_N q_N = f_N. Remember that m_N and k_N are respectively the modal mass and modal stiffness of the mode we are considering. And f_N(t) in the projection on the mode of the loading on the system. Of course, these quantities are going to be quite different from cases to cases. Remember also that dot means time derivation so that we have actually m_N d^2 q_N / dt^ 2+k_N q_N = f_N. Certainly, we can write this in a simpler way, using rescaled quantities. Let us define a new time, a dimensionless time T_N as the product of the modal frequency omega_N t = sqrt(k_N/m_N) t. Then by a simple change of variable in time we can re write our equation above as m_N omega_N^2 d^2q_N / dT_N^2 + k_N q_N = f_N. Here is the equation again. Now, because omegan is sqrt(k_N/m_N), this reads simply k_N d^2q_N/dT_N^ 2 + k_N q_N = f_N, and by deviding by k_N and upon using F_N=f_N/k_N, we have only d^2 q_N/dT_N^2 + q_N = F_N. This is much simpler! There is something else, quite important, to consider. Dynamics always includes initial conditions, on the state of the system at the origin of time for us, time t equals zero. In the modal space, this implies initial conditions on each and every mode, as we found last week. Remember, that these initial conditions can be computed from the initial conditions in the real domain, and for each mode we have noted them q_N at time zero equals q_N^0, a quantity that we know and (q_N)dot at time zero equals (q_N)dot^0 a quantity that we know. We may now write these same initial conditions, using our new time scale, the first one is unchanged, of course, q_N at time = 0 is q_N^0. the second one is a time derivative. So, if we change the time from t to Tn, again we have the new initial condition d q_N/d T_N at time 0 = sqrt(m_N/k_N)(q_N)dot^0. These are our two initial conditions considering the new dimensionless time. Let us summarize. Each and every modal equation had the form of m_N (q_N)ddot + k_N q_N = f_N. After using appropriate changes of variables, one different for each mode, all these equations take exactly the simple common form of (q)ddot + q = f, with well defined initial conditions. Of course, for each mode, the time reference is different, but if we can solve this one and only equation, we can solve everything. In other terms, to solve all the modal equations we only need to solve this simple one. This is a major result in dynamics. Let us summarize. Last week by using modes, we could split the general problem of dynamics to a series of independent and simpler problems of modal dynamics. That was nice, because we could hope to solve these modal problems one by one and then rebuild the whole solution. Schematically, I would say that we had individual stones, which are easier to handle than the whole stone wall. These individual stones were the modal equations, each simple enough, but different for every mode. But we just found that these stones are actually all the same, in the sense that with appropriate rescaling the modal equations reduce all to the same elementary oscillator equation. So, in a sense, these stones are actually bricks, and all identical. We just have to solve one problem, that of the oscillator equation, and everything will be solved. What is a bit surprising is that this would apply to problems of dynamics of all kinds. It would work for the dynamics of tensionned cables, here in a bridge but also for sloshing of a fluid with a free surface, here in a glass or even the acoustics of a musical instrument such as a saxophone. For all of them the one and only problem to solve is that of the elementary brick , the oscillator equation. Do you know what we should do now? Certainly, we should solve this elementary oscillator equation which is the key to all dynamics. This is exactly what you will do in the next video.