# Simple Harmonic Motion

## Homework Statement

Two-dimensional SHM: A particle undergoes simple harmonic motion in both the x and y directions
simultaneously. Its x and y coordinates are given by
x = asin(ωt)
y = bcos(ωt)

Show that the quantity x$\dot{y}$-y$\dot{x}$ is also constant along the ellipse, where here the dot means the derivative with respect to time. Show that this quantity is related to the angular momentum of the system.

L = mv x r

## The Attempt at a Solution

Hi, so for the first part it is pretty simple and my answer is -abω, unless i made a dumb mistake which I don't think I did.

It's the second part that is giving me issues. How do I show that it is related to angular momentum? I tried doing this

L = $\sqrt{L^{2}_{x}+L^{2}_{y}}$

then L$_{x}$ = m$\frac{∂x}{∂t}$ x r

where r = $\sqrt{x^{2}+ y^{2}}$

and then plugging everything in. I was hoping all the cos and sin were going to cancel out but it got really huge and messy. I didn't think it was supposed to be that hard so can anyone tell me if I am going in the right direction or if there is something I am missing?

Thanks

TSny
Homework Helper
Gold Member
Use the vector definition of angular momentum of a particle: ##\vec{L} = \vec{r} \times \vec{p}= m\; \vec{r} \times \vec{v}##, where ##\vec{r} = x\hat{i} + y\hat{j}##. (Note, the order of the cross product is important. Thus, ##\vec{L} = m\;\vec{v} \times \vec{r}## is not correct.)

Right whoops.

so then v = dx/dt i + dy/dt j

and when we cross them we get m(xdy/dt - ydx/dt) = L correct?

TSny
Homework Helper
Gold Member
Yes. Good.