| Topic: |
Science > Physics |
| User: |
"Euh" |
| Date: |
16 Jan 2007 02:09:34 PM |
| Object: |
Need help for solving a problem |
I have a differential equation of the form
d^2x/dt^2 = (2x+3)/(12+x)*9.81
(it's from an application of Newton's 2nd law in which the mass is
increasing with x)
I figured out the initial conditions:
x(t=0)=0
dx/dt(t=0)=0
Now I need to find "t" when x will be equal to 8.
It's a nonlinear second order D.E. But given the relative simplicity of
the problem, it must be possible to solve it by hand ?
If I set: v=dx/dt
I get:
v dv/dx = (2x+3)/(12+x)*9.81
which gives me:
v^2/2 = (2x-24*ln(12+x) +C)*9.81
I can find C from one of my initial condition.
Then knowing v = dx/dt = f(x), I can in theory find x=f(t) by
integrating once more.
Is this the right approach ?
I'm stuck because my f(x) seems to be a fairly complicated expression
to integrate.
any help would be appreciated.
.
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| User: "Robert Israel" |
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| Title: Re: Need help for solving a problem |
16 Jan 2007 03:36:21 PM |
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In article <1168978174.482342.314090@51g2000cwl.googlegroups.com>,
Euh <EuhSCQ@hotmail.com> wrote:
I have a differential equation of the form
d^2x/dt^2 = (2x+3)/(12+x)*9.81
(it's from an application of Newton's 2nd law in which the mass is
increasing with x)
I figured out the initial conditions:
x(t=0)=0
dx/dt(t=0)=0
Now I need to find "t" when x will be equal to 8.
It's a nonlinear second order D.E. But given the relative simplicity of
the problem, it must be possible to solve it by hand ?
If I set: v=dx/dt
I get:
v dv/dx = (2x+3)/(12+x)*9.81
which gives me:
v^2/2 = (2x-24*ln(12+x) +C)*9.81
I can find C from one of my initial condition.
Then knowing v = dx/dt = f(x), I can in theory find x=f(t) by
integrating once more.
Is this the right approach ?
I'm stuck because my f(x) seems to be a fairly complicated expression
to integrate.
Yes, the integral can't be done in closed form. You could use
numerical integration. Or you could use numerical methods on the
original DE. Thus in Maple 10:
Digits:= 20:
de:= diff(x(t),t$2) = 9.81*(2*x(t)+3)/(12+x(t));
xproc:= subs(
dsolve({de, x(0)=0, D(x)(0)=0}, x(t), numeric, output=listprocedure),
x(t)):
fsolve(xproc = 8, 0..3);
2.0794321715194089016
Robert Israel
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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| User: "Euh" |
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| Title: Re: Need help for solving a problem |
16 Jan 2007 07:57:08 PM |
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Yes, the integral can't be done in closed form. You could use
numerical integration. Or you could use numerical methods on the
original DE. Thus in Maple 10:
Digits:= 20:
de:= diff(x(t),t$2) = 9.81*(2*x(t)+3)/(12+x(t));
xproc:= subs(
dsolve({de, x(0)=0, D(x)(0)=0}, x(t), numeric, output=listprocedure),
x(t)):
fsolve(xproc = 8, 0..3);
2.0794321715194089016
I made a booboo.
The equation should read:
d^2x/dt^2 = (2*x+3)*9.81/20
It's a bit easier to integrate by hand.
I end up with t=2.6. Could you confirm this result with Maple please ?
.
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| User: "Robert Israel" |
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| Title: Re: Need help for solving a problem |
16 Jan 2007 11:47:05 PM |
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In article <1168999027.269556.131210@a75g2000cwd.googlegroups.com>,
Euh <EuhSCQ@hotmail.com> wrote:
Yes, the integral can't be done in closed form. You could use
numerical integration. Or you could use numerical methods on the
original DE. Thus in Maple 10:
Digits:= 20:
de:= diff(x(t),t$2) = 9.81*(2*x(t)+3)/(12+x(t));
xproc:= subs(
dsolve({de, x(0)=0, D(x)(0)=0}, x(t), numeric,
output=listprocedure),
x(t)):
fsolve(xproc = 8, 0..3);
2.0794321715194089016
I made a booboo.
The equation should read:
d^2x/dt^2 = (2*x+3)*9.81/20
It's a bit easier to integrate by hand.
More than "a bit" easier, but much less interesting...
I end up with t=2.6. Could you confirm this result with Maple please ?
Actually it's sqrt(10/9.81) ln((19 - 4 sqrt(22))/3).
Robert Israel
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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