The best explanation I can offer is found on the Cardas website here:

http://www.cardas.com/insights_golden_section.php
http://www.cardas.com/insights_golden_ratio.php

Note these passages in particular:

Alternating current can shake a wire like a guitar string. The audio signal in a stereo system is seen as alternating current. The audio signal, be it digital or analog, through tube or solid state, is always alternating current. The signal’s cyclic effect, causes all the wire in the system to vibrate and ring. This ring becomes a song sung to the resonance of the electrical and mechanical components of the cable. Every interconnect, every speaker cable, every chassis and speaker wire has its own song. The stored mechanical and electrical tension and time delay characteristics of the cable determine what song is sung. Each and every strand in a cable has its own note or beat. Two or more wires of the same mass and tension have common mechanical resonating points and share the same note. Two or more wires or bundles of wires, differing in size, each have their own resonant points. When combined, wires find new points of interaction, creating yet another note. The sound produced by any stereo system depends on the purity of the audio signal. When the cable that links all the components together imparts its own sound the audio signal is corrupted.

Other related articles on their site include:

http://www.cardas.com/room_setup_golden_trapagon.php
http://www.cardas.com/room_setup_golden_cuboid.php

Thank you. But how does the golden section affects the resonance in the wire?

Thank you very much.

The 3-dimensional ratio is .618 x 1 x 1.618 so height is A, width is B, and length is A+B.

The golden ratio is in the diameters of the various wire sizes bundled in the wire cable. I’ve added an extra image above to show this more clearly.

IDK, but Arthur C. Clarke talked about a 2x4x9 monolith… seems there might be something in there… the dude was some kind of genius.

That is true. In fact F(2n-1)=F(n)^2+F(n-1)^2, where F(n) is the nth Fibonacci number.
F(0)=0, F(1)=1, F(2)=1, F(3)=2, …

Another relation is that F(2n)=F(n+1)^2-F(n-1)^2.

An induction proof follows:

1) For n = 2, both statements are true.

2) Assume that both statements are true for n=k, if we can prove that the statements are true for n=k+1, we are done. From the definition of the fibonacci sequence, we have

F(2k+1)=F(2k)+F(2k-1)

Now we use the assumption that the statements are true for n=k. That is

F(2k+1)=(F(k+1)^2-F(k-1)^2)+(F(k)^2+F(k-1)^2)
That is

F(2k+1)=F(k+1)^2+F(k)^2 (1)

It remains to show that F(2k+2)=F(k+3)^2-F(k+1)^2.
I leave that as an exercise.

I have not checked the details but my first guess would be
W/H=L/W=phi would be a good candidate for the three-dimensional case.
L/H=1+phi which is almost as irrational as phi.

phi=1+1/(1+1/(1+1/(1+1/(1+1/(1+1/(…

while

1+phi=2+1/(1+1/(1+1/(1+1/(1+1/(1+1/(…