Carbon dioxide as an oscillator; normal modes.

[novop]

Homework Statement

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

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 m_1 = m_3 = 16 units and m_2 = 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)
m_3\frac{d^2x_3}{dt^2} = -k(x_3-x_2)
m_2\frac{d^2x_2}{dt^2} = -k(x_2-x_1) -k(x_2-x_3)
m_1\frac{d^2x_1}{dt^2} = -k(x_1-x_2)

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

w^2C_1m_1=k(x_1-x_2)
w^2C_2m_2=k(x_2-x_1) + k(x_2-x_3)
w^2C_3m_3=k(x_3-x_2)

Where do I go from here?

 

[ehild]

Plug in the assumed solution for al xi-s.

ehild

 

[novop]

w^2C_1m_1=k(C_1-C_2)
w^2C_2m_2=k(2C_2-C_1-C_3)
w^2C_3m_3=k(C_3-C_2)

So that's just replacing the x's with C's, considering the cos(wt) terms cancel.

 

[novop]

Anyone?

 

[ehild]

You have a system of linear equations with the unknowns C1, C2, C3. Group the unknowns and arrange the equations in matrix form. w^2 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

 

[novop]

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:

(w^2m_1-k)(w^2m_2-2k)(w^2m_3-k)=0

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.

 

[RoyalCat]

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:

(w^2m_1-k)(w^2m_2-2k)(w^2m_3-k)=0

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.

 

[novop]

So then my two modes are w_1^2=\frac{k}{m_1}=w_3^2 and w_2^2=\frac{2k}{m_2}?

If this is the case, the ratio between these two frequencies isn't giving me what the textbook says.

 

[ehild]

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:

(w^2m_1-k)(w^2m_2-2k)(w^2m_3-k)=0

Some terms are missing.

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

m1=m3=m, so

D=(w^2m-k)(w^2m_2-2k)(w^2m-k)-2k^2(w^2m-k)=0

Factor out w^2m-k.

D=(w^2m-k)((w^2m_2-2k)(w^2m-k)-2k^2)=0

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

The other possibilities come from the condition that

(w^2m_2-2k)(w^2m-k)-2k^2=0

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

 

[novop]

Thanks that's perfect.