CODEWITHSUNDEEP
canimationmathwolfram-mathematicadata-visualization
Ricardo Sanchez-Saez
Ricardo Sanchez-Saez

Reputation: 9566

Animate a sine wave from the center

I want to animate a sine wave as if it's generated from the center and moves outwards (to the left and right).

I started with this library: SISinusWaveView and made some adjustments. Currently I have this naive code for calculating the Y position of the curve depending on the X, phase and frequency: 

float width = view.width; float mid = width / 2;
float adjustedX = x < mid ? x : width - x;
float y = maxAmplitude * sinf(2 * M_PI *(adjustedX / mid) * frequency + phase);
// phase increases every frame

Quite obviously, this causes a sharp angle at the middle of the sine wave, as seen below:

sine-wave

I would like to make it so the horizontal center of the animation is a smooth curve rather than a sharp angle, while keeping the animation horizontally symmetric. How would I approach this? Any mathematical insight for achieving this is appreciated.

EDIT

I tried to implement @TheBigH suggestion, but the parabola section is not seamlessly continuing the sine curve. Here's what I tried (implemented on Mathematica for quick visualization):

amp = 10;
freq = 1.5;
phase = 0.5;
Z = 1;
plotSine = Plot[amp*Sin[freq*x + phase], {x, Z, 2 Pi}];
aPara = amp*freq*Cos[phase]/(2 Z);
bPara = 0;
cPara = amp*Sin[c] - aPara*Z^2;
plotPara = 
  Plot[aPara*x^2 + bPara*x + cPara, {x, -Z, Z }, 
   PlotRange -> {{-Z, Z}, {-20, 20}}];
Show[plotPara, plotSine, PlotRange -> {{-2 Pi, 2 Pi}, {-20, 20}}

Which results in this:

enter image description here

Changing the sign of the parabola didn't quite work either:

enter image description here

EDIT 2

I see now that the problem was assuming s(0) = p(Z) and s'(0) = p'(Z); instead of s(z) = p(Z) and s'(Z) = p'(Z). Moving the sine wave so it would start exactly at the end of the parabole would fix the problem, but it's more convenient solving the parabola such as s(z) = p(Z) and s'(Z) = p'(Z) as that would simplify the implementation. How to do this?

enter image description here

EDIT 3

See this math.stackexchange.com answer for the final solution.

Upvotes: 4

Views: 1183

Answers (1)

TheBigH
TheBigH

Reputation: 554

Since OP has asked me to elaborate, here's my take as an answer.

Most generally, you're plotting the function s(x) = a sin(bx + c), where a, b and c come from the original problem. Later we will shift the sine curve by some offset Z but I'll leave it out of the sine curve for now as it will complicate the mathematics.

The new parabolic section will have equation p(x) = Ax^2 + Bx + C (A, B and C are different variables than a,b and c).

You want the two equations to join up cleanly, which means s(0) = p(Z). You also want the slopes to join up nicely so that there are no corners. That means you also need s'(0) = p'(Z). Also, since the parabola is centered about the origin, B = 0.

Thus you have two simultaneous equations for A, C given that you already know a, b, c and Z

a sin( c ) = A Z^2 + C
ab cos( c ) = 2AZ

or

A = ab cos( c ) / (2Z)
C = a sin (c) - A Z^2

This gives you the equation of the parabola, which you plot between -Z and Z. Then all you have to do is plot the sine curves, now adding in that offset. Let me know if any of this is unclear.

EDIT: I see there is a vertical shift of the sine wave as well. This does not pose any problems; just leave it out to begin with and add it to the parabola and sine waves at the end.

Upvotes: 2

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