• Support PF! Buy your school textbooks, materials and every day products Here!

Carbon dioxide as an oscillator; normal modes.

  • Thread starter novop
  • Start date
  • #1
124
0

Homework Statement



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?

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?
 

Answers and Replies

  • #2
ehild
Homework Helper
15,477
1,854
Plug in the assumed solution for al xi-s.

ehild
 
  • #3
124
0
[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:
  • #4
124
0
Anyone?
 
  • #5
ehild
Homework Helper
15,477
1,854
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
 
  • #6
124
0
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.
 
  • #7
671
2
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.
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.
 
  • #8
124
0
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:
  • #9
ehild
Homework Helper
15,477
1,854
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]
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
 
  • #10
124
0
Thanks that's perfect.
 

Related Threads on Carbon dioxide as an oscillator; normal modes.

Replies
5
Views
3K
Replies
0
Views
2K
Replies
7
Views
907
  • Last Post
Replies
6
Views
691
Replies
1
Views
974
Replies
5
Views
23K
Replies
3
Views
2K
Replies
24
Views
8K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
22
Views
6K
Top