Damped Simple Harmonic Motion — Question 2

Question 2

Release from rest requires a sine term in the decaying motion

An underdamped oscillator has $\beta=1\,\mathrm{s^{-1}}$ and $\omega_d=3\,\mathrm{rad/s}$. It is released from $x_0=.10\,\mathrm{m}$ with zero velocity. Write its motion as $x=e^{-\beta t}(C\cos\omega_dt+D\sin\omega_dt)$.

1. Find $C,D$ and the first time it crosses equilibrium.

2. Explain why $x=x_0e^{-\beta t}\cos\omega_dt$ is incompatible with release from rest. Check the initial acceleration of the correct solution.

Question 2 - Solution

Release from rest requires a sine term in the decaying motion

(a) Derivation and calculation

$$x(0)=C=x_0=.100\,\mathrm m,\quad\dot x(0)=-\beta C+\omega_dD=0.$$$$\boxed{D=\beta C/\omega_d=.033333\,\mathrm m}.$$$$\boxed{x=e^{-t}(.100\cos3t+.033333\sin3t)\,\mathrm m}.$$

(b) Result and interpretation

$$\cos3t+\tfrac13\sin3t=0\Rightarrow\tan3t=-3.$$$$\boxed{t_{\mathrm{first}}=\frac{\pi-\arctan3}{3}=.63085\,\mathrm s}.$$$$\dot x_{\mathrm{cosine-only}}(0)=-\beta x_0=-.100\,\mathrm{m/s}\ne0.$$$$\ddot x(0)=-2\beta\dot x(0)-(\omega_d^2+\beta^2)x_0=\boxed{-1.00\,\mathrm{m/s^2}}.$$

Physical check

The first root lies in the second phase quadrant; using the negative principal arctangent would give an unphysical negative crossing time.