Lowest freq mode of fixed-end linear chain oscillations

[iibewegung]

Hi,

(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).
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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!

 

[pam]

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.

 

[charng]

1. The reason for the lowest-frequency mode being the symmetric one in a fixed-end linear chain oscillation system is due to the conservation of energy and the equilibrium position of the masses. In the symmetric mode, the masses are all moving in the same direction, which means that the potential energy stored in the springs is at its minimum. This also means that the kinetic energy of the masses is at its maximum, resulting in the lowest frequency of oscillation.

2. This principle holds true for any number of masses between any number of springs in a fixed-end linear chain oscillation system. The masses in the middle may seem to have less interaction with the springs, but they still play a crucial role in maintaining the equilibrium and energy conservation of the system. In fact, as the number of masses increases, the symmetric mode becomes even more dominant, as the masses on the ends have more springs to interact with and distribute the stress of driving the motion.