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Mechanics }Normal Modes
Because there are two degrees of freedom in this problem, there are two normal mode frequencies.
Because there is no external torque acting on the system, the center of mass of the system stays the same through time.
From common sense, one deduces that  has to do with the outer masses moving perfectly out of phase, i.e., masses A and C moving either towards the left and right (away from each other), respectively, or right and left (towards each other), respectively---and B being perfectly stationary, thereby ``conserving" the center of mass.
The other angular frequency,  , has to do with either masses A and C moving in phase and mass B out of phase.
is actually equivalent to having a single mass on a string, since because the middle mass doesn't move, it acts as a sort of support for the spring.  , which would correspond to choice (B). (Of course, one should recall the obvious, that  .)
(Incidentally, one can derive without having to resort to the formalism of matrix mechanics: Since the center of mass remains 0, one has  . Solving, one gets  . The displacement of the middle mass, mass B, is thus  , while the displacements of the smaller masses, masses A and C, are both  . The displacement of each spring is  . Potential energy is thus ^2=\frac{1}{2}8kx^2) . The kinetic energy is ^2+\frac{1}{2}(2m)(\dot{x})^2=\frac{1}{2}4m(\dot{x})^2) . The normal mode frequency is deduced by  )
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