| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
12 May 2007 08:41:34 PM |
| Object: |
Interesting question that I'm stuck on |
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: "Igor Khavkine" |
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| Title: Re: Interesting question that I'm stuck on |
13 May 2007 11:51:02 AM |
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On 2007-05-13, <> 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.
First, to fix the notation, take x to be the coordinate pointing
vertically down from the point of suspension, X(t) to be the position of
the mass m at time t. Also, consider the spring to be made up of lots of
tiny springs that can be displaced, can be stretched or compressed, but
do not loose their identity (i.e. they keep their mass). Let x(s,t)
denote the position of such an infinitesimal spring element, each of
mass m' ds, where s is a continuous parameter ranging, say, from 0 to 1.
As is, this system has an infinite number of degrees of freedom, since
the position x(s,t) of each individual infinitesimal spring element can
be considered an independent generalized coordinate. But for the
purposes of this problem, you reduce the system to a single degree of
freedom, say X(t). Now, you have to express x(s,t) in terms of X(t)
alone. The book tells you to consider the case of linear stretching,
that is, x(s,t) is proportional to s. Since we already know that x(1,t)
= X(t), the unique possibility is x(s,t) = s X(t).
Here's the gist of what you need to keep in mind to build the
Lagrangian. The kinetic energy of each infinitesimal spring element
is (m' ds)/2 [(d/dt) x(s,t)]^2, you'll need to integrate that over s.
The question only asks you to find the frequency of oscillations, which
means that you need to consider only terms quadratic in X(t) in the
potential energy. Terms linear in X(t) will not affect the frequency,
only the equilibrium position. The book tells you that the overall
stiffness of the spring is k, which means that this k takes into account
the potential energy that can be stored in the entire spring (assuming
linear stretching), already including the contributions from all
infinitesimal spring elements.
Finally, the expressions for the total kinetic and potential energies
are:
m /1 m'
T = - [d/dt X(t)]^2 + | ds - [d/dt s X(t)]^2 = < evaluate this >
2 /0 2
k
U = - X(t)^2 + < terms linear in X(t) >
2
From here, finding the oscillation period is straight forward.
If you want to understand this problem in more detail and see why these
assumptions about the effective behavior of the spring are correct, it's
a good exercise to derive the full dynamical equations for the spring
continuum x(s,t). This more general setup can be reduced to the problem
at hand, again, assuming linear stretching. I think the most intuitive
approach there is to build a discrete bead-spring model of the massive
spring (...-*-+++-*-+++-*-..., where * is a massive bead and -+++- is a
massless spring), and then take the continuum limit. You can do this by
following, for instance, the example in sections 11.5-6 of your book.
Hope this helps.
Igor
.
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| User: "" |
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| Title: Re: Interesting question that I'm stuck on |
13 May 2007 05:27:24 PM |
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m /1 m'
T = - [d/dt X(t)]^2 + | ds - [d/dt s X(t)]^2 = < evaluate this >
2 /0 2
k
U = - X(t)^2 + < terms linear in X(t) >
2
I'm not quite comprehending your notation.
for T are you saying (m/2)*(dx/dt)^2+(integral of ds from 2 to m' )
+s*(dx/dt)^2 and th we should evaluate the first term from 0 to 1?
.
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| User: "Igor Khavkine" |
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| Title: Re: Interesting question that I'm stuck on |
13 May 2007 07:40:39 PM |
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On 2007-05-13, <> wrote:
m /1 m'
T = - [d/dt X(t)]^2 + | ds - [d/dt s X(t)]^2 = < evaluate this >
2 /0 2
k
U = - X(t)^2 + < terms linear in X(t) >
2
I'm not quite comprehending your notation.
for T are you saying (m/2)*(dx/dt)^2+(integral of ds from 2 to m' )
+s*(dx/dt)^2 and th we should evaluate the first term from 0 to 1?
I fear you may have been looking at my formulas in a variable width
font. With a fixed with font, all integral signs and fractions would
align. Here they are again in inline notation
T_mass = (m/2) [d/dt X(t)]^2
T_spring = (integral from 0 to 1) ds (m'/2) s^2 [d/dt X(t)]^2
T = T_mass + T_spring
U = (k/2) X(t)^2 + < terms linear in X(t) >
Hope this helps.
Igor
.
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| User: "" |
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| Title: Re: Interesting question that I'm stuck on |
14 May 2007 02:54:33 PM |
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On May 13, 8:40 pm, Igor Khavkine <igor...@gmail.com> wrote:
On 2007-05-13, <> wrote:
m /1 m'
T = - [d/dt X(t)]^2 + | ds - [d/dt s X(t)]^2 = < evaluate this >
2 /0 2
k
U = - X(t)^2 + < terms linear in X(t) >
2
I'm not quite comprehending your notation.
for T are you saying (m/2)*(dx/dt)^2+(integral of ds from 2 to m' )
+s*(dx/dt)^2 and th we should evaluate the first term from 0 to 1?
I fear you may have been looking at my formulas in a variable width
font. With a fixed with font, all integral signs and fractions would
align. Here they are again in inline notation
T_mass = (m/2) [d/dt X(t)]^2
T_spring = (integral from 0 to 1) ds (m'/2) s^2 [d/dt X(t)]^2
T = T_mass + T_spring
U = (k/2) X(t)^2 + < terms linear in X(t) >
Hope this helps.
Igor
it at least tgto me pointed in the correct direction.
.
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| User: "" |
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| Title: Re: Interesting question that I'm stuck on |
13 May 2007 05:03:38 PM |
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On May 13, 12:51 pm, Igor Khavkine <igor...@gmail.com> wrote:
On 2007-05-13, <> 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.
First, to fix the notation, take x to be the coordinate pointing
vertically down from the point of suspension, X(t) to be the position of
the mass m at time t. Also, consider the spring to be made up of lots of
tiny springs that can be displaced, can be stretched or compressed, but
do not loose their identity (i.e. they keep their mass). Let x(s,t)
denote the position of such an infinitesimal spring element, each of
mass m' ds, where s is a continuous parameter ranging, say, from 0 to 1.
As is, this system has an infinite number of degrees of freedom, since
the position x(s,t) of each individual infinitesimal spring element can
be considered an independent generalized coordinate. But for the
purposes of this problem, you reduce the system to a single degree of
freedom, say X(t). Now, you have to express x(s,t) in terms of X(t)
alone. The book tells you to consider the case of linear stretching,
that is, x(s,t) is proportional to s. Since we already know that x(1,t)
= X(t), the unique possibility is x(s,t) = s X(t).
Here's the gist of what you need to keep in mind to build the
Lagrangian. The kinetic energy of each infinitesimal spring element
is (m' ds)/2 [(d/dt) x(s,t)]^2, you'll need to integrate that over s.
The question only asks you to find the frequency of oscillations, which
means that you need to consider only terms quadratic in X(t) in the
potential energy. Terms linear in X(t) will not affect the frequency,
only the equilibrium position. The book tells you that the overall
stiffness of the spring is k, which means that this k takes into account
the potential energy that can be stored in the entire spring (assuming
linear stretching), already including the contributions from all
infinitesimal spring elements.
Finally, the expressions for the total kinetic and potential energies
are:
m /1 m'
T = - [d/dt X(t)]^2 + | ds - [d/dt s X(t)]^2 = < evaluate this >
2 /0 2
k
U = - X(t)^2 + < terms linear in X(t) >
2
From here, finding the oscillation period is straight forward.
If you want to understand this problem in more detail and see why these
assumptions about the effective behavior of the spring are correct, it's
a good exercise to derive the full dynamical equations for the spring
continuum x(s,t). This more general setup can be reduced to the problem
at hand, again, assuming linear stretching. I think the most intuitive
approach there is to build a discrete bead-spring model of the massive
spring (...-*-+++-*-+++-*-..., where * is a massive bead and -+++- is a
massless spring), and then take the continuum limit. You can do this by
following, for instance, the example in sections 11.5-6 of your book.
Hope this helps.
Igor
much obliged sir, i'll see where this gets me.
.
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| User: "" |
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| Title: Re: Interesting question that I'm stuck on |
13 May 2007 08:44:04 AM |
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On May 12, 9:41 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 ???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
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