= Solution
Let $\theta$ and $\varphi$ be the lower and upper azimuthal angles, measured from a common fixed direction. The squared spring length is
$$
d^2=(a\cos\theta-A\cos\varphi)^2+(a\sin\theta-A\sin\varphi)^2+h^2=a^2+A^2+h^2-2aA\cos(\theta-\varphi).
$$
By <Hooke's law>, a spring with zero natural length has <potential energy> $kd^2/2$. Therefore
$$
\boxed{L=\frac12ma^2\dot\theta^2+\frac12MA^2\dot\varphi^2-\frac k2[a^2+A^2+h^2-2aA\cos(\theta-\varphi)].}
$$
Constant gravitational energies can be omitted. It is useful to define the <moments of inertia> $I_1=ma^2$, $I_2=MA^2$ and the angular coupling $\kappa=kaA$. The part of the potential that affects motion is simply $-\kappa\cos(\theta-\varphi)$. In particular, $h$ changes the constant spring energy but not the angular equations, because the spring has zero natural length.
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