Simple Harmonic Motion and Uniform Circular Motion — Question 5

Question 5

Counterrotating circles can generate motion along one line

Two mathematical vectors of equal length $a$ rotate about the same origin. Their angles are $\omega t+\alpha$ and $-\omega t+\beta$. Their vector sum is $\mathbf r(t)$.

1. Find the direction of the line traced by the sum and the scalar displacement along it.

2. For $a=.020\,\mathrm{m}$, $\alpha=\pi/3$ and $\beta=-\pi/3$, find the amplitude and initial displacement. What changes if the two lengths are unequal?

Question 5 - Solution

Counterrotating circles can generate motion along one line

(a) Derivation and calculation

Put $\gamma=(\alpha+\beta)/2$ and $\eta=\omega t+(\alpha-\beta)/2$. Then$$\cos(\gamma+\eta)+\cos(\gamma-\eta)=2\cos\gamma\cos\eta,$$$$\sin(\gamma+\eta)+\sin(\gamma-\eta)=2\sin\gamma\cos\eta.$$$$\boxed{\mathbf r=2a\cos\eta\,(\cos\gamma,\sin\gamma)}.$$

(b) Result and interpretation

$$a=.0200,\quad\alpha=\pi/3,\quad\beta=-\pi/3:\quad\gamma=0,\quad\eta=\omega t+\pi/3.$$$$\boxed{A=2a=.0400\,\mathrm m},\qquad\boxed{x(0)=.0400\cos(\pi/3)=.0200\,\mathrm m}.$$For lengths $a_1\ne a_2$, coordinates along and across the bisector are$$u=(a_1+a_2)\cos\eta,\quad w=(a_1-a_2)\sin\eta,$$$$\frac{u^2}{(a_1+a_2)^2}+\frac{w^2}{(a_1-a_2)^2}=1.$$

Physical check

The equal-length limit sets the perpendicular coordinate to zero, collapsing the ellipse to a line.