Damped Simple Harmonic Motion — Question 9

Question 9

A contracting state path has an energy-based radius

For a viscously damped oscillator with $\omega_0^2=k/m$, define state coordinates $X=x$ and $Y=v/\omega_0$. Let $\rho^2=X^2+Y^2$.

1. Derive $d(\rho^2)/dt$ and interpret it physically.

2. At a turning point and at an equilibrium crossing, compare the instantaneous contraction of $\rho^2$. Does the state remain on an ellipse of fixed area?

Question 9 - Solution

A contracting state path has an energy-based radius

(a) Derivation and calculation

$$\rho^2=x^2+v^2/\omega_0^2,\qquad\dot v=-\omega_0^2x-2\beta v.$$$$\frac{d\rho^2}{dt}=2xv+\frac{2v\dot v}{\omega_0^2}=\boxed{-4\beta\frac{v^2}{\omega_0^2}=-4\beta Y^2\le0}.$$$$E=\tfrac12k\rho^2\Rightarrow\dot E=-2k\beta v^2/\omega_0^2=-bv^2.$$

(b) Result and interpretation

$$\text{turning point: }Y=0\Rightarrow\boxed{d\rho^2/dt=0};$$$$\text{equilibrium crossing: }X=0\Rightarrow\boxed{d\rho^2/dt=-4\beta\rho^2}.$$The state follows an inward spiral rather than a fixed ellipse. Its radial contraction varies around a cycle even when the displacement envelope is exponential.

Physical check

When $\beta=0$, the radius becomes constant and the conservative state circle is recovered.