Solving Tricky ODE System Homework

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Homework Help Overview

The discussion revolves around a system of ordinary differential equations (ODEs) involving three masses connected by springs, with a driving force applied to the first mass. The original poster attempts to derive the equation of motion for the last mass using the Lagrangian method but encounters difficulties in solving for one of the variables.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore various approaches, including the use of general solutions and transformations of variables to simplify the equations. There are attempts to express the equations in terms of differences between the masses and discussions about the implications of the center of mass.

Discussion Status

Participants are actively engaging with the problem, providing suggestions for variable substitutions and questioning the original poster's setup. Some guidance has been offered regarding the physical significance of certain combinations of variables, but no consensus has been reached on how to proceed with the solution.

Contextual Notes

The original poster has specified initial conditions for the masses, indicating that they start from rest in equilibrium before the driving force is applied. There is also an acknowledgment of potential errors in previous attempts to manipulate the equations.

s.g.g
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Homework Statement


I have 3 masses in 1-D connected by two springs. A driving force is exerted on the first mass and i need to derive the equation of motion of the last mass. I have worked out the Lagrangian to determine the equations of motion but cannot solve for z.

Homework Equations


The equations are
d^2(x)/dt^2 = A(y + B*cos(omega*t) - x)
d^2(y)/dt^2 = A(z+x-2y)
d^2(x)/dt^2 = A(y-z)

x,y,and z are the positions of the masses in 1-D and A,B are constants

The Attempt at a Solution


I tried general solutions of x= Csin(kt) +Dcos(kt) ect and also x= Ce^(ikt) but could not work it through.
Where am i supposed to start?
 
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Hi s.g.g! :smile:

(try using the X2 tag just above the Reply box :wink:)
s.g.g said:
d^2(x)/dt^2 = A(y + B*cos(omega*t) - x)
d^2(y)/dt^2 = A(z+x-2y)
d^2(x)/dt^2 = A(y-z)

Where am i supposed to start?

(i assume the last one should start d2z/dt2? :wink:)

You seem to have selected unhelpful variables …

variables like (x - y) might make it easier …

and what is d2(x+y+z)/dt2 ? :smile:
 
Sorry, yeah the last equation should be
d^2z/dt2 = A(y-z)

x, y and z are all functions of time.

d^2(x+y+z)/dt^2 = d^2x/dt^2 + d^2y/dt^2 + d^2z/dt^2 isn't it?
 
(please use the X2 tag just above the Reply box :wink:)
s.g.g said:
d^2(x+y+z)/dt^2 = d^2x/dt^2 + d^2y/dt^2 + d^2z/dt^2 isn't it?

Yes! :smile:

and that equals … ? :wink:
 
Ha I'm starting to think the answer is so easy i can't see it.

That equals the LHS of the sum of the equations. So therefor

d2/dt2(x+y+z) = AB*cos (omega*t) after cancelling,

integrating twice with respect to t gives

x(t)+y(t)+z(t) = -(AB/omega2)cos(omega*t) + C

How do i solve this for x(t) y(t) and z(t)?
I know that the masses are originally at rest in their equilibrium position before mass x is subject to a driving force, so I have the initial conditions that
x(0) = x0 , dx(0)/dt=0
y(0) = y0 , dy(0)/dt=0
z(0) = z0 , dz(0)/dt=0
 
Hi s.g.g! :smile:

(have an omega: ω :wink:)
s.g.g said:
x(t)+y(t)+z(t) = -(AB/omega2)cos(omega*t) + C

good so far :smile:

but you haven't yet tried my other suggestion, of using (x - y) etc.

Note that (x+y+z) gives you the physical centre of mass, and you would expect that to have an easy equation …

similarly you would expect the difference of any two of x y and z to have a better physical significance than x y or z on its own, and therefore to have a simpler equation. :wink:
 
I still can't do it, I am trying to compute d2(x-y) /dt2
and letting (x-y) =gamma. But i cannot solve for gamma
 
s.g.g said:
I still can't do it, I am trying to compute d2(x-y) /dt2
and letting (x-y) =gamma. But i cannot solve for gamma

You need to show us how far you got. :smile:

(and did you try other combinations, like (y-z)?)

(and don't use gamma! … use something easy and sensible, like p or q or u! :rolleyes:)
 
Ha sorry:smile:, Heres what iv done.

say y-x=p & z-y=q
hence
d2x/dt2 = d2p/dt2+d2y/dt2,
d2y/dt2 = d2p/dt2+d2x/dt2 = d2q/dt2 + d2z/dt2 &
d2z/dt2= d2q/dt2+d2y/dt2

the original equations then become
d2p/dt2=A(2p-q-Bcos(omega*t))
d2q/dt2=A(2q-p)

but I am then stuck with the same problem. how do i solve for p and q
 
  • #10
sorry tim, disregard that rubbish last remark. it is completely incorrect.

i say that x-y=p & y-z=q

i then find that
d2p/dt2=Bcos(omega*t) +Aq
& d2q/dt2=A(p-2q)

But same as before, i cannot solve for p and q, what am i missing?
 
  • #11
s.g.g said:
sorry tim, disregard that rubbish last remark. it is completely incorrect.

i say that x-y=p & y-z=q

i then find that
d2p/dt2=Bcos(omega*t) +Aq
& d2q/dt2=A(p-2q)

But same as before, i cannot solve for p and q, what am i missing?
Differentiate the second equation twice more:
[itex]d^4q/dt^4= A(d^2p/dt^2- 2 d^2q/dt^2)[itex]. <br /> <br /> Now, from the first equation, [itex]d^2p/dt^2= Bcos(omega t)+ Aq[itex]so you have<br /> [itex]d^4 q/d^4= A(Bcos(omega t)+ Aq)- 2d^2q/dt^2[/itex] or<br /> <br /> [itex]d^4q/dt^4+ 2d^2q/dt^2- Aq= ABcos(omega t)[/itex]<br /> <br /> Once you have found q, integrate [itex]d^2p/dt^2= Bcos(omegat t) + Aq[/itex] to find p.[/itex][/itex][/itex][/itex]
 
  • #12
s.g.g said:
x-y=p & y-z=q

d2p/dt2=Bcos(omega*t) +Aq
& d2q/dt2=A(p-2q)

Assuming this is right (I haven't checked it) …

try looking for an eigenvector (ie. a combination so that d2/dt2(Bp + q) is a multiple of Bp +q)
 

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