Damped Simple Harmonic Motion — Question 5

Question 5

Critical damping does not forbid an equilibrium crossing

A critically damped oscillator has $\omega_0=\beta=4\,\mathrm{s^{-1}}$. Its general motion is $x=(C+Dt)e^{-4t}$. Initially $x_0=.10\,\mathrm{m}$ and $v_0=-.60\,\mathrm{m/s}$.

1. Find the motion and decide whether it crosses equilibrium.

2. For general $x_0>0$, derive the initial-velocity condition for a future crossing. Explain why a single crossing is not an oscillation.

Question 5 - Solution

Critical damping does not forbid an equilibrium crossing

(a) Derivation and calculation

$$x=(C+Dt)e^{-4t},\quad C=x_0=.100,\quad v_0=D-4C.$$$$D=-.600+4(.100)=-.200\,\mathrm{m/s}.$$$$\boxed{x=(.100-.200t)e^{-4t}\,\mathrm m},\quad\boxed{t_{\mathrm{cross}}=.500\,\mathrm s}.$$

(b) Result and interpretation

For general $x_0>0$,$$D=v_0+\omega_0x_0,\quad x=0\iff x_0+Dt=0.$$$$t_{\mathrm{cross}}=-x_0/D>0\iff\boxed{v_0<-\omega_0x_0}.$$The positive exponential never changes sign and the linear prefactor has at most one root, so a single crossing does not constitute repeated oscillation.

Physical check

Critical damping specifies a repeated real characteristic root; initial conditions still determine overshoot.