I think that your model is correct, but I think that it's possible to add a few simplifications.
If the speed of sound is ~5m/s and the feather length is 0,2m then the time to for a signal to travel from one tip to the other is ~1/25s. That is approximately one frame in a standard camera (PAL/NTSC). So if they are using a standard camera, it's possible to assume that the drop signal travels almost instantly.
The system is totally overdamped, so I think that it's better to model it as a movement in a fluid with hight viscosity (like a spring inside honey). If you drop the acceleration term and you get
2ζω.D[x] + ω²x = 0
where 2ζω/ρ is something like the apparent viscosity that see a barbule moving inside the "fluid" of the other barbules.
The solution is
X = A exp(-ω/2ζ t)
So the relaxation time is ω/2ζ. When it's overdamped the system almost don't oscillates. Watching at the video, after it falls 1-ball-diameter the barbules reach the final position. I guess the ball is ~0,3m, so I guess that ω/2ζ ~= Sqrt(2h/g) = 0,8 seconds ~= 20 frames in a standard camera. But I have not idea about a sensible value for ζ.
If the speed of sound is ~5m/s and the feather length is 0,2m then the time to for a signal to travel from one tip to the other is ~1/25s. That is approximately one frame in a standard camera (PAL/NTSC). So if they are using a standard camera, it's possible to assume that the drop signal travels almost instantly.
The system is totally overdamped, so I think that it's better to model it as a movement in a fluid with hight viscosity (like a spring inside honey). If you drop the acceleration term and you get
2ζω.D[x] + ω²x = 0
where 2ζω/ρ is something like the apparent viscosity that see a barbule moving inside the "fluid" of the other barbules.
The solution is
X = A exp(-ω/2ζ t)
So the relaxation time is ω/2ζ. When it's overdamped the system almost don't oscillates. Watching at the video, after it falls 1-ball-diameter the barbules reach the final position. I guess the ball is ~0,3m, so I guess that ω/2ζ ~= Sqrt(2h/g) = 0,8 seconds ~= 20 frames in a standard camera. But I have not idea about a sensible value for ζ.