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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.

Write $\gamma =b/(2m)$ and $\nu =\sqrt{{\omega }^2-{\gamma }^2}$, where $\omega =\sqrt{k/m}$ is the undamped angular frequency. The linear drag gives the vector damped harmonic oscillator equationFor the original circular initial conditions, ${\mathbf{r}}_0=(c,c)$ and ${\mathbf{v}}_0=\omega (-c,c)$, the exact underdamped solution isUsing $|{\mathbf{r}}_0{|}^2=2c^2$, $|{\mathbf{v}}_0{|}^2=2{\omega }^2{c}^2$ and ${\mathbf{r}}_0\cdot {\mathbf{v}}_0=0$ givesThis exact potential energy is useful because it satisfies the prescribed initial conditions, including $\dot{U}(0)=k{\mathbf{r}}_0\cdot {\mathbf{v}}_0=0$. In the stated weak-damping limit $\gamma \ll \omega$, $\nu \simeq \omega$ and the square bracket differs from one by a small oscillatory term. The leading approximation is thereforeThe radius is approximately $\sqrt{2}c{e}^{-bt/(2m)}$, so the trajectory is approximately a shrinking logarithmic spiral. The exactly exponential result would require slightly adjusted initial velocity; with the specified undamped circular velocity, the small modulation above is real. This is the initially circular damped planar oscillator.

The parameters giveThe coordinate expressions for the initially circular damped planar oscillator areThe damped oscillation period $2\pi /\nu =1.00052\mathrm{s}$ gives almost four anticlockwise turns in four seconds. The radius envelope falls by $e^{-\gamma (4\mathrm{s})}=e^{-1}=0.3679$. Substitution into the exact coordinates givesThe left panel annotates the initial tangent, time marks and endpoint. Because $c$ is an unspecified starting coordinate, the axes are normalized by $c$ rather than assigned numerical lengths.