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UnkleRhaukus

  • one year ago

\[a(x) = \frac{\mathrm d^2x}{\mathrm dt^2} = -4\pi^2x+x^2\] \[x_0 = \pi^2,\qquad v_0=\left.\frac{\mathrm dx}{\mathrm dt}\right|_0=0\] Numeric Integration (or any other method that works )

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  1. UnkleRhaukus
    • one year ago
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    *second order*

  2. Empty
    • one year ago
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    Now having two initial conditions actually makes sense haha

  3. Empty
    • one year ago
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    So what are you looking for exactly, a program or a specific result? Without a step size for the time and without an ending time there's not a lot I can really do for you. I can help walk you through the leapfrog method though if you like.

  4. UnkleRhaukus
    • one year ago
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    the program i have wrote has a step size of 0.01, and goes t = [0:5]

  5. UnkleRhaukus
    • one year ago
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    doesn't look right , i doubt that the leap frog method is stable for this equation

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  6. UnkleRhaukus
    • one year ago
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    it should decay

  7. UnkleRhaukus
    • one year ago
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  8. Empty
    • one year ago
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    Well it's not terribly bad for an approximation, I guess it's all a matter of what percent error you're willing to tolerate. You can make a smaller step size and it should make the error smaller.

  9. Empty
    • one year ago
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    Ohhh it should decay ok let me see.

  10. UnkleRhaukus
    • one year ago
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    i was expecting something like this dotted line here

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  11. UnkleRhaukus
    • one year ago
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    i think the leap frog method dosen't handle the sharp corner very well

  12. Empty
    • one year ago
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    I was looking through your code for bugs and couldn't find anything, just out of curiosity what happens when you change b=1 to something else like b=10 or b=100 or something ridiculous.

  13. UnkleRhaukus
    • one year ago
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    bad things happen

  14. Empty
    • one year ago
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    Hahaha I guess I just wanted to make sure it was still working. I think you're probably right about leapfrogging over those narrow curves, I don't think it's able to quite turn fast enough so instead of decaying it's rotated slightly off and is instead decreasing your frequency which is sorta what it looks like.

  15. UnkleRhaukus
    • one year ago
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    and A = x_0 should be less than 4pi^2

  16. Astrophysics
    • one year ago
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    \[x''= -w_0^2x+2 \beta x'\] I thought if b >0 then the force would be resisting for this at least

  17. Astrophysics
    • one year ago
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    err that should be \[x''= -w_0^2x-2 \beta x'\]

  18. Empty
    • one year ago
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    @Astrophysics Think of it like this, an undamped spring will have this equation: x''=-x That means the acceleration is larger the further away you are from equilibrium (x=0). This is what pushes it back. So if you have some other factor getting bigger, then you're damping the spring force to come back. Of course here it's not quite the same situation, equilibrium is not at x=0, it's shifted up so that we don't end up with a greater restoring force than normal on the other side of the swing.

  19. UnkleRhaukus
    • one year ago
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    equilibrium is not at x = 0?

  20. Empty
    • one year ago
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    Woah my bad I don't know what I'm saying, I don't know why I said that part

  21. Empty
    • one year ago
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    I was thinking about the initial position being at \(\pi^2\) and for some reason I said that nonsense lol

  22. UnkleRhaukus
    • one year ago
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    is there some other ( stable ) simple second order numeric method, i can compare with ?

  23. Empty
    • one year ago
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    Hmm I think the only other options I know of is midpoint method and runge-kutta, but not very well. I can't really say how accurate they'll be.

  24. UnkleRhaukus
    • one year ago
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    RK4 won't work, it's a second order DE

  25. Empty
    • one year ago
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    I don't know, what happens when you make the step size smaller on your current program? Does it make the wavelength shorter and dampen it? Try d=0.001 or d=0.00005 maybe and see what happens.

  26. UnkleRhaukus
    • one year ago
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  27. UnkleRhaukus
    • one year ago
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    the shape of the green leapfrog approximation is unconvincing,

  28. anonymous
    • one year ago
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    $$x''=-4\pi^2 x+x^2\\x''=(x-2\pi^2)^2-4\pi^4\\y''=y^2-k$$now consider $$2y'y''=2y^2y'-ky'\\\frac{d}{dt}(y')^2=\frac{d}{dt}\left(\frac23y^3-ky+C\right)\\(y')^2=\frac23y^3-ky+C$$which looks like a Weierstrass function

  29. IrishBoy123
    • one year ago
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    no decay at dt = 0.00001

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