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Math Help - Lissajou Figures

  1. #1
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    Lissajou Figures

    Hi, i'm new to Maths Forum and bit of a problem.

    it's to do with two basic functions but i just don't understand the question.

    X= 10sin(2.pi.500t)
    Y= 10sin(2.pi.1000t)

    i am asked to write Y as a time independant function of X.

    Another thing. If in Radians, wouldn't both equations = 0 :S

    Nick
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  2. #2
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    Hello, Nick!

    \begin{array}{ccc}x &=& 10\sin(2\pi\cdot 500t) \\ \\[-4mm] y &=& 10\sin(2\pi\cdot1000t) \end{array}

    Write y as a function of x.

    We have: . \begin{array}{cccc} x &=& 10\sin\theta & {\color{blue}[1]} \\ y &=& 10\sin2\theta & {\color{blue}[2]} \end{array}


    From {\color{blue}[2]}\!:\;\;y \:=\:10(2\sin\theta\cos\theta) \quad\Rightarrow\quad y \:=\:20\sin\theta\cos\theta\;\;{\color{blue}[3]}


    From {\color{blue}[1]}\!:\;\;\sin\theta \:=\:\frac{x}{10}\;\;{\color{blue}[4]}

    Since \cos^2\theta \:=\:1-\sin^2\theta, we have: . \cos\theta \:=\:\pm\sqrt{1 - \left(\frac{x}{10}\right)^2} \quad\Rightarrow\quad \cos\theta \:=\:\pm\frac{\sqrt{100-x^2}}{10}\;\;{\color{blue}[5]}


    Substitute {\color{blue}[4]}\text{ and }{\color{blue}[5]} \text{ into }{\color{blue}[3]}\!:\;\;y \;=\;20\left(\frac{x}{10}\right)\left(\pm\frac{\sq  rt{100-x^2}}{10}\right)


    Therefore: . y \;=\;\pm\tfrac{1}{5}x\sqrt{100-x^2}

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  3. #3
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    Thanks very much for that. Helped me a lot.

    Nick
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