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Carbon dioxide as an oscillator; normal modes.

  1. Mar 31, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the CO2 molecule as a system made of a central mass [tex]m_2[/tex] connected by equal springs of spring constant k to two masses [tex]m_1[/tex] and [tex]m_3[/tex]

    a) set up and solve the equations for the two normal modes in which the masses oscillate along the line joining their centers (the x-axis).

    b) putting [tex]m_1[/tex] = [tex]m_3[/tex] = 16 units and [tex]m_2[/tex] = 12 units, what would be the ratio of the frequencies of the two normal modes?

    3. The attempt at a solution

    I made x_1,x_2 and x_3 the displacement to the right from equilibrium position.

    a)
    [tex]m_3\frac{d^2x_3}{dt^2} = -k(x_3-x_2)[/tex]
    [tex]m_2\frac{d^2x_2}{dt^2} = -k(x_2-x_1) -k(x_2-x_3)[/tex]
    [tex]m_1\frac{d^2x_1}{dt^2} = -k(x_1-x_2)[/tex]

    Assuming [tex]x_1 = C_1cos(wt), x_2=C_2cos(wt)[/tex] etc...
    solving for the double time derivatives and plugging them in above gives:

    [tex]w^2C_1m_1=k(x_1-x_2) [/tex]
    [tex]w^2C_2m_2=k(x_2-x_1) + k(x_2-x_3) [/tex]
    [tex]w^2C_3m_3=k(x_3-x_2)[/tex]

    Where do I go from here?
     
  2. jcsd
  3. Mar 31, 2010 #2

    ehild

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    Plug in the assumed solution for al xi-s.

    ehild
     
  4. Mar 31, 2010 #3
    [tex]w^2C_1m_1=k(C_1-C_2) [/tex]
    [tex]w^2C_2m_2=k(2C_2-C_1-C_3)[/tex]
    [tex]w^2C_3m_3=k(C_3-C_2)[/tex]

    So that's just replacing the x's with C's, considering the cos(wt) terms cancel.
     
    Last edited: Mar 31, 2010
  5. Mar 31, 2010 #4
    Anyone?
     
  6. Apr 1, 2010 #5

    ehild

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    You have a system of linear equations with the unknowns C1, C2, C3. Group the unknowns and arrange the equations in matrix form. [itex]w^2[/itex] is the parameter you have to find so the system of equations has non-zero solution. For that, the determinant of the matrix has to vanish.

    ehild
     
  7. Apr 2, 2010 #6
    This is pretty foreign to me; if I make a matrix with columns C1, C2 and C3, for my determinant to be zero, the following must hold true:

    [tex](w^2m_1-k)(w^2m_2-2k)(w^2m_3-k)=0[/tex]

    How do I solve this? Am I on the right track? I should have added that I need to solve for "w" before plugging in the masses.
     
  8. Apr 2, 2010 #7
    For that equation to hold true, the LHS has to equal the RHS, 0.

    If A*B*C=0 then A=0 or B=0 or C=0.
     
  9. Apr 2, 2010 #8
    So then my two modes are [tex]w_1^2=\frac{k}{m_1}=w_3^2[/tex] and [tex]w_2^2=\frac{2k}{m_2}[/tex]?

    If this is the case, the ratio between these two frequencies isn't giving me what the textbook says.
     
    Last edited: Apr 2, 2010
  10. Apr 2, 2010 #9

    ehild

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    Some terms are missing.

    [tex]D=(w^2m_1-k)(w^2m_2-2k)(w^2m_3-k)-k^2(w^2m_1-k)-k^2(w^2m_3-k)=0[/tex]

    m1=m3=m, so

    [tex]D=(w^2m-k)(w^2m_2-2k)(w^2m-k)-2k^2(w^2m-k)=0[/tex]

    Factor out w^2m-k.

    [tex]D=(w^2m-k)((w^2m_2-2k)(w^2m-k)-2k^2)=0[/tex]

    w^2m-k=0 gives the first angular frequency.

    The other possibilities come from the condition that

    [tex](w^2m_2-2k)(w^2m-k)-2k^2=0[/tex]

    Simplify and solve. One root is w^2=0. This corresponds to a translation of the molecule as a whole. You need the other frequency.

    ehild
     
  11. Apr 2, 2010 #10
    Thanks that's perfect.
     
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