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[H] Advanced Mechanics

Blogs > deconduo
Post a Reply
deconduo
Profile Blog Joined January 2008
Ireland4122 Posts
Last Edited: 2009-10-30 23:17:37
October 30 2009 23:16 GMT
#1
This is technically a homework thread, but I'm not asking for someone to do all my work for me and solve it completely, all I need is a nudge in the right direction because I'm fairly lost and can't find help anywhere else. If even this is not allowed, then I'm sorry


The question:

+ Show Spoiler +
Find the one dimensional particle motion in the Morse Potential

[image loading]


What I have so far:

+ Show Spoiler +
[image loading]


I now need to change the variables to get something integrable, but I'm unsure what to use.

For example:

[image loading]


doesn't work because I need the du expression to be free of the variable x.
I think I need to divide by something but I just can't work out what that is. Any tips or hints would be greatly appreciated, thanks.


Oh, and http://thornahawk.unitedti.org/equationeditor/equationeditor.php is what I use for the equations.

IntoTheWow
Profile Blog Joined May 2004
is awesome32274 Posts
Last Edited: 2009-10-30 23:32:33
October 30 2009 23:29 GMT
#2
nvm this doens't work ;_;
Moderator<:3-/-<
aqui
Profile Blog Joined January 2008
Germany1023 Posts
Last Edited: 2009-10-31 00:39:53
October 31 2009 00:13 GMT
#3
use the lagrangefunction.
L=T-U , T beeing kinetic and U potential energy.
this
[image loading]
gives you directly the differential equation for the motion.
q beeing x in your case.
The solution of the resulting motion equation depends on your initial values ofc. depending on them you could solve at least the linear part.
You could also expand tp the 2nd order your inital potential around its minimum and approximate the tail to its right. use the solution of one as boundary condition for the other.
deconduo
Profile Blog Joined January 2008
Ireland4122 Posts
October 31 2009 00:30 GMT
#4
On October 31 2009 09:13 aqui wrote:
use the lagrangefunction.
L=T-U , T beeing kinetic and U potential energy.
this
[image loading]
gives you directly the differential equation for the motion.
q beeing x in your case.


Thats pretty much what I used to get to the first step I've written down, its from there that the problem I have is.

I know how to get the EoM, its just that specific integral I'm stuck on. Thanks though.
micronesia
Profile Blog Joined July 2006
United States24666 Posts
October 31 2009 00:38 GMT
#5
I'm so glad I don't have to do this anymore lol

What course is this exactly and what program is it for? When I learned lagrangians in mechanics I don't think I ever was given this potential to use.
ModeratorThere are animal crackers for people and there are people crackers for animals.
deconduo
Profile Blog Joined January 2008
Ireland4122 Posts
October 31 2009 00:57 GMT
#6
On October 31 2009 09:38 micronesia wrote:
I'm so glad I don't have to do this anymore lol

What course is this exactly and what program is it for? When I learned lagrangians in mechanics I don't think I ever was given this potential to use.


Its just called advanced mechanics, its part of 2nd year in my degree. (theoretical physics) This question is part of a section on integrating the equations of motion.
Luddite
Profile Blog Joined April 2007
United States2315 Posts
October 31 2009 01:08 GMT
#7
You can just use computer software to evaluate that integral, or look it up in a table of integrals. If you really have to do it by hand, I think you would have to substitute some sort of trig function.
Can't believe I'm still here playing this same game
aqui
Profile Blog Joined January 2008
Germany1023 Posts
Last Edited: 2009-10-31 01:12:36
October 31 2009 01:09 GMT
#8
On October 31 2009 09:30 deconduo wrote:
Show nested quote +
On October 31 2009 09:13 aqui wrote:
use the lagrangefunction.
L=T-U , T beeing kinetic and U potential energy.
this
[image loading]
gives you directly the differential equation for the motion.
q beeing x in your case.


Thats pretty much what I used to get to the first step I've written down, its from there that the problem I have is.

I know how to get the EoM, its just that specific integral I'm stuck on. Thanks though.

thought you just solved t=x/v. because thats what stands there^^ the integral looks like a bitch though^^ no idea how to solve it, sry:p
Baytuts
Profile Joined January 2009
Brazil101 Posts
Last Edited: 2009-10-31 02:18:34
October 31 2009 02:15 GMT
#9
I dont know how to solve that integral... but:

http://www93.wolframalpha.com/input/?i=integrate [1%2F%28%28k-%281-e^%28-ax%29%29%29%29^0.5%2C+x]%2C+a%3D{-infinite%2Cinfinite}%2Ck%3D{-infinite%2Cinfinite}

dont know if everything is set up correctly, so give a look before follow it blindly

edit: sorry.. thats the best way to put the link.. copy all of it
...
datscilly
Profile Blog Joined November 2007
United States528 Posts
October 31 2009 02:37 GMT
#10
Let a program integrate for you? wolfram integrator of the above expression
DefMatrixUltra
Profile Blog Joined June 2009
Canada1992 Posts
October 31 2009 02:38 GMT
#11
If you have some idea that 'a' is very small or that 'x' will be very small, then you can use the binomial expansion or the Taylor expansion to simplify the expression to something integrable. Without any paper immediately on hand, the obvious substitution looks like u = ax. Then you can probably do something like partial fractions.

Other than that, just try the Integrator.

http://integrals.wolfram.com/index.jsp?expr=1/sqrt(1-(1-exp(-3x))^2)&random=false
deconduo
Profile Blog Joined January 2008
Ireland4122 Posts
October 31 2009 03:25 GMT
#12
Thanks for the integrator thing, but I have to do it by hand and show the work unfortunately. I'll see if I can work backwards from the result though.
DefMatrixUltra
Profile Blog Joined June 2009
Canada1992 Posts
October 31 2009 15:44 GMT
#13
Now that I look at it more closely, I don't see where you got that integral at all.

If you use the Lagrangian (or in this case, just say F = -grad(u), which I think is faster), you should get the expression

mx'' = -2Aa(exp(-ax)-exp(-2ax))

From there you should be able to separate variables and integrate twice. There is a substitution needed (for the x side) and some algebraic manipulation, then you will end up with two integrals for the x side which should solvable. Then just backtrack your substitution carefully after you've solved.
15vs1
Profile Joined November 2007
64 Posts
Last Edited: 2009-11-02 14:29:51
November 02 2009 14:25 GMT
#14
[image loading]

after the substitution
[image loading]

we have
[image loading]


the next substitution is
[image loading]


[image loading]

where
[image loading]

if E>A (infinite motion) then the equation in the denominator has 2 different real roots, if E<1 this equation has no real roots. These two cases should be considered separatly.

Let us consider the first case.
[image loading]

where
[image loading]

[image loading]

where
[image loading]


The second case is similar but there will be arctan. It looks messy, so maybe some simplifications are available.
Luddite
Profile Blog Joined April 2007
United States2315 Posts
November 02 2009 16:08 GMT
#15
impressive... most impressive.
Can't believe I'm still here playing this same game
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