Simple Harmonic Motion and Uniform Circular Motion — Question 10

Question 10

A phase relation determines both path shape and circulation

A planar motion has $x=A\cos\omega t$ and $y=A\cos(\omega t+\delta)$, with $A,\omega>0$. Consider $0<\delta<\pi$.

1. Eliminate time to find the path equation and identify when it is a circle.

2. Compute $x\dot y-y\dot x$ and infer the circulation direction. Describe the limits $\delta\to0$ and $\delta\to\pi$.

Question 10 - Solution

A phase relation determines both path shape and circulation

(a) Derivation and calculation

$$y/A=(x/A)\cos\delta-\sin\omega t\sin\delta.$$$$\left(\frac{y-x\cos\delta}{A\sin\delta}\right)^2+\frac{x^2}{A^2}=1.$$$$\boxed{x^2+y^2-2xy\cos\delta=A^2\sin^2\delta}.$$$$\delta=\pi/2\Rightarrow\boxed{x^2+y^2=A^2}.$$

(b) Result and interpretation

$$x\dot y-y\dot x=A^2\omega[\sin\omega t\cos(\omega t+\delta)-\cos\omega t\sin(\omega t+\delta)]$$$$\boxed{x\dot y-y\dot x=-A^2\omega\sin\delta<0}.$$The motion circulates clockwise for $0<\delta<\pi$.$$\delta\to0:\ y=x;\qquad\delta\to\pi:\ y=-x.$$

Physical check

Equal coordinate amplitudes yield a circle only in quadrature; the endpoint phases give degenerate straight paths.