# Carbon dioxide as an oscillator; normal modes.

## 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?

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ehild
Homework Helper
Plug in the assumed solution for al xi-s.

ehild

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

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

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Anyone?

ehild
Homework Helper
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

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.

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.

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.

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ehild
Homework Helper
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

Thanks that's perfect.