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Consider the following example of a coupled oscillator. Let two identical pendulums, each of length $\ell$ and mass $m$ be connected by a spring of force constant $k$. The system has two normal modes with frequencies $\omega_1=\sqrt{g/\ell}$ and $\omega_2=\sqrt{(g/\ell)+(2k/m)}$.

Why is any arbitrary motion of this coupled oscillator writable as a linear combination of the two normal modes? I mean how does the conclusion follow?

Solidification
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  • Relevant: https://physics.stackexchange.com/questions/266889/eigenvalue-equation-for-kinetic-and-potential-energy?noredirect=1&lq=1, https://physics.stackexchange.com/questions/478923/why-are-all-solutions-to-this-system-of-pendulum-differential-equations-a-linear?rq=1 – Andrew May 01 '22 at 18:44
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    Because it's a system of linear differential equations is the simplest explanation. – Triatticus May 01 '22 at 19:41
  • This is the spectral theorem isn't it? – John Rennie May 02 '22 at 09:00

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