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theEric Group Title

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 \(\alpha=\sqrt{\frac km~~}\)

  • 6 months ago
  • 6 months ago

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  1. theEric Group Title
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    I know I should use \(F(x)=m\ddot x\) somwhere...

    • 6 months ago
  2. theEric Group Title
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    somewhere*

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

    • 6 months ago
  4. theEric Group Title
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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.

    • 6 months ago
  5. Kainui Group Title
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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 months ago
  6. theEric Group Title
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    That might do it, thanks!

    • 6 months ago
  7. theEric Group Title
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    I'll look into it..

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

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

    • 6 months ago
  10. Kainui Group Title
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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. =)

    • 6 months ago
  11. theEric Group Title
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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!

    • 6 months ago
  12. theEric Group Title
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    Thanks! :)

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

    • 6 months ago
  14. Kainui Group Title
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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?

    • 6 months ago
  15. Vincent-Lyon.Fr Group Title
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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.

    • 6 months ago
  16. theEric Group Title
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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 \(x\), \({\large\int_0^\dot x}\dfrac{d\dot x}{dx}\dot x\ dx={\large\int_{x_0}^x}\alpha^2x\ dx\) \({\large\int_0^\dot x}\dot x\ d\dot x=\alpha^2{\large\int_{x_0}^x}x\ dx\) \(\frac12\dot x^2=\alpha^2\left(\frac12x^2-\frac12x_0^2\right)\) \(\frac12\dot x^2=\frac12\alpha^2\left(x^2-x_0^2\right)\) \(\dot x^2=\alpha^2\left(x^2-x_0^2\right)\) \(\dfrac{\dot x^2}{\alpha^2}=x^2-x_0^2\) \(\dfrac{\dot x^2}{\alpha^2}+x_0^2=x^2\) \(x(t)=\pm\sqrt{\dfrac{\dot x^2}{\alpha^2}+x_0^2~~}\)

    • 6 months ago
  17. theEric Group Title
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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...

    • 6 months ago
  18. theEric Group Title
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    They should be the same, I think.

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

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

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

    • 6 months ago
  22. Vincent-Lyon.Fr Group Title
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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.

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

    • 6 months ago
  24. Kainui Group Title
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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: \[x(t)=x_0 \frac{e^{\alpha t}+e^{- \alpha t}}{2}=x_0 \cosh( \alpha t)\]

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

    • 6 months ago
  26. theEric Group Title
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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 :)

    • 6 months ago
  27. Vincent-Lyon.Fr Group Title
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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 \(\ddot x -\alpha ^2\,x=0\)

    • 6 months ago
  28. theEric Group Title
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    Ohhh, okay! Thanks! :)

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

    • 6 months ago
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