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theEric
Group Title
This is for mechanics II.
We're working with a particle of mass \(m\) in onedimensional motion.
The potential energy is given by
\(V(x)=\frac12kx^2\)
and the force is antirestoring 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 \(\alpha=\sqrt{\frac km~~}\)
 8 months ago
 8 months ago
theEric Group Title
This is for mechanics II. We're working with a particle of mass \(m\) in onedimensional motion. The potential energy is given by \(V(x)=\frac12kx^2\) and the force is antirestoring 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 \(\alpha=\sqrt{\frac km~~}\)
 8 months ago
 8 months ago

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theEric Group TitleBest ResponseYou've already chosen the best response.1
I know I should use \(F(x)=m\ddot x\) somwhere...
 8 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
somewhere*
 8 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
The \(e^{stuff}\) will probably come from integration with \(x\)'s in the denominator.
 8 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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.
 8 months ago

Kainui Group TitleBest ResponseYou've already chosen the best response.1
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.
 8 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
That might do it, thanks!
 8 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
I'll look into it..
 8 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
I'll start by integrating with respect to \(t\) :)
 8 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Or not.. I'll see...
 8 months ago

Kainui Group TitleBest ResponseYou've already chosen the best response.1
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. =)
 8 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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!
 8 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Thanks! :)
 8 months ago

Kainui Group TitleBest ResponseYou've already chosen the best response.1
When I said trig substitution I mean hyperbolic trig function btw haha.
 8 months ago

Kainui Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{d^2x}{dt^2}=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v\] Where along this are you stuck, or are you already past this and it makes sense now?
 8 months ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.0
\(\ddot x\alpha ^2 x=0\) simply leads to solution : \(x(t)=A\;e^{\alpha x}+B\;e^{+\alpha x}\) where A and B are determined by initial conditions.
 8 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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 \(x\), \({\large\int_0^\dot x}\dfrac{d\dot x}{dx}\dot x\ dx={\large\int_{x_0}^x}\alpha^2x\ dx\) \({\large\int_0^\dot x}\dot x\ d\dot x=\alpha^2{\large\int_{x_0}^x}x\ dx\) \(\frac12\dot x^2=\alpha^2\left(\frac12x^2\frac12x_0^2\right)\) \(\frac12\dot x^2=\frac12\alpha^2\left(x^2x_0^2\right)\) \(\dot x^2=\alpha^2\left(x^2x_0^2\right)\) \(\dfrac{\dot x^2}{\alpha^2}=x^2x_0^2\) \(\dfrac{\dot x^2}{\alpha^2}+x_0^2=x^2\) \(x(t)=\pm\sqrt{\dfrac{\dot x^2}{\alpha^2}+x_0^2~~}\)
 8 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
@VincentLyon.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...
 8 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
They should be the same, I think.
 8 months ago

Kainui Group TitleBest ResponseYou've already chosen the best response.1
I suppose I'm really taking the long way, sorry! Haha. Yes, they're the same though.
 8 months ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.0
You can alternatively use \(x(t)=C\cosh (\alpha x) + D \sinh (\alpha x)\)
 8 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
I'd really rather not use the hyperbolic trigonometric functions, haha!!
 8 months ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.0
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.
 8 months ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.0
If initial velocity is zero, then the hyperbolic functions are simpler.
 8 months ago

Kainui Group TitleBest ResponseYou've already chosen the best response.1
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: \[x(t)=x_0 \frac{e^{\alpha t}+e^{ \alpha t}}{2}=x_0 \cosh( \alpha t)\]
 8 months ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.0
Sorry, I meant : \(x(t)=C\cosh (\alpha t) + D \sinh (\alpha t)\)
 8 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
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 :)
 8 months ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.0
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 \(\ddot x \alpha ^2\,x=0\)
 8 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Ohhh, okay! Thanks! :)
 8 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
And I got it with the initial conditions :) Thank you both! :)
 8 months ago
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