1. 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

2. Hello, Nick!

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

Write $\displaystyle y$ as a function of $\displaystyle x.$

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

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

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

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

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

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

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

Nick