I can reconcile the idea that a termination point is a node with the idea that the bridge moves. If the termination point moves, it's not a node. But you still get modes of vibration on the string. I have done the calculation for the following problem: A string is fixed at one end. At the other end, it is attached to a ring that slides freely (i.e. no friction) up and down a pole. In other words, the non-fixed end can move freely up and down. The ring has a mass, and there is a constant tension in the string. This is not really a model of what happens in a piano string, but it demonstrates a scenario where a termination point moves. The result is that if the mass is infinitely large (which corresponds to two fixed termination points), you get the result you'd expect for two fixed termination points: the fundamental has one node at each termination point. If the mass is zero, for the fundamental you get a node at the fixed end of the string, and then the free end is where the string has its maximum amplitude (which is like the "thumb piano" in the link given below). In the next higher mode for this case, there is a node at the fixed termination point and another node 2/3 the speaking length away from the fixed termination point. The string has its maximum amplitude at the freely moving termination point, as well as the point 1/3 the speaking length away from the fixed termination point. As the mass is increased from zero, the node at 2/3 the speaking length away from the fixed termination point (in the overtone described above) moves towards the free termination point. Eventually, as the mass becomes infinite, this mode becomes the fundamental mode of the case with two fixed termination points. (What happens to the fundamental of the case where the mass is zero? Well, as the mass increases, the wavelength gets longer and longer and eventually becomes infinitely long as the mass approaches infinity, corresponding to a zero frequency mode for the case with two fixed termination points.) So, when you have a big but not infinite mass (i.e. the "free" termination point moves only a little bit), you have a node very near the free termination point, but not actually AT it. An interesting thing is that in this model, the more the bridge moves the sharper the string gets. But also as the brige moves more, I think there's a little less tension on the string in real life (but not in this simple model), so that would make the string flatter. More investigation is needed here. A while ago we were discussing the phenomenon where three strings produce a tone which is flatter than each individual string. It was concluded (I have some links to the archives if anyone is interested) that three strings vibrating in phase cause the bridge to move more. This movement makes the speaking length "seem" longer, and thus the sound is flatter. That could mean that the node moves from just about on the bridge to just beyond the bridge. (Alternatively, I wonder if the movement of the bridge relieves tension in the string. Something I've been thinking about.) Also, here's an interesting link on standing waves. I refered to this above. http://id.mind.net/~zona/mstm/physics/waves/standingWaves/standingWaves.html Again, the model I described above is not intended to model a piano string exactly. But it does demonstrate that a string can have vibrational modes where one termination point can move. Charles P.S. I guess it's time I "came out" as having a significant physics background. However, that doesn't mean I'm an expert in acoustics, so I haven't had much to say before.
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