Its kx vs mg until they hit. Hence wave equation. But because it's a wave equation there's a finite propagation speed. That means there's a finite time before the coils can hit each other. And since the propagation is from the top where the slinky is released there'll be a finite time before the bottom moves. So your worry doesn't kick in for a while and doesn't change the effect seen in the video!

ok,.. so the wave equation comes from local forces (2nd derivative) having to cancel out? I prefer thinking in terms of symmetries or conservations instead of equations but I guess it's the same thing?No, it has become trivial! Once you know it's a wave equation everything follows very simply! All the results about the bottom, gravity, etc. Had it been a diffusion equation, or elliptic, then the bottom would have fallen (out of this brain teaser) immediately!

And that's in slinky coordinates, right?

At any point h on the slinky (measures from the bottom along the coil) there is h*m mass below it. So the gravitational force components grow linear along the slinky going up. In *our* coordinates the slinky get's stretched -also linear growing from bottom to top assuming an ideal slinky which gives hooks law-. The coils at the top will have more space between them then at the bottom.

Yes, the coordinate system relative to compressed slinky, like how many coils along.

But don't think in terms of weight above or below. Just think locally.

Use [$]d(x,t)[$] as distance from fixed point for point [$]x[$] along the slinky. In fully compressed state you'd have [$]d(x,t)=x[$].

Then look at how much slinky is stretched each side of a point. On one side it's [$]d(x+\delta x,t)-d(x,t)[$], so that gives force one direction. (Easy to assume force linear in that.) And similar other direction. Net force gives second [$]x[$] derivative. Which then balances second [$]t[$] derivative. Result similar to wave on string but different reason.

*

Similar for conditions at the ends.

Throw in gravity. Solve for initial/steady state.

Then change boundary condition at top to represent release.

But to get the result that the bottom doesn't move for a while you don't need anything from (*) on. The wave equation is enough.

But don't think in terms of weight above or below. Just think locally.

Use [$]d(x,t)[$] as distance from fixed point for point [$]x[$] along the slinky. In fully compressed state you'd have [$]d(x,t)=x[$].

Then look at how much slinky is stretched each side of a point. On one side it's [$]d(x+\delta x,t)-d(x,t)[$], so that gives force one direction. (Easy to assume force linear in that.) And similar other direction. Net force gives second [$]x[$] derivative. Which then balances second [$]t[$] derivative. Result similar to wave on string but different reason.

*

Similar for conditions at the ends.

Throw in gravity. Solve for initial/steady state.

Then change boundary condition at top to represent release.

But to get the result that the bottom doesn't move for a while you don't need anything from (*) on. The wave equation is enough.

All clear! .. except that you need to solve things.. The crucial part from the wave equation view on it is that the release needs time to propagate.

- Traden4Alpha
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Hmmm... The strange thing is that the actual frame-by-frame dynamics are more complex than just a simple collapse from the top down. During the first 1/3 to 1/2 of the drop, the top portion is only partially compressed. It's only at about frame #36 that the top of the slinky is in a fully compressed state that is rushing down collapsing all the loops below it.

- Cuchulainn
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Is a wave equation needed for this? Is it not just a classic quickest descent Brachistochrone problem.

The time to slide along the

What's the question, really? Shortest transit time?

Last edited by Cuchulainn on September 12th, 2017, 11:01 am, edited 2 times in total.

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argh.. you're talking about the balls now, right? Paul is applying the wave eq to the slinky.

The track is not a Brachistochrone

The track is not a Brachistochrone

The wave equation has to be solved, as outrun says, to find out what happens where all the action is. In the solution I found it looked like it was initially not fully compressed at the top. I didn't believe my result but if your observation are correct maybe it's not as bad as I thought!Hmmm... The strange thing is that the actual frame-by-frame dynamics are more complex than just a simple collapse from the top down. During the first 1/3 to 1/2 of the drop, the top portion is only partially compressed. It's only at about frame #36 that the top of the slinky is in a fully compressed state that is rushing down collapsing all the loops below it.

Looks like people at Princeton worked on it:

http://www.princeton.edu/~rvdb/tex/slinky/slinky.pdf

http://www.princeton.edu/~rvdb/tex/slinky/slinky.pdf

- Cuchulainn
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Good to know. This is not the same as your post of July 23.. where u state ..argh.. you're talking about the balls now, right? Paul is applying the wave eq to the slinky.

The track is not a Brachistochrone

Have the requirements changed?

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The angled lines are the characteristics, think of them as lines along which information flows.

There are *two* brainteasers in this thread. That in itself turns out to too complex for some to handle!Good to know. This is not the same as your post of July 23.. where u state ..argh.. you're talking about the balls now, right? Paul is applying the wave eq to the slinky.

The track is not a Brachistochrone

Yes, the question is: why is the wavy one faster then the flat one?

Have the requirements changed?

You are referring to the first one. In that one you can see 3 bumbs ("wavy")

I think we see a different coordinate that x? x is the number of cycles? It need to be converted to stretched factor (=force) and then integrated to get actual distances between points?image1.JPG

The angled lines are the characteristics, think of them as lines along which information flows.

Yes, x is the distance measured in hoops. So 0.5 means paint the middle hoop red and follow its path. It won't, of course, remain in the "middle"! The distance of the point x=0.5 from the point at which the top was being held is d(0.5,t) and it's d(x,t) that you solve for. No integrating as such.I think we see a different coordinate that x? x is the number of cycles? It need to be converted to stretched factor (=force) and then integrated to get actual distances between points?image1.JPG

The angled lines are the characteristics, think of them as lines along which information flows.

x and t are the independent variables, x is from 0 to 1, t is from 0. d(x,t) is the dependent variable.

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