Coupled oscillator: 2 masses and 3 different springs

AI Thread Summary
The discussion revolves around solving a problem involving two coupled harmonic oscillators with masses m and spring constants kA, kB, and kC. The equations of motion are derived, leading to a complex quadratic equation for the normal frequencies ω' and ω''. Participants express difficulty in simplifying the quadratic form to match the expected solutions from the textbook. There is a consensus that the book's answer may be overly complicated, particularly regarding the dependence on kC. Ultimately, the conversation highlights the challenges of deriving the correct normal modes of oscillation and the importance of careful algebraic manipulation.
x24759
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Homework Statement


Two harmonic oscillators A and B , of mass m and spring constants kA and kB are coupled together by a spring of spring constant kC .Find the normal frequencies ω' and ω'' and describe the normal modes of oscillation if (k C)2= kAkB)

Homework Equations

The Attempt at a Solution


The system I have drawn looks like this
wall|---kA---m---kc---m---kb---|wall

with axes XA and XB at the equilibrium position of the left mass and right masses, the differential equations for the motion are:

mx¨A+kAxA+kC(xA-xB)=0

mx¨B+kBxB+kC(xB-xA)=0

After rearranging and ωi=√(ki/m)
A +xA[(ωA)2+(ωC)2]-xBC)2 = 0
and x¨B+xB[(ωB)2)+(ωC)2]-xAC)2=0

(I can't find anywhere on the site how to make a proper x double dot, anyone know?)displacement of each mass with have the form:
xA=Acos(ωt) and xB=Bcos(ωt)

plugging into the differential equations and solving for A/B on each and setting them equal to each other:

ωC2/(ωA2C22)=(ωb2C22)/ωC2

and here is where I get stuck. I have not been able to get the quadratic for ω2 into a form I can work with, so I think I have done something wrong, or there is a better way to go about this problem
 
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x24759 said:
(I can't find anywhere on the site how to make a proper x double dot, anyone know?)
If you use LaTeX (recommended) it's \ddot as a prefix to the character(s).
x24759 said:
I have not been able to get the quadratic for ω2 into a form I can work with,
Multiply out your equation, and think of ω2 as the unknown.
 
haruspex said:
Multiply out your equation, and think of ω2 as the unknown.

That is what I did, not getting anywhere. x=ω2 and A, B, C= ω2A B or Cx2 - x(A+B+C) +[AC+AB+BC]
doing the quadratic formula
x = (A+B+C)/2 ± ½√(A2+B2+C2-3AB-3BC-3AC)

I don't think that is moving in the right directing judging by the answer in the back (ω'2 = (kA+kB+kC)/m and ω''2 = kC/m)
 
x24759 said:
That is what I did, not getting anywhere. x=ω2 and A, B, C= ω2A B or Cx2 - x(A+B+C) +[AC+AB+BC]

Check the coefficient of C in the middle term.
 
TSny said:
Check the coefficient of C in the middle term.

I found the error, thanks.

the equation is x2 - x(A+B+2C) +[AC+AB+BC] and the roots are x = (A+B+2C)/2 ± ½√(A2+B2+4C2-2AB), which still does not get me to the right answer..
 
x24759 said:
I found the error, thanks.

the equation is x2 - x(A+B+2C) +[AC+AB+BC] and the roots are x = (A+B+2C)/2 ± ½√(A2+B2+4C2-2AB), which still does not get me to the right answer..
Yes, that's where I got to. I find the book answer doubtful, that there should be a mode that depends only on kC.
 
Remember, C can be written in terms of A and B.
 
TSny said:
Remember, C can be written in terms of A and B.
I read that as only pertaining to the last part of the question. If it is supposed to apply to the first part as well then the book answer is unnecessarily complicated.
 
I got it. Thanks for the help.
 
  • #10
haruspex said:
I read that as only pertaining to the last part of the question. If it is supposed to apply to the first part as well then the book answer is unnecessarily complicated.
How would you simplify the book answer?
 
  • #11
TSny said:
How would you simplify the book answer?
You're right - I was looking at x24759's answer.
 
  • #12
ok.
 
  • #13
haruspex said:
Yes, that's where I got to. I find the book answer doubtful, that there should be a mode that depends only on kC.

The problem provides kC=kAkB

After the eigenvalue equation is simplified into quadratic form, simplification is trivial, albeit a pain in the a**.
 
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