Two Simple Harmonic Motion (S.H.M.) in Perpendicular Direction

Question

Suppose a particle is moving under the superposition of two S.H.M in the perpendicular direction... The general equation for the trajectory for the resultant motion arising due to the two component S.H.Ms(with phase difference $/delta$ and amplitude $A_1$ and $A_2$) is known to be a ellipse.. But how i may determine if the particle will be moving clockwise or anticlockwise in the ellipse. When i researched about it, I found that it depends on which component leads the other but Iam not able to understand it

Please explain in simple manner

Answer 1

Consider an example of a circular trajectory:

$$\mathbf{r}(t)=\cos(t)\hat{\mathbf{x}}+\sin(t)\hat{\mathbf{y}}$$

We can rewrite this purely in terms of cosines, by remembering that $\sin(t)=\cos(t-\pi/2)$:

$$\mathbf{r}(t)=\cos(t)\hat{\mathbf{x}}+\cos\left(t-\frac{\pi}{2}\right)\hat{\mathbf{y}}$$

This corresponds to counterclockwise motion. If we take $t\to-t$, then clearly we will be left with clockwise motion:

$$\begin{align*}\mathbf{s}(t)&=\cos(-t)\hat{\mathbf{x}}+\sin(-t)\hat{\mathbf{y}}\\\sin(-t)=-\sin(t)\implies&=\cos(-t)\hat{\mathbf{x}}-\sin(t)\hat{\mathbf{y}}\\\cos(-t)=\cos(t)\implies&=\cos(t)\hat{\mathbf{x}}-\sin(t)\hat{\mathbf{y}}\\&=\cos(t)\hat{\mathbf{x}}+\cos\left(t+\frac{\pi}{2}\right)\hat{\mathbf{y}}\end{align*}$$

You can see that the two forms are the same, except in the clockwise case the $\hat{\mathbf{y}}$ component has a phase shift which brings it "ahead" of the $\hat{\mathbf{x}}$ component, and vice versa for the clockwise case. This is what is meant by one component "leading" the other.

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So, in your case for a general elliptical trajectory

$$\mathbf{r}(t)=A_1\cos(t+\delta_1)\hat{\mathbf{x}}+A_2\cos(t+\delta_2)\hat{\mathbf{y}}$$ with $\delta=\delta_2-\delta_1$, we can determine the directionality of the trajectory based on $\delta$. Based on our observations above, we can see that $\delta<0$ implies the trajectory will be counterclockwise, and for $\delta>0$ it will be clockwise. If we wrap that into the typical $[0,2\pi)$ interval, we can say that a clockwise trajectory will have $0<\delta<\pi$, and a counterclockwise one will have $\pi<\delta<2\pi$. A directionality is not really defined for $\delta=n\pi:n\in\mathbb{Z}$ because the trajectory collapses to a line.

Answer 2

It is perhaps easiest to look at a non-generalised example?

A position in the x-y plane, $(\sin(2\pi f t)\,,\,\sin(2\pi f t\pm \phi)$ which depends on time $t$ and where $f$ is the frequency and $\phi$ is the phase angle between the two motions.

First consider the left-hand diagram with position $(\sin(2\pi f t)\,,\,\sin(2\pi f t+\pi/2)$, ie the y-coordinate is leading the x-coordinate by a quarter of a period, $T/4$

You will note that whatever value the y-coordinate, that value is the x-coordinate later in time.
This results in a clockwise motion of the position.

In the right hand diagram, whatever value the x-coordinate, that value is the y-coordinate later in time.
This results in a counterclockwise motion of the position.

So all you need to do is to look at the phase difference between the two oscillations with the magnitude of the phase angle less than $180^\circ$.

This link to Desmos allows you to change the parameters and see the effect on the position as a function of time.