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

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?

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.

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

Very clear, that makes sense, ..thanks!

So how did you solve it and saw strange answers? With a numerical tool like Matlab, or Mathematica?

Re: Can you motivate this speed difference??

Posted: September 12th, 2017, 12:21 pm

by Paul

Pencil and paper, I'm afraid. V low tech! It's really easy. That's why I think it's a perfect exam question. You've got the modelling to get the equation of motion, then the boundary conditions. The initial, steady-state condition. And then solving using characteristics. Still can't believe I haven't seen this before, or just done it myself for fun.

Re: Can you motivate this speed difference??

Posted: September 16th, 2017, 11:23 am

by katastrofa

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!

But this particular scenario is NOT governed by the wave equation because it's a slinky, not a general spring with an open-coil rest state. The wave equation involves only simple elasticity, not the highly non-linear effects of the slinky coils collapsing to a solid cylinder.

Thus, this it is not a simple kx vs. mg in that k has extremely low value for some parts of the system and at some times but then becomes vastly greater when the coils collapse and contact each other.

It cannot be anything else because the only forces in the system are the spring tension T and gravity. Slinky is a bit pretentious (Tp - is the pretension force what you meant was missing?), so stretching it creates an opposing tension, T = Tp - kx. When you hold the slinky by the topmost coil, it stretches out until the force of gravity is balanced by the tension. When you release the top, the consecutive coils start to collapse onto each other from the top as the "deformation" propagates through the spring in a wave-like manner: [$]\partial^2 \psi/\partial t^2 = T_p/\rho\ \partial^2 \psi/\partial x^2[$]. I think the beauty of the experiment is in how the centre of mass of the slinky experiences a free fall, while the slinky as a whole is in zero-g. It's the "opposite" situation to popular experiments demonstrating free fall with water in a punched bottle - when it falls, the water doesn't leak, because it's in zero-g.

Re: Can you motivate this speed difference??

Posted: September 16th, 2017, 3:17 pm

by Cuchulainn

It is highly unusual to use upper case Greek letters as the dependent variable in PDEs. What's the rationale?

Re: Can you motivate this speed difference??

Posted: September 17th, 2017, 12:15 am

by katastrofa

It is highly unusual to use upper case Greek letters as the dependent variable in PDEs. What's the rationale?

No rationale. I was possibly thinking about Schroedinger equation with broken time-reversal symmetry because of the gravity doing the same in this problem, which is a stupid connotation. I'm simply getting out of practice - or was it a subconscious association with cats... BTW, they started to sleep on my desk while I'm working:

Re: Can you motivate this speed difference??

Posted: September 17th, 2017, 10:24 am

by Traden4Alpha

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!

But this particular scenario is NOT governed by the wave equation because it's a slinky, not a general spring with an open-coil rest state. The wave equation involves only simple elasticity, not the highly non-linear effects of the slinky coils collapsing to a solid cylinder.

Thus, this it is not a simple kx vs. mg in that k has extremely low value for some parts of the system and at some times but then becomes vastly greater when the coils collapse and contact each other.

It cannot be anything else because the only forces in the system are the spring tension T and gravity. Slinky is a bit pretentious (Tp - is the pretension force what you meant was missing?), so stretching it creates an opposing tension, T = Tp - kx. When you hold the slinky by the topmost coil, it stretches out until the force of gravity is balanced by the tension. When you release the top, the consecutive coils start to collapse onto each other from the top as the "deformation" propagates through the spring in a wave-like manner: [$]\partial^2 \psi/\partial t^2 = T_p/\rho\ \partial^2 \psi/\partial x^2[$]. I think the beauty of the experiment is in how the centre of mass of the slinky experiences a free fall, while the slinky as a whole is in zero-g. It's the "opposite" situation to popular experiments demonstrating free fall with water in a punched bottle - when it falls, the water doesn't leak, because it's in zero-g.

It depends on what we mean by "the" wave equation. In terms of the basic set-up, sure it's all about an interaction of kx and g in the distributed mass of the spring. But there's two values of k in the system, the very very low uncompressed one that creates waves traveling on the order of a few meters a second and a very very high value of k for the compressed spring that creates waves traveling within the compressed section on the order of a several thousand meters a second.

There are combinations of g, k, L, and Tp that create springs that don't fully collapse when dropped in this manner. They'd show different dynamics with a more modest compression wave traveling down and up the spring as it fell.

Re: Can you motivate this speed difference??

Posted: September 18th, 2017, 8:55 am

by Cuchulainn

Could you guess the shape of the curves if you could hear the noise the ball bearing makes?

Re: Can you motivate this speed difference??

Posted: September 18th, 2017, 12:06 pm

by Traden4Alpha

Could you guess the shape of the curves if you could hear the noise the ball bearing makes?

Tricky!

I'm sure one can estimate the ball's speed (overall tone) and acceleration (volume and more subtle changes in tone caused by nonlinear changes in the ball-track contact point). From there, one can deduce (with some redundancy between the speed and acceleration estimate), the vertical path of the track. It may be hard to distinguish vertical vs. horizontal acceleration although one large difference between the two is that vertical acceleration would be associated with significant changes in speed but horizontal acceleration would not (assuming friction is low). I doubt one could hear the difference between a left vs. right curve without two or more microphones.