= Solution
For the <gyrocompass oscillation>, use the same local basis and define $\theta$ about the stable orientation, with spin <angular momentum> directed north at $\theta=0$. Because the vertical is perpendicular to the horizontal axle, the extra vertical <angular momentum> from the slow azimuthal motion is $I_2\dot\theta$. To leading order in <Earth>'s small <angular velocity>,
$$
\mathbf J\simeq J_s\mathbf u+I_2\dot\theta\,\mathbf z,\qquad J_s=I_1\omega.
$$
A constant vertical contribution $I_2\Omega_E\sin\lambda$ from <Earth's rotation> has zero terrestrial time <derivative>; the transverse terrestrial contributions affect the following equation only at order $\Omega_E^2$. Taking the vertical component of the <torque> equation gives
$$
0=I_2\ddot\theta+J_s\Omega_E\cos\lambda\sin\theta.
$$
The terrestrial <derivative> of $J_s\mathbf u$ is horizontal, as is $\boldsymbol\Omega_E\times(I_2\dot\theta\mathbf z)$, so neither contributes to this component. Thus the small <angle> satisfies, using the <small-angle approximation> $\sin\theta\simeq\theta$,
$$
\ddot\theta+\frac{I_1}{I_2}\omega\Omega_E\cos\lambda\,\theta\simeq0,
\qquad
\boxed{\omega_{\rm shm}^2\simeq\frac{I_1}{I_2}\omega\Omega_E\cos\lambda.}
$$
This is <simple harmonic motion>, $\theta=A\cos(\omega_{\rm shm}t+\phi)$, with positive squared <angular frequency> away from the poles. The approximation requires rapid spin, small angular displacement and negligible bearing <friction>. Near the opposite orientation, putting $\theta=\pi+\delta$ instead yields $\ddot\delta-\omega_{\rm shm}^2\delta\simeq0$, which is unstable. Consequently, the paper's oscillation claim applies to the stable <equilibrium>, not both north--south spin directions. Without <damping> the ideal <gyrocompass> continues to oscillate rather than settling.
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