theEric
  • theEric
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 \(\alpha=\sqrt{\frac km~~}\)
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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theEric
  • theEric
I know I should use \(F(x)=m\ddot x\) somwhere...
theEric
  • theEric
somewhere*
theEric
  • theEric
The \(e^{stuff}\) will probably come from integration with \(x\)'s in the denominator.

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theEric
  • theEric
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.
Kainui
  • Kainui
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.
theEric
  • theEric
That might do it, thanks!
theEric
  • theEric
I'll look into it..
theEric
  • theEric
I'll start by integrating with respect to \(t\) :)
theEric
  • theEric
Or not.. I'll see...
Kainui
  • Kainui
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. =)
theEric
  • theEric
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!
theEric
  • theEric
Thanks! :)
Kainui
  • Kainui
When I said trig substitution I mean hyperbolic trig function btw haha.
Kainui
  • Kainui
\[\frac{d^2x}{dt^2}=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v\] Where along this are you stuck, or are you already past this and it makes sense now?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
\(\ddot x-\alpha ^2 x=0\) simply leads to solution : \(x(t)=A\;e^{-\alpha x}+B\;e^{+\alpha x}\) where A and B are determined by initial conditions.
theEric
  • theEric
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 \(x\), \({\large\int_0^\dot x}\dfrac{d\dot x}{dx}\dot x\ dx={\large\int_{x_0}^x}\alpha^2x\ dx\) \({\large\int_0^\dot x}\dot x\ d\dot x=\alpha^2{\large\int_{x_0}^x}x\ dx\) \(\frac12\dot x^2=\alpha^2\left(\frac12x^2-\frac12x_0^2\right)\) \(\frac12\dot x^2=\frac12\alpha^2\left(x^2-x_0^2\right)\) \(\dot x^2=\alpha^2\left(x^2-x_0^2\right)\) \(\dfrac{\dot x^2}{\alpha^2}=x^2-x_0^2\) \(\dfrac{\dot x^2}{\alpha^2}+x_0^2=x^2\) \(x(t)=\pm\sqrt{\dfrac{\dot x^2}{\alpha^2}+x_0^2~~}\)
theEric
  • theEric
@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...
theEric
  • theEric
They should be the same, I think.
Kainui
  • Kainui
I suppose I'm really taking the long way, sorry! Haha. Yes, they're the same though.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
You can alternatively use \(x(t)=C\cosh (\alpha x) + D \sinh (\alpha x)\)
theEric
  • theEric
I'd really rather not use the hyperbolic trigonometric functions, haha!!
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
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.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
If initial velocity is zero, then the hyperbolic functions are simpler.
Kainui
  • Kainui
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: \[x(t)=x_0 \frac{e^{\alpha t}+e^{- \alpha t}}{2}=x_0 \cosh( \alpha t)\]
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Sorry, I meant : \(x(t)=C\cosh (\alpha t) + D \sinh (\alpha t)\)
theEric
  • theEric
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 :)
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
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 \(\ddot x -\alpha ^2\,x=0\)
theEric
  • theEric
Ohhh, okay! Thanks! :)
theEric
  • theEric
And I got it with the initial conditions :) Thank you both! :)

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