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[GIF] Drumbeat (Vibration mode of a circular drum)

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Vibration mode of a circular drum

Drumbeat

Inspired by a comment made by @J. M. on Square Up, this is an animation showing the (1,2) vibrational mode of a circular drum. I've added a (quite possible completely unphysical) exponential decay term, just because.

As you can see on the Wikipedia page, the $(m,n)$ vibrational mode of a circular membrane is given, as a function whose input is the position and whose output is the (vertical) displacement from the rest state, by

$ u_{mn} (r,\theta,t) = (A \cos c \lambda_{mn} t + B \sin c \lambda_{mn} t) J_m(\lambda_{mn} r) (C \cos m\theta + D \sin m \theta)$,

where $\lambda_{mn}$ is the $n$th root of the Bessel function $J_m$, $c$ is a constant giving the speed of wave propogation, and $A$, $B$, $C$, and $D$ are constants.

This is of course easy enough to define, especially since Mathematica has a handy BesselJZero function:

BesselZeros = N@Table[BesselJZero[m, n], {m, 0, 5}, {n, 1, 5}];
λ[m_, n_] := BesselZeros[[m + 1, n]];
u[m_, n_, r_, θ_, t_, c_, A_, B_, C_, D_] :=
  (A Cos[c λ[m, n] t] + B Sin[c λ[m, n] t]) 
    BesselJ[m, λ[m, n] r] (C Cos[m θ] + D Sin[m θ]); 

So then I just took an (automatically generated) triangulation of the disk and evaluated the above function u (with various more or less arbitrary choices of parameters) on the vertices. Here's the code:

DynamicModule[
 {m = 1,
  n = 2,
  k = 3,
  c = 5,
  cols, mesh, planarverts, drumverts, lines, cl},
 cols = RGBColor /@ {"#08D9D6", "#FF2E63", "#252A34"};
 mesh = DiscretizeRegion[Disk[], MaxCellMeasure -> .009];
 planarverts = MeshCoordinates[mesh];
 lines = MeshCells[mesh, 1];
 Manipulate[
  drumverts = Append[#, 1.5 E^(-k t) u[m, n, Norm[#], ArcTan[#[[1]], #[[2]]], t, c, 1/2, 0, 1/2, 0]] & /@ planarverts;
  cl = Blend[cols[[;; 2]], Abs[#[[3]]]/.2] & /@ drumverts;
  Graphics3D[
   GraphicsComplex[
    drumverts,
    {Thickness[.004], Append[#, VertexColors -> Automatic] & /@ lines},
    VertexColors -> cl
    ],
   PlotRange -> 1, Boxed -> False, Axes -> None, 
   ImageSize -> {540, 405}, ViewAngle -> π/10, 
   ViewCenter -> {1/2, 1/2, .46}, Background -> cols[[-1]]
   ],
  {t, 0., 2 π/λ[m, n]}]
 ]
POSTED BY: Clayton Shonkwiler
Answer
1 year ago

I'm glad you used the idea. :) Normally I'd use a polar mesh to depict this, but I think the triangulated circle looks great here. Makes it look like a funky trampoline.

POSTED BY: J. M.
Answer
1 year ago

Yes, I definitely started with a polar mesh, but it wasn't quite working for me, so I went for something a little more asymmetrical.

POSTED BY: Clayton Shonkwiler
Answer
1 year ago

enter image description here - another post of yours has been selected for the Staff Picks group, congratulations !

We are happy to see you at the top of the "Featured Contributor" board. Thank you for your wonderful contributions, and please keep them coming!

POSTED BY: Moderation Team
Answer
1 year ago

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