- Cuchulainn
**Posts:**62046**Joined:****Location:**Amsterdam-
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It's not too late to repent!

- Traden4Alpha
**Posts:**23951**Joined:**

Tanks for the compliment!That's deserve a Patton the back.Sher, man.Like this ion engine?

- Cuchulainn
**Posts:**62046**Joined:****Location:**Amsterdam-
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Must be very exhausting!Tanks for the compliment!That's deserve a Patton the back.Sher, man.

Enough!

So the governing equation for longitudinal motion in a slinky is a wave equation? Plain vanilla wave equation or something else? In what coordinate system, fixed in space or distance along the slinky? Can't believe I haven't seen this before!

So the governing equation for longitudinal motion in a slinky is a wave equation? Plain vanilla wave equation or something else? In what coordinate system, fixed in space or distance along the slinky? Can't believe I haven't seen this before!

Most talks you've probably read is about the balls, ..not the slinky?

The slinky can be modelled as an elastic rope with weight.

The slinky can be modelled as an elastic rope with weight.

- Traden4Alpha
**Posts:**23951**Joined:**

It's only the wave equation if the slinky remains somewhat stretched in all parts of the spring and at all times. But in this case, the slinky hits full compression which creates a HUGE nonlinear increase in the spring constant. The phenomenon in the video is more of a shockwave propagation in which some dynamical phenomenon is being limited by the "speed of sound" in the medium.

It would be interesting to know if all slinkies in all gravitational fields behave the same or if some combinations of k, m, & g parameters show different dynamics (e.g., the bottom of the slinky starts dropping "immediately" after the top is released and the slinky never fully compresses).

It would be interesting to know if all slinkies in all gravitational fields behave the same or if some combinations of k, m, & g parameters show different dynamics (e.g., the bottom of the slinky starts dropping "immediately" after the top is released and the slinky never fully compresses).

- Traden4Alpha
**Posts:**23951**Joined:**

Actually, it's not the stiffness so much as the rest-state helix dimensions that matter. The rest-state of a standard new slinky has the loops in contact with each other which means the slinky loops hit each other during the top-drop dynamics. Such a slinky instantly transitions from elastic to inelastic dynamics to form a shockwave.A stiff slinky would

To have pure wave equation motion, the rest-state of the slinky must be expanded enough so that when the slinky is further stretched by gravity and then released, the internal compression wave can oscillate to either side of the rest-state condition without the loops hitting each other to create a non-linear incompressible shockwave. That is, the largest loop-to-loop stretched state deflection must be less than twice the rest-state loop-to-loop distance.

If you were to take a new slinky and over-stretch it so that the loops permanently deform in the stretched state, then the dynamics of the top-dropped slinky would be very different.

Under simple and obvious assumptions you get wave equation with source term (gravity), boundary conditions independent of gravity, and initial parabolic displacement. Independent variables are initial position of each hoop and time. A little bit of characteristics and Bob's your uncle.

You can make it nonlinear v easily if you like.

You can make it nonlinear v easily if you like.

- Cuchulainn
**Posts:**62046**Joined:****Location:**Amsterdam-
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I suppose the independent variables are (x,t)?Under simple and obvious assumptions you get wave equation with source term (gravity), boundary conditions independent of gravity, and initial parabolic displacement. Independent variables are initial position of each hoop and time. A little bit of characteristics and Bob's your uncle.

You can make it nonlinear v easily if you like.

Then z = (x+t), w = (x-t) gives a nonlinear Goursat PDE(z,w) that can be solved by the Riemann method or the FDM solver I used for BVN.

Since it's a wave equation, by releasing the top of the slinky it takes a while before the "information" gets to the bottom. The wave speed depends on the properties of the slinky, and not on gravity. All gravity does is determine the initial stretch in the slinky.

Should be a second-year maths undergrad question.

Should be a second-year maths undergrad question.

I like T4s answer better, it all getting needless complicated.

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!

- katastrofa
**Posts:**9184**Joined:****Location:**Alpha Centauri

Omg, pls, it rly is as simple as -kx vs mg, ok?

- Traden4Alpha
**Posts:**23951**Joined:**

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.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!

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.

GZIP: On