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Lowest freq mode of fixed-end linear chain oscillations

  1. Jan 29, 2008 #1

    (All oscillations I'll be talking about here are longitudinal.)

    For coupled oscillations of 2 masses between 3 identical springs (ends held fixed by walls), I think it was a standard textbook mechanics problem to show that the lowest-frequency mode is the symmetric one (where the masses move in the same direction).

    1. Is there a higher-level reason for this?
    (by "higher-level" I mean knowing the answer without doing the calculation ie. the "physics" answer; by "lower-level" I mean actually going through the tedious coupled ODE-solving to find out)

    2. Will this be true for N masses between N+1 springs (again, ends fixed)?
    I'm asking this because in the N=2 case, both masses have at least one spring to interact with. But for the "symmetric" mode (all masses moving together) in a larger-N system, the masses close to the middle barely seem to interact with any spring at all. In this case, it seems natural to imagine that the 2 masses at the ends of the chain bear all the stress of driving the motion...

    Any insights much appreciated!
    Last edited: Jan 29, 2008
  2. jcsd
  3. Jan 30, 2008 #2


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    For two masses, the symmetrilc mode only stretches one spring for each mass, so the effective k is least.
    For larger N, the equations are like those of transverse waves on a string. The lowest mode will have no nodes, which only happens for the symmetric mode.
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