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theEric

  • 9 months ago

This is for mechanics II. We're working with a particle of mass \(m\) in one-dimensional motion. The potential energy is given by \(V(x)=-\frac12kx^2\) and the force is anti-restoring at \(F(x)=kx\) It is also given that we see unstable equilibrium at \(x=0\). We want to consider the initial conditions \(\quad\)\(t=0\\x=x_0\\\dot x=0\) And we want to show that the motion is an "exponential 'runaway'" \(x(t)=\frac12x_0(e^{\alpha t}+e^{-\alpha t})\) where \(\alpha=\sqrt{\frac km~~}\)

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  1. theEric
    • 9 months ago
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    I know I should use \(F(x)=m\ddot x\) somwhere...

  2. theEric
    • 9 months ago
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    somewhere*

  3. theEric
    • 9 months ago
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    The \(e^{stuff}\) will probably come from integration with \(x\)'s in the denominator.

  4. theEric
    • 9 months ago
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    I looked at integrating \(F(x)\) from \(x_0\) to \(x\) so I could equate it to \(-V(x)\) to see if that would help. But I found that \(x_0=0\), I think because the interval of integration was meaningless.

  5. Kainui
    • 9 months ago
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    I think simply just setting the two equal and solving the differential equation should do it. \[m x'' = k x\] I haven't solved it yet.

  6. theEric
    • 9 months ago
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    That might do it, thanks!

  7. theEric
    • 9 months ago
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    I'll look into it..

  8. theEric
    • 9 months ago
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    I'll start by integrating with respect to \(t\) :)

  9. theEric
    • 9 months ago
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    Or not.. I'll see...

  10. Kainui
    • 9 months ago
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    So here's what I did to get you started: \[\frac{d^2x}{dt^2}=\frac{k}{m} x\]\[\frac{dv}{dt}=\frac{k}{m} x\] Chain rule here\[\frac{dv}{dx}\frac{dx}{dt}=\frac{k}{m} x\]but wait, dx/dt is v \[\frac{dv}{dx}v=\frac{k}{m} x\] Now you can integrate with respect to x and v separately and solve for your constant of integration by applying your initial conditions. Then continue onwards. You will get a trig substitution and I can help you get through that too and it works out pretty nicely. =)

  11. theEric
    • 9 months ago
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    The answer I found has the line \(F(x)=-\dfrac{dV(x)}{dx}=kx=m\ddot x=m\dot x\dfrac{d\dot x}{dx}\) Right now I can't see how the last two expressions are equivalent. I'm looking at your post now!

  12. theEric
    • 9 months ago
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    Thanks! :)

  13. Kainui
    • 9 months ago
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    When I said trig substitution I mean hyperbolic trig function btw haha.

  14. Kainui
    • 9 months ago
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    \[\frac{d^2x}{dt^2}=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v\] Where along this are you stuck, or are you already past this and it makes sense now?

  15. Vincent-Lyon.Fr
    • 9 months ago
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    \(\ddot x-\alpha ^2 x=0\) simply leads to solution : \(x(t)=A\;e^{-\alpha x}+B\;e^{+\alpha x}\) where A and B are determined by initial conditions.

  16. theEric
    • 9 months ago
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    Well... From \(\frac{dv}{dx}v=\frac{k}{m} x\) I put in my variables to make it: \(\dfrac{d\dot x}{dx}\dot x=\alpha^2x\) Integrating with respect to \(x\), \({\large\int_0^\dot x}\dfrac{d\dot x}{dx}\dot x\ dx={\large\int_{x_0}^x}\alpha^2x\ dx\) \({\large\int_0^\dot x}\dot x\ d\dot x=\alpha^2{\large\int_{x_0}^x}x\ dx\) \(\frac12\dot x^2=\alpha^2\left(\frac12x^2-\frac12x_0^2\right)\) \(\frac12\dot x^2=\frac12\alpha^2\left(x^2-x_0^2\right)\) \(\dot x^2=\alpha^2\left(x^2-x_0^2\right)\) \(\dfrac{\dot x^2}{\alpha^2}=x^2-x_0^2\) \(\dfrac{\dot x^2}{\alpha^2}+x_0^2=x^2\) \(x(t)=\pm\sqrt{\dfrac{\dot x^2}{\alpha^2}+x_0^2~~}\)

  17. theEric
    • 9 months ago
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    @Vincent-Lyon.Fr Using my initial conditions, that should give me the answer. Thanks! I'll see what I can do! It's the same equation that @Kainui , to which I found a different result than wolframalpha.com...

  18. theEric
    • 9 months ago
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    They should be the same, I think.

  19. Kainui
    • 9 months ago
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    I suppose I'm really taking the long way, sorry! Haha. Yes, they're the same though.

  20. Vincent-Lyon.Fr
    • 9 months ago
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    You can alternatively use \(x(t)=C\cosh (\alpha x) + D \sinh (\alpha x)\)

  21. theEric
    • 9 months ago
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    I'd really rather not use the hyperbolic trigonometric functions, haha!!

  22. Vincent-Lyon.Fr
    • 9 months ago
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    What you are trying to so is actually a maths problem, proving the nature of the solution of the differential equation. Usually, the mathematicians input a solution that fits, then simply prove there can be no other one.

  23. Vincent-Lyon.Fr
    • 9 months ago
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    If initial velocity is zero, then the hyperbolic functions are simpler.

  24. Kainui
    • 9 months ago
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    It's sort of obvious when you look at it actually haha. Since the hyperbolic sine and cosine functions are their own second derivatives, subtracting them from themselves will give you 0. Here you really should use the hyperbolic trig functions. For instance: \[x(t)=x_0 \frac{e^{\alpha t}+e^{- \alpha t}}{2}=x_0 \cosh( \alpha t)\]

  25. Vincent-Lyon.Fr
    • 9 months ago
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    Sorry, I meant : \(x(t)=C\cosh (\alpha t) + D \sinh (\alpha t)\)

  26. theEric
    • 9 months ago
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    So \(\ddot x(t)=x(t)=C\cosh (\alpha x) + D \sinh (\alpha x)\) ? Thanks! I'll look into that later! I mean, it's cool, but I have other problems to do and was never introduced to the hyperbolic trigonometric functions. I think I'm getting somewhere since I got that general solution \(x(t)=c_1e^{\alpha t}+c_2e^{-\alpha t}\) and am using the initial conditions :)

  27. Vincent-Lyon.Fr
    • 9 months ago
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    Not exactly: \(x(t)=C\cosh (\alpha t) + D \sinh (\alpha t)\) and \(\ddot x(t)=\alpha ^2 C\cosh (\alpha t) + \alpha ^2 D \sinh (\alpha t)\) Hence \(\ddot x -\alpha ^2\,x=0\)

  28. theEric
    • 9 months ago
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    Ohhh, okay! Thanks! :)

  29. theEric
    • 9 months ago
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    And I got it with the initial conditions :) Thank you both! :)

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