= Initially circular damped planar oscillator
{title2=$\ddot{\mathbf r}+2\gamma\dot{\mathbf r}+\omega^2\mathbf r=0$}
For an isotropic <harmonic oscillator> with <mass> $m$, restoring constant $k$ and <linear drag> coefficient $b$, put $\omega=\sqrt{k/m}$ and $\gamma=b/(2m)$. Choose initial <position> $\mathbf r_0=(c,c)$ and tangential <velocity> $\mathbf v_0=\omega(-c,c)$, which would give anticlockwise <circular motion> without <damping>. For <underdamping>, $\nu=\sqrt{\omega^2-\gamma^2}$ gives
$$
\mathbf r(t)=e^{-\gamma t}\left[\mathbf r_0\cos\nu t+\frac{\mathbf v_0+\gamma\mathbf r_0}{\nu}\sin\nu t\right].
$$
Taking the squared norm and multiplying by $k/2$ gives the exact <potential energy>
$$
U(t)=kc^2e^{-2\gamma t}\left[1+\frac\gamma\nu\sin2\nu t+\frac{2\gamma^2}{\nu^2}\sin^2\nu t\right].
$$
Only to leading order in $\gamma/\omega\ll1$ does this reduce to $kc^2e^{-2\gamma t}$. The exact expression has $\dot U(0)=0$, consistent with the initially tangential <velocity>, while the leading approximation does not reproduce this short-time derivative. Angular momentum obeys $L_z(t)=L_z(0)e^{-2\gamma t}$ by <angular momentum decay under uniform linear drag>.
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