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2 Perpendicular oscillations

  1. Jul 22, 2015 #1
    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). :mad:

    Thank you in advance for reading this wall of text.
     
    Last edited: Jul 22, 2015
  2. jcsd
  3. Jul 22, 2015 #2
    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)?
     
  4. Jul 22, 2015 #3
    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)...
     
  5. Jul 22, 2015 #4

    Ray Vickson

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    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
  6. Jul 22, 2015 #5
    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
  7. Jul 22, 2015 #6

    Ray Vickson

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    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##.
     
  8. Jul 22, 2015 #7
    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)

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

    and where I got

    $$\frac {x^{2}} {A^{2}} +\frac {y^{2}} {B^{2}}-2\frac {xy} {AB}\cos\Delta\phi=sin^{2}\Delta\phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (3)$$
     
  9. Jul 22, 2015 #8

    Ray Vickson

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    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.
     
  10. Jul 22, 2015 #9
    Ur wrong in the sense that saying what you just said is not getting me closer to solving the problem. But thank you anyway.
     
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