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My friend linked this .gif to me tonight, and asked me if I knew of any equations that might model these bottom two waves (the blue and green waves). Unfortunately, I am not far enough in my education to recognize if any such model exists. Are these waves modeled after some equation, or is this just some piece of eye candy?

Waves

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2 Answers 2

up vote 5 down vote accepted

All can be modeled by some $2 \pi$ periodic function $r:\mathbb{R} \to \mathbb{R}$, then the equation is $t \mapsto r(t) \sin t$.

For the red wave, use $r(t) = 1$.

For the blue wave, use $r(t) = \sqrt{1+\sin^2t}$, for $t \in [-\frac{\pi}{4}, \frac{\pi}{4})$, and let $r$ by $\frac{\pi}{2}$ periodic.

For the green wave, use the same formula as for the blue wave, except for the domain $[-\frac{\pi}{6}, \frac{\pi}{6})$, and let $r$ by $\frac{\pi}{3}$ periodic.

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So, for the fist one obviously, it's sine function. $$ h(t) = r \sin \omega t $$ As for the second one, $$ h(t) = \frac a2 \cdot \left \{ \begin{array}{lcc} \tan \left (-\frac \pi 4 + \omega t \right )& \text{if} & 0 \le t \le \frac T4 \\ 1 & \text{if} & \frac T4 \le t \le \frac T2 \\ \tan \left ( \frac {3\pi}4 + \omega t \right ) & \text{if} & \frac T2 \le t \le \frac {3T}4 \\ -1 & \text{if} & \frac {3T}4 \le t \le T \end{array}\right . $$ For the last one, $$ h(t) = a \cdot \left \{ \begin{array}{lcc} \frac {\sqrt 3}2 \tan \left ( -\frac \pi 6 + \omega t\right) & \text{if} & 0 \le t \le \frac T6 \\ \frac 12 + \frac 12 \left [\frac 12 + \frac {\sqrt 3}2 \tan \left( -\frac \pi 6 + \omega t - \frac \pi 3\right) \right ] & \text{if} & \frac T6 \le t \le \frac T3 \\ 1 - \frac 12 \left [\frac 12 + \frac {\sqrt 3}2 \tan \left( -\frac \pi 6 + \omega t - \frac {2\pi} 3\right) \right ] & \text{if} & \frac T3 \le t \le \frac T2 \\ \frac {\sqrt 3}2 \tan \left( \pi - \omega t + \frac \pi 6 \right) & \text{if} & \frac T2 \le t \le \frac {2T}3 \\ -\frac 12 - \frac 12 \left [ \frac 12 + \frac {\sqrt 3}2 \tan \left ( -\frac \pi 6 + \omega t - \frac {4 \pi}3 \right )\right ] & \text{if} & \frac {2T}3 \le t \le \frac {5T}6 \\ -1 + \frac 12 \left [ \frac 12 + \frac {\sqrt 3}2 \tan \left ( -\frac \pi 6 + \omega t - \frac {5 \pi}3\right )\right ] & \text{if} & \frac {5T}6 \le t \le T \end{array}\right . $$

PS: For last two cases $a$ is a square or hexagon side. $T = \frac {2\pi}\omega$ Initial time angles are chosen according to this

initial

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It might be simplified by choosing different initial time moment and grouping. –  Kaster Feb 18 '13 at 10:01

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