| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
12 May 2007 11:18:53 AM |
| Object: |
Analytical Mechanics Question |
Hello. I have a problem that I've been jammed up on for like five
days now. I've worked the problem six times with six varying lines of
thought. I've gotten six varying answers and none of them are the
correct one.
I'll write the basics of the problem out and then give my thoughts on
it. It's out of the seventh edition of Fowles and Cassiday's
Analytical Mechanics.
A heavy elastic spring with uniform density supports mass,m. m' is the
mass of the spring and k is it's stiffness.
Show the period of oscillation is: 2*Pi*Sqrt[(m+(m'/3))/k]
Set up the Lagrange equation and assume the velocity of any portion of
m' is proportional to it's distance from the point of suspension.
Here's the gist of what I see:
I chose x as my single generalized coordinate. I thought it a fair
assumption to treat the mass as point mass at it's center of mass and
place that at the immediate end of the spring. I also thought I could
treat them as a single object since they are attached and both masses
appear in the answer.
So, T=.5*m*(dx/dt)^2+.5*dm' *(dx'/dt)^2 but the velocity dx'/dt
is proportional to cx' and dm'=(lambda)dx'
Then, T=.5*m*(dx/dt)^2+.5*[(lambda)dx' ]*(cx' )^2
meaning that potential of the spring is the integral of .
5*(lambda)*(c^2)*(x'^2)*dx' from x'=0 to x'=x
Meaning, T=(1/2)*m*(dx/dt)^2+(1/6)*(lambda)*(c^2)*(x^3)
=(1/2)*m*(dx/dt)^2+(1/6)*(m' )*(c^2)*(x^2)
=(1/2)*m*(dx/dt)^2+(1/6)*(m' )*(c^2)*(dx/dt)^2
Also, V=m*g*x+g*dm' *x'+(1/2)*k*x' *dx' But dm'=(lambda)*dx' and
dx'=dm' /(lambda) and x'=m' /(lambda)
so the potential energy of the spring is the sum of two integrals.
the first is g*(lambda)*x' *dx' from 0 to x.
and the second is (1/2)*k*[m' /(lambda)]*dm' from 0 to m-m'
meaning that V=m*g*x+(1/2)*g*(lambda)*(x^2)+(1/2)*k*[(m-m' )^2/
(lambda)]
setting up the lagrangrian getting to the dif eq is no problem for
me. I think that my jam originates is my derivations of the kinetic
energy and potential due to stiffness of the spring.
I've also tried a couple of other ways but this derivation obtains the
closest form I've gotten. Any thoughts?
.
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| User: "" |
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| Title: Re: Analytical Mechanics Question |
12 May 2007 02:08:20 PM |
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On May 12, 12:18 pm, "rehamkcir...@gmail.com" <rehamkcir...@gmail.com>
wrote:
Hello. I have a problem that I've been jammed up on for like five
days now. I've worked the problem six times with six varying lines of
thought. I've gotten six varying answers and none of them are the
correct one.
I'll write the basics of the problem out and then give my thoughts on
it. It's out of the seventh edition of Fowles and Cassiday's
Analytical Mechanics.
A heavy elastic spring with uniform density supports mass,m. m' is the
mass of the spring and k is it's stiffness.
Show the period of oscillation is: 2*Pi*Sqrt[(m+(m'/3))/k]
Set up the Lagrange equation and assume the velocity of any portion of
m' is proportional to it's distance from the point of suspension.
Here's the gist of what I see:
I chose x as my single generalized coordinate. I thought it a fair
assumption to treat the mass as point mass at it's center of mass and
place that at the immediate end of the spring. I also thought I could
treat them as a single object since they are attached and both masses
appear in the answer.
So, T=.5*m*(dx/dt)^2+.5*dm' *(dx'/dt)^2 but the velocity dx'/dt
is proportional to cx' and dm'=(lambda)dx'
Then, T=.5*m*(dx/dt)^2+.5*[(lambda)dx' ]*(cx' )^2
meaning that potential of the spring is the integral of .
5*(lambda)*(c^2)*(x'^2)*dx' from x'=0 to x'=x
Meaning, T=(1/2)*m*(dx/dt)^2+(1/6)*(lambda)*(c^2)*(x^3)
=(1/2)*m*(dx/dt)^2+(1/6)*(m' )*(c^2)*(x^2)
=(1/2)*m*(dx/dt)^2+(1/6)*(m' )*(c^2)*(dx/dt)^2
Also, V=m*g*x+g*dm' *x'+(1/2)*k*x' *dx' But dm'=(lambda)*dx' and
dx'=dm' /(lambda) and x'=m' /(lambda)
so the potential energy of the spring is the sum of two integrals.
the first is g*(lambda)*x' *dx' from 0 to x.
and the second is (1/2)*k*[m' /(lambda)]*dm' from 0 to m-m'
meaning that V=m*g*x+(1/2)*g*(lambda)*(x^2)+(1/2)*k*[(m-m' )^2/
(lambda)]
setting up the lagrangrian getting to the dif eq is no problem for
me. I think that my jam originates is my derivations of the kinetic
energy and potential due to stiffness of the spring.
I've also tried a couple of other ways but this derivation obtains the
closest form I've gotten. Any thoughts?
??
.
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