= Free two-mass spring
{title2=$\omega=\sqrt{k/\mu}$}
Take two <point masses> joined by a massless <spring> of <spring constant> $k>0$ and equilibrium length $\ell>0$. Assume one-dimensional motion along the <spring>, no external <force>, no <damping>, and an extension $s=x_2-x_1-\ell$ satisfying <Hooke's law>; keep $x_2>x_1$. For a real <spring> this is a model within its elastic range. Its <potential energy> is $U=ks^2/2$, so
$$
\mu\ddot s=-ks,\qquad s(t)=A\cos(\omega t+\phi),\qquad
\omega=\sqrt{\frac{k}{\mu}}=\sqrt{k\left(\frac1{m_1}+\frac1{m_2}\right)},\qquad \nu=\frac\omega{2\pi}.
$$
Here $\omega$ is <angular frequency> and $\nu$ is <frequency>. Choose $|A|<\ell$ so the model bodies do not cross. The internal <normal mode> has <displacements> $u_1=-m_2s/M$ and $u_2=m_1s/M$ about equilibrium offsets from the <centre of mass>; its <amplitude> ratio is $|u_1|/|u_2|=m_2/m_1$. The other <normal mode> is collective translation with zero <frequency> and no <spring> extension. This free pair has different boundary conditions from <normal modes of two equal masses between three springs>.
For $k=10\,\mathrm{N/m}$ and two $1\,\mathrm{kg}$ <masses>, $\omega=\sqrt{20}\,\mathrm{s^{-1}}\approx4.47214\,\mathrm{s^{-1}}$ and $\nu\approx0.711763\,\mathrm{Hz}$. Replacing the second <mass> by $100\,\mathrm{kg}$ gives $\omega=\sqrt{10.1}\,\mathrm{s^{-1}}\approx3.17805\,\mathrm{s^{-1}}$ and $\nu\approx0.505802\,\mathrm{Hz}$. Fixing the second endpoint instead would give $\sqrt{10}\,\mathrm{s^{-1}}$: the free heavy partner still moves. The free pair's <angular frequency> exceeds that fixed-end limit by the factor $\sqrt{1+1/q}$, about $1.004988$ at $q=100$.
\Image[/reduced-mass-ratio-spring-frequency.png]
{title=<Reduced mass> and <free two-mass spring> <angular frequency> as the partner becomes heavier}
{id=image-reduced-mass-ratio-spring-frequency}
{description=Original dimensionless model with $m_1$ and $k$ held fixed, $q=m_2/m_1>0$, $\mu/m_1=q/(1+q)$ and $\omega/\sqrt{k/m_1}=\sqrt{1+1/q}$. The marks at $q=1$ and $q=100$ correspond to the computed examples; curves are predictions, not measured data.}
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