- #1
allshaks
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I'm currently reading class notes from an introductory waves course, written by the professor himself. I'm stuck in the Fourier analysis part, because he gives the formulas for the nth mode amplitude of a standing wave with fixed ends and then states some properties which I can't really make sense of. The formula given is:
$$B_p=\frac{2}{L}\left\{ \left[ \int_{0}^{L}{\psi (x,0) \sin{k_p x} dx} \right] ^2 + \frac{1}{\omega _p^2}\left[ \int_0^L{\frac{\partial \psi(x,0)}{\partial t} \sin{k_p x} dx} \right] ^2 \right\} ^{\frac{1}{2}}$$
No problem with this. But then one of the properties listed is (translated from spanish):
"If the initial velocity of every point of the rope is zero and if the initial shape has a node that coincides with a mode's node, the rope will continue oscillating in a superposition of modes that also have a node at the same place."
The next one is:
"In analogy, when the initial shape has non-zero displacement at the place that corresponds to a harmonic's node, the resultant effect is the suppression of such harmonic."
Is this really true? I find it extremely puzzling. I mean, I'm quite sure that they are wrong. For instance, the second one. I found counterproof, there are lots of examples. ##-|x-\frac{L}{2}|+\frac{L}{2}## has plenty of harmonics, for starters. Not just the first mode. What I have seen, but not proven, is that it seems to be that when the initial shape has not only non-zero but a maximum in the amplitude at the place that corresponds to a harmonic's node, then indeed the harmonic seems to be suppressed.
The first statement doesn't seem to be true either. Playing with Mathematica, I found functions with nodes that did have some harmonics without that node. Again, I have seen something without giving a proof. That is, that if the function has a certain symmetry, then indeed the modes with no nodes seem to be getting suppressed. I find that this has to do with the following: odd harmonics are "mirrored" in the middle of the rope, whereas even harmonics are not only mirrored but also inverted. They have a different kind of symmetry. This would mean, I guess, that functions with the first kind of symmetry will only be composed of odd harmonics and those with the second kind of even harmonics. But a general type of function will indeed be composed of harmonics with both sorts of symmetries.
I've also related this, in my mind, with energy flow through the string. If the shape of the string is symmetric in any way, there won't be a flow of energy from one side to the other. That's simply because you can rotate the frame of reference, and that shouldn't make a difference into which side the energy is flowing. Nodes are places where there is no net exchange of energy, right? I mean, if there were some energy going from one side of the string to the other, the node should move because there's energy being transfered. So my guess has something to do with this: if the zero is not in a kind of symmetric position, that is, if the energy is not distributed symmetrically in some way (I'm not sure what that would be in a most general case), then it will not remain at position 0, it will not be a node, because there will be some energy going from one place to the other through that point. This means that there has to be, necessarily, a harmonic with no node in that point.
Anyway, this is all pure intuition, extremely prone to be flawed. But because I have to take an exam, I would like to know exactly what parts of this reasoning are wrong, why, and maybe were there anything correct about it, how I can justify it with more convincing arguments. Thank you very much.
$$B_p=\frac{2}{L}\left\{ \left[ \int_{0}^{L}{\psi (x,0) \sin{k_p x} dx} \right] ^2 + \frac{1}{\omega _p^2}\left[ \int_0^L{\frac{\partial \psi(x,0)}{\partial t} \sin{k_p x} dx} \right] ^2 \right\} ^{\frac{1}{2}}$$
No problem with this. But then one of the properties listed is (translated from spanish):
"If the initial velocity of every point of the rope is zero and if the initial shape has a node that coincides with a mode's node, the rope will continue oscillating in a superposition of modes that also have a node at the same place."
The next one is:
"In analogy, when the initial shape has non-zero displacement at the place that corresponds to a harmonic's node, the resultant effect is the suppression of such harmonic."
Is this really true? I find it extremely puzzling. I mean, I'm quite sure that they are wrong. For instance, the second one. I found counterproof, there are lots of examples. ##-|x-\frac{L}{2}|+\frac{L}{2}## has plenty of harmonics, for starters. Not just the first mode. What I have seen, but not proven, is that it seems to be that when the initial shape has not only non-zero but a maximum in the amplitude at the place that corresponds to a harmonic's node, then indeed the harmonic seems to be suppressed.
The first statement doesn't seem to be true either. Playing with Mathematica, I found functions with nodes that did have some harmonics without that node. Again, I have seen something without giving a proof. That is, that if the function has a certain symmetry, then indeed the modes with no nodes seem to be getting suppressed. I find that this has to do with the following: odd harmonics are "mirrored" in the middle of the rope, whereas even harmonics are not only mirrored but also inverted. They have a different kind of symmetry. This would mean, I guess, that functions with the first kind of symmetry will only be composed of odd harmonics and those with the second kind of even harmonics. But a general type of function will indeed be composed of harmonics with both sorts of symmetries.
I've also related this, in my mind, with energy flow through the string. If the shape of the string is symmetric in any way, there won't be a flow of energy from one side to the other. That's simply because you can rotate the frame of reference, and that shouldn't make a difference into which side the energy is flowing. Nodes are places where there is no net exchange of energy, right? I mean, if there were some energy going from one side of the string to the other, the node should move because there's energy being transfered. So my guess has something to do with this: if the zero is not in a kind of symmetric position, that is, if the energy is not distributed symmetrically in some way (I'm not sure what that would be in a most general case), then it will not remain at position 0, it will not be a node, because there will be some energy going from one place to the other through that point. This means that there has to be, necessarily, a harmonic with no node in that point.
Anyway, this is all pure intuition, extremely prone to be flawed. But because I have to take an exam, I would like to know exactly what parts of this reasoning are wrong, why, and maybe were there anything correct about it, how I can justify it with more convincing arguments. Thank you very much.