= Solution
Let $x$ be signed displacement along the rails, measured from the centre. The diagram in the original paper shows a guide extending across the centre, so the signed coordinate can pass through zero. <Hooke's law> gives the radial <force> $-kx$. In the <rotating reference frame>, the radial <centrifugal force> is $+m\Omega^2x$. The <Coriolis force> is perpendicular to the rails and is balanced by their reaction; it has no radial component. Therefore <Newton's second law> gives
$$
m\ddot x=-kx+m\Omega^2x,\qquad\ddot x+\left(\frac km-\Omega^2\right)x=0.
$$
Equivalently, the radial component of <acceleration in polar coordinates> in the <inertial frame> is $\ddot x-\Omega^2x$. The <radial spring oscillator on a rotating guide> has <simple harmonic motion> only when $k/m-\Omega^2>0$, with
$$
\boxed{\omega_{\rm osc}=\sqrt{\frac km-\Omega^2},\qquad f_{\rm osc}=\frac1{2\pi}\sqrt{\frac km-\Omega^2}.}
$$
The <angular frequency> is in $\mathrm{rad\,s^{-1}}$ and the <frequency> in $\mathrm{Hz}$. The critical <angular speed> is
$$
\boxed{\Omega_{\rm crit}=\sqrt{k/m},\qquad 0\le\Omega<\Omega_{\rm crit}.}
$$
Strictly this is an upper bound, not an attained maximum for <oscillations>: at equality $\ddot x=0$, so a displaced object released from rest stays at its displaced position relative to the rails; above it, the central <equilibrium> is unstable.
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