# Tricky ODE system

1. Sep 4, 2009

### s.g.g

1. The problem statement, all variables and given/known data
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.

2. Relevant 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

3. 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?

2. Sep 5, 2009

### tiny-tim

Hi s.g.g!

(try using the X2 tag just above the Reply box )
(i assume the last one should start d2z/dt2? )

You seem to have selected unhelpful variables …

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

and what is d2(x+y+z)/dt2 ?

3. Sep 5, 2009

### s.g.g

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 isnt it?

4. Sep 5, 2009

### tiny-tim

Yes!

and that equals … ?

5. Sep 5, 2009

### s.g.g

Ha I'm starting to think the answer is so easy i cant 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

6. Sep 6, 2009

### tiny-tim

Hi s.g.g!

(have an omega: ω )
good so far

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.

7. Sep 6, 2009

### s.g.g

I still cant do it, im trying to compute d2(x-y) /dt2
and letting (x-y) =gamma. But i cannot solve for gamma

8. Sep 6, 2009

### tiny-tim

You need to show us how far you got.

(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! )

9. Sep 6, 2009

### s.g.g

Ha sorry, 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 im then stuck with the same problem. how do i solve for p and q

10. Sep 6, 2009

### s.g.g

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. Sep 6, 2009

### HallsofIvy

Staff Emeritus
Differentiate the second equation twice more:
$d^4q/dt^4= A(d^2p/dt^2- 2 d^2q/dt^2)[itex]. Now, from the first equation, [itex]d^2p/dt^2= Bcos(omega t)+ Aq[itex] so you have [itex]d^4 q/d^4= A(Bcos(omega t)+ Aq)- 2d^2q/dt^2$ or

$d^4q/dt^4+ 2d^2q/dt^2- Aq= ABcos(omega t)$

Once you have found q, integrate $d^2p/dt^2= Bcos(omegat t) + Aq$ to find p.

12. Sep 6, 2009

### tiny-tim

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)