Simple Harmonic Motion and Uniform Circular Motion — Question 2

Question 2

The correct auxiliary circle uses a scaled velocity coordinate

An oscillator has angular frequency $\omega=5\,\mathrm{rad/s}$. At an instant, $x=.060\,\mathrm{m}$ and $v=-.40\,\mathrm{m/s}$. Define an auxiliary coordinate $y=-v/\omega$.

1. Find the auxiliary circle radius and the polar angle of the state.

2. Show that $(x,y)$ rotates counterclockwise uniformly. Explain why plotting unscaled velocity against displacement does not generally draw a circle.

Question 2 - Solution

The correct auxiliary circle uses a scaled velocity coordinate

(a) Derivation and calculation

$$y=-v/\omega=-(-.400)/5=.0800\,\mathrm m.$$$$\boxed{A=\sqrt{x^2+y^2}=\sqrt{.0600^2+.0800^2}=.100\,\mathrm m}.$$$$\boxed{\psi=\operatorname{atan2}(.0800,.0600)=.92730\,\mathrm{rad}}.$$

(b) Result and interpretation

$$\dot x=v=-\omega y,\qquad\dot y=-\dot v/\omega=\omega x.$$$$\frac{d}{dt}(x^2+y^2)=0,\quad\dot\psi=\frac{x\dot y-y\dot x}{x^2+y^2}=\boxed\omega.$$$$\frac{x^2}{A^2}+\frac{v^2}{(\omega A)^2}=1.$$The scaled $(x,-v/\omega)$ coordinates form a counterclockwise circle; unscaled velocity has different units and amplitude.

Physical check

Negative velocity places this auxiliary point above the horizontal axis, fixing the phase quadrant.