Discussion
The figure shows a system of two equal masses \(m\) and three massless horizontal springs with spring constants \(k_{1}\), \(k_{2}\), \(k_{1}\). Ignore gravity. The masses can move only in the horizontal direction and there is no dissipation. If \(m = 1\), \(k_{1} = 2\) and \(k_{2} = 3\) (all in appropriate units), the frequencies of the normal modes of the system in the same system of units are
Answer: A
The normal-mode frequencies are \(\sqrt{2}\) and \(\sqrt{8}\).
With displacements \(x_{1}, x_{2}\), the equations of motion are
\[ \begin{aligned} m\ddot{x}_{1} &= -k_{1}x_{1} - k_{2}(x_{1} - x_{2}) \\ m\ddot{x}_{2} &= -k_{1}x_{2} - k_{2}(x_{2} - x_{1}) \end{aligned} \]
In the symmetric mode (\(x_{1} = x_{2}\)) the middle spring is not stretched, so \(\omega^{2} = \frac{k_{1}}{m} = 2\). In the antisymmetric mode (\(x_{1} = -x_{2}\)) the middle spring is stretched twice as much, so \(\omega^{2} = \frac{k_{1} + 2k_{2}}{m} = 2 + 6 = 8\).
\(\sqrt{6}\) forgets the outer springs in the second mode (it is \(2k_{2}\) alone), and the options with \(\sqrt{3}\) use \(k_{2}\) for the in-phase mode, where the middle spring plays no part.
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