Initially circular damped planar oscillator ($\ddot{\mathbf{r}}+2\gamma \dot{\mathbf{r}}+{\omega }^2\mathbf{r}=0$)

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, $\nu =\sqrt{{\omega }^2-{\gamma }^2}$ givesTaking the squared norm and multiplying by $k/2$ gives the exact potential energyOnly to leading order in $\gamma /\omega \ll 1$ does this reduce to $k{c}^2{e}^{-2\gamma t}$. 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.

For the initially circular damped planar oscillator with overdamping $\gamma >\omega$, put $\alpha =\sqrt{{\gamma }^2-{\omega }^2}$. The exact solution replaces $\cos \nu t,\sin \nu t/\nu$ by $\cosh \alpha t,\sinh \alpha t/\alpha$. Resolving along the starting radius and its anticlockwise tangent yieldsIts limiting angle is the title value. Both coordinates remain positive when $c>0$, so the trajectory returns to the origin without repeated revolutions. It initially moves tangentially and bends toward its slow decaying eigenmode, rather than following a radial straight line from the start.