# 2 Perpendicular oscillations

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1. Jul 22, 2015

### Xsnac

1. The problem statement, all variables and given/known data

I have 2 perpendicular oscilations and I have to find the trajectory equation.

$$x=A\cos\omega t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (1)$$
$$y=B\cos(\omega t+\Delta\phi)$$

2. Relevant equations

$$\cos (x+y) =\cos x\cos y -\sin x\sin y$$
$$\cos^{2} x+\sin^{2} x =1$$

and from (1)
$$\cos\omega t =\frac {x}{A}$$

3. The attempt at a solution

I basicaly spent 2 hours trying to algebraicaly manipulate the equation

The closes I could get was:

$$\frac {y^{2}} {B^{2}} -2\frac {xy} {AB}\cos\Delta\phi + \frac {x^{2}} {A^{2}}\cos^{2}\Delta\phi = (1 - \frac{x^{2}} {A^{2}} )sin^{2}\Delta\phi\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (2)$$

The end result should be:
$$\frac {x^{2}} {A^{2}} +\frac {y^{2}} {B^{2}}-2\frac {xy} {AB}\cos\Delta\phi=sin^{2}\Delta\phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (3)$$

I have no ideea how to get from (2) to (3).

Last edited: Jul 22, 2015
2. Jul 22, 2015

### Dr. Courtney

I'm not sure what you want. You already have parametric trahectory equations: x(t) = stuff; y(t) = other stuff

Do you want a coordinate equation y(x)?

3. Jul 22, 2015

### Xsnac

I don't know how to get from (2) to (3). I want to know the steps I have to make to get from (2) to (3)...

4. Jul 22, 2015

### Ray Vickson

We need to see the details: what are the steps you took to get to your (2)? Perhaps you made some errors, but we cannot tell until you show us.

For the record: I get the result (3).

BTW: your equation $\cos x+y =\cos x\cos y -\sin x\sin y$ is almost certainly false, as it says $y + \cos x =\cos x\cos y -\sin x\sin y$, which is untrue for most $x,y$. However, an equation such as $\cos (x+y) =\cos x \cos y -\sin x \sin y$ is true for all real $x,y$.

Last edited: Jul 22, 2015
5. Jul 22, 2015

### Xsnac

sorry it was a typo ... well the stepps I took are the following
I expanded the cos function of $$y=B\cos(\omega t+\Delta\phi)$$
and replaced all the $$\cos\omega t$$ with $$\frac {x}{A}$$
then I transformed the $$\sin\omega t$$ into $$\sqrt{1 - \frac{x^{2}} {A^{2}}}$$
In the end I divided everything by B and squared everything to get rid of the root.

Last edited: Jul 22, 2015
6. Jul 22, 2015

### Ray Vickson

Transforming $\sin(\omega t)$ into $\sqrt{1 - \cos^2(\omega t)}$ is a mistake, as it eliminates the parts where $\sin(\omega t) < 0$. The point is that, by definition, $\sqrt{ \cdot } \geq 0$ for any real argument, but sometimes you want to have $\sin(\omega t) = - \sqrt{1 - \cos^2(\omega t)}$. It is a lot easier to just use $u = \cos(\omega t), v = \sin(\omega t)$ and then express $x(t), y(t)$ in terms of $u,v$. Of course, $u^2 + v^2 = 1$ for all $t$.

7. Jul 22, 2015

### Xsnac

Ur wrong. I figured it out with a little help from someone else. Had to multiply the right parenthesis with $$\sin^{2} \Delta \phi$$ , move the second term to the left side, factorise out $$\frac {x^{2}} {A ^{2}}$$ and then the remaining equation is identical to (3)

$$\frac {y^{2}} {B^{2}} -2\frac {xy} {AB}\cos\Delta\phi + \frac {x^{2}} {A^{2}}\cos^{2}\Delta\phi = (1 - \frac{x^{2}} {A^{2}} )sin^{2}\Delta\phi\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (2)$$

and where I got

$$\frac {x^{2}} {A^{2}} +\frac {y^{2}} {B^{2}}-2\frac {xy} {AB}\cos\Delta\phi=sin^{2}\Delta\phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (3)$$

8. Jul 22, 2015

### Ray Vickson

No, I am NOT WRONG. Writing $\sin(r) = \sqrt{1 - \cos^2(r)}$ can be incorrect, and by definition of the function $\sqrt{ \cdot }$, it is $\geq 0$ for all real arguments. Those are just plain facts.

These issues have been explained over and over again, many times, in numerous discussions in this forum. I did not make these things up.

Aside from that, whether of not expressing $x,y$ in terms of $u,v$ and then using $1 = u^2 + v^2$ is easier or harder than some other way is largely a matter of opinion; I found it easier.

9. Jul 22, 2015

### Xsnac

Ur wrong in the sense that saying what you just said is not getting me closer to solving the problem. But thank you anyway.