We have found that a tensioned cable with boundary conditions at both extremities could support what we called modes of vibrations. This was a rather general feature because we could find it everywhere, when the system was dispersive or not dispersive. And it also existed in acoustics, and you will see other examples with surface waves, beams, and so on. Modes are nice oscillatory solutions that satisfy both the equation inside the domain and the boundary conditions. Among all the possible ways that the cable can move they are a bit particular because they stay the same in time - just a shape that oscillates. We could consider them just as rather odd cases. Symbolically, in the landscape of dynamics, like some kind of pearls. And because we have many modes for a given system Many pearls. Nice, but not much useful. But there is much more to it! You are now going to enter a bit further into an essential and very useful property of modes, what we call modal orthogonality. This is a bit mathematical, but it is an essential step in dynamics. Let get in there step by step. What do I mean by orthogonality? You know what orthogonal or perpendicular directions are. If you have two vectors, u and v, they are orthogonal if their inner product is zero, u.v = 0. This is pure geometry. But in Mechanics we also use the inner product, when we talk about power of a force F. The power P is actually the inner product of F with the velocity V of the point where the force is applied. More generally, if we have a continuous system, of dimension one, with deformation in one direction, like ours, we can define the power P of a force f(x,t). It is the sum over the whole domain of the force times the local velocity v(x,t). This inner product is noted (f,v). Fine, but what about modes? Imagine that we have a free motion on mode N, say y_N(x,t) = phi_N(x) sin(omega_N t). This motion satisfies the equation for the cable on elastic foundation, in the general case, m (y_N)ddot - T (y_N)'' + a y_N = 0. In this equation, the three terms have the dimension of forces - inertia force, tension induced stiffness force, foundation stiffness force. I can compute the power of such forces in a given motion. Let us compute the power of the inertia force m (y_N)ddot in a motion on a different mode, say P , y_P(x,t) = phi_P(x) sin(omega_P t), The power will be given by the inner product between the inertia of mode N and the velocity of mode P, (m (y_N)ddot , (y_P)dot), which is just a mix of cosines and sines times (m phi_N , phi_P). The core of this quantity is the inner product (m phi_N, phi_P). What is that? A product involving the two mode shapes. Can we compute it? Yes, and we know its value. Remember that the motion according to mode N satisfied the equation of motion, of course m (y_N)ddot - T (y_N)'' + a y_N = 0. Which means that the modal shape phin satisfies -omega_N^2 m phi_N - T (phi_N)''+a phi_N = 0. So, my quantity of interest, (m phi_N, phi_P) is also (1/omega_N^2) ((-T (phi_N)''+a phi_N), phi_P). What is that? We have a ((phi_N)'', phi_P) term and a (phi_N,phi_P) one. The ((phi_N)'', phi_P) is sum from zero to L of (phi_N)'' phi_P. I can integrate this by part twice and get the sum from zero to L of (phi_P)'' phi_N plus something which involves the values at the boundaries and that is equal to zero because the mode shapes satisfy the boundary conditions. So my ((phi_N)'', phi_P) is also ((phi_P)'', phi_N) On the other side, the <phi_N,phi_P> is of course (phi_P,phi_N). We are almost there, my quantity of interest (m phi_N , phi_P) also reads (1/omega_n^2) ( (-T (phi_N)''+a phi_N), phi_P). I have permuted the N and P's. Here is it again. But phi_P satisfies the equation dynamic equation with the frequency omega_P. As a result we have simply (m phi_N, phi_P) = omega_P^2/omega_N^2 (m phi_N,phi_P). That was a long derivation, but we have obtained something very important here. Look ! Because the two frequencies are different, their ratio is not one. The only solution for this to be satisfied is that our quantity of interest is exactly zero. (m phi_N, phi_P)= 0. This is a major result of dynamics. What does it say, physically? Remember, we took the inertia of the motion of mode N and computed the power by mixing with the velocity of mode P. That power is zero. It means that a motion on a mode does not interplay mechanically with the others. This is why they are modes! It you have a motion on a mode it will continue as such. They are mechanically orthogonal when I used the mass, like here. But I can derive the same results using the stiffness (-T (phi_N)''+a phi_N,phi_P) = 0. Modes are mechanically orthogonal in terms of mass and in terms of stiffness. You may think all this is because the modal shapes are sinusoidal. Not at all. I never used that property in the derivation. Of course, in simple cases where the modal shapes are sin(N pi x/L), the modal shapes are orthogonal in some way. But this does not mean anything mechanically. If we take a cable where the mass per unit length is not uniform like here, the modal shapes would not be sinusoidal any more but orthogonality, mechanical orthogonality through mass, still holds (m(x) phi_N, phi_P)= 0. Surprising! You will see other examples of this later. Let us summarize. We tried to work on the properties of modes, and we found something a bit surprising - they are mechanically orthogonal. What I called pearls, in the sense of some very special solutions are actually not just individuals. They have some relations between them, orthogonality, which I symbolize here. Can we do anything with this? We can do a lot. Let us see that next.