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theEric
 2 years ago
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 \(\alpha=\sqrt{\frac km~~}\)
theEric
 2 years ago
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 \(\alpha=\sqrt{\frac km~~}\)

This Question is Closed

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1I know I should use \(F(x)=m\ddot x\) somwhere...

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

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1I 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
 2 years ago
Best ResponseYou've already chosen the best response.1I 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
 2 years ago
Best ResponseYou've already chosen the best response.1That might do it, thanks!

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1I'll start by integrating with respect to \(t\) :)

Kainui
 2 years ago
Best ResponseYou've already chosen the best response.1So 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
 2 years ago
Best ResponseYou've already chosen the best response.1The 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!

Kainui
 2 years ago
Best ResponseYou've already chosen the best response.1When I said trig substitution I mean hyperbolic trig function btw haha.

Kainui
 2 years ago
Best 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?

anonymous
 2 years ago
Best 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.

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

theEric
 2 years ago
Best 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...

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1They should be the same, I think.

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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0You can alternatively use \(x(t)=C\cosh (\alpha x) + D \sinh (\alpha x)\)

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1I'd really rather not use the hyperbolic trigonometric functions, haha!!

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0What 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.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0If initial velocity is zero, then the hyperbolic functions are simpler.

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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry, I meant : \(x(t)=C\cosh (\alpha t) + D \sinh (\alpha t)\)

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1So \(\ddot x(t)=x(t)=C\cosh (\alpha x) + D \sinh (\alpha x)\) ? 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 :)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Not 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
 2 years ago
Best ResponseYou've already chosen the best response.1And I got it with the initial conditions :) Thank you both! :)
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