- #51
rude man
Science Advisor
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EDIT: possible approach iff v = (const)/√x. I should have looked at all the prior posts so the following is wheel-spinning ... I don't read Russian but post 40 sure looks like v = 0.55 something/s i.e. constant.
1. Let length of pendula = L instead of 1. I don't lke losing the ability to check dimensions as I stumble along.
2. Change v = a/√x to c/√x. "a" should be reserved for acceleration.
So one mass starts at x = -x0/2 and the other at x = x0/2.
With these mods, taking the mass at x= x0/2,
Σ Fx/m = ax = kq2(x)/mx2 - gx/2L,
vx = ∫ ax dt = ∫ ax/vx dx = cx-1/2
winding up after some grief with
q2(x) = (m/2k)(gx3/L - c2)
So we have q(x).
To get q(t) solve dx/dt = -cx-1/2 giving x(t) = {x03/2 - (3/2)ct}2/3.
Then substitute for x in q(x) and that's it.
Still very laborious. Lots of opportunities for mistakes!
I wonder if we could cheat and ignore gravity?
1. Let length of pendula = L instead of 1. I don't lke losing the ability to check dimensions as I stumble along.
2. Change v = a/√x to c/√x. "a" should be reserved for acceleration.
So one mass starts at x = -x0/2 and the other at x = x0/2.
With these mods, taking the mass at x= x0/2,
Σ Fx/m = ax = kq2(x)/mx2 - gx/2L,
vx = ∫ ax dt = ∫ ax/vx dx = cx-1/2
winding up after some grief with
q2(x) = (m/2k)(gx3/L - c2)
So we have q(x).
To get q(t) solve dx/dt = -cx-1/2 giving x(t) = {x03/2 - (3/2)ct}2/3.
Then substitute for x in q(x) and that's it.
Still very laborious. Lots of opportunities for mistakes!
I wonder if we could cheat and ignore gravity?
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