Feel free to test your latex phrases here. I think that things enclosed with a slash-paren will be interpreted as latex now. Anyone want to write a short tutorial?
I have a few words on the debacle presented earlier (regarding a nonzero velocity for our purely translational mode). I may not have the full story here, but I can give background as to how we can safely say that \[x_1\] and \[x_2\] in general can depend on \[vt\] explicitly. Unless there is some math trick I am neglecting, I don’t believe the required solution can be manipulated out of our assumed sinusoidal solution, which can only give us a constant term in position. We must go back and look at our equations of motion:
We started here by assuming a sinusoidal solution, which fits fine, of course. However, with our zero frequency mode, it is not the only one that fits. In fact, plugging in \[\omega = 0\] and \[x_1 = x_2\], we get a trivial \[0 = 0\] in both cases. But, if we start with \[x_1 = x_2\], we have:
\[ m\ddvec{x_1} = 0 \\ m\ddvec{x_2} = 0 \]
which can be satisfied if \[x_1 = x_2\] is of the form \[A + Bt\], where \[A\] and \[B\] are determined by initial condition. Then, differentiating, we have a potentially constant, nonzero velocity as we wanted.
However, this seems a little bit roundabout. After all, the entire process that brought us to this was assuming that x $was$ sinusoidal. It would be nice if we could get this result from more general reasoning; we can do so by considering the center of mass coordinate, \[X = (x_1+x_2)/2\]. Given no external forces, \[m\dd{X}= 0\] which we can verify quickly by rearranging our 2 E.O.M.s into one in terms of our CM coordinate. Then, we integrate down to \[X = A + Bt\] as before. This does not explicitly assert that \[x_1 = x_2\], but if we only consider CM motions such that the distance between the two masses does not change (i.e. normal to spring oscillation) that relation follows.
Anyway, I think that’s on the right track. I’d love to hear corrections/elaborations.
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ReplyDelete\[\omega\] test
ReplyDeleteI have a few words on the debacle presented earlier (regarding a nonzero velocity for our purely translational mode). I may not have the full story here, but I can give background as to how we can safely say that \[x_1\] and \[x_2\] in general can depend on \[vt\] explicitly. Unless there is some math trick I am neglecting, I don’t believe the required solution can be manipulated out of our assumed sinusoidal solution, which can only give us a constant term in position. We must go back and look at our equations of motion:
ReplyDelete\[ m\ddvec{x_1} = k(x_2-x_1) \\
m\ddvec{x_2} = k(x_1-x_2) \]
We started here by assuming a sinusoidal solution, which fits fine, of course. However, with our zero frequency mode, it is not the only one that fits. In fact, plugging in \[\omega = 0\] and \[x_1 = x_2\], we get a trivial \[0 = 0\] in both cases. But, if we start with \[x_1 = x_2\], we have:
\[ m\ddvec{x_1} = 0 \\
m\ddvec{x_2} = 0 \]
which can be satisfied if \[x_1 = x_2\] is of the form \[A + Bt\], where \[A\] and \[B\] are determined by initial condition. Then, differentiating, we have a potentially constant, nonzero velocity as we wanted.
However, this seems a little bit roundabout. After all, the entire process that brought us to this was assuming that x $was$ sinusoidal. It would be nice if we could get this result from more general reasoning; we can do so by considering the center of mass coordinate, \[X = (x_1+x_2)/2\]. Given no external forces, \[m\dd{X}= 0\] which we can verify quickly by rearranging our 2 E.O.M.s into one in terms of our CM coordinate. Then, we integrate down to \[X = A + Bt\] as before. This does not explicitly assert that \[x_1 = x_2\], but if we only consider CM motions such that the distance between the two masses does not change (i.e. normal to spring oscillation) that relation follows.
Anyway, I think that’s on the right track. I’d love to hear corrections/elaborations.
formatting is pretty bad here uhhh maybe ill fix it later
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ReplyDeleteThis comment has been removed by the author.
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