So I have my differential equations for 2 bodies coupled with 3 springs.

 

[LATEX]\ddot{x_{a}}+2\omega_{o}^2x_{a}-\omega_{o}^2x_{b}=0[/LATEX] and

 

[LATEX]\ddot{x_{b}}+2\omega_{o}^2x_{b}-\omega_{o}^2x_{a}=0[/LATEX]

 

I use 2 trial functions [LATEX]q_{1}=Acos(\omega t)[/LATEX]and [LATEX] q_{2}=Bcos(\omega t)[/LATEX]

 

and get two solutions for the normal mode frequencies:

 

[LATEX]\omega_{-}=\omega_{o}[/LATEX]and [LATEX] \omega_{+}=\omega_{o}\sqrt{3}[/LATEX]

 

 

 

Now to get the motion for [LATEX]x_{a},x_{b}[/LATEX], I am supposed to take:

 

[LATEX]\frac{q_{1}+q_{2}}{2}[/LATEX]and [LATEX] \frac{q_{1}-q_{2}}{2}[/LATEX]

 

 

 

My questions:

 

-Why do I have to divide by 2 ? Since the solution is a superposition of normal modes, why not just add q1 and q2. Shouldn't the constants just take care of any initial conditions I impose?

 

-If there were 3 bodies, and therefore, 3 trial solutions [LATEX]q_{1}=Acos(\omega t), q_{2}=Bcos(\omega t), q_{3}=Ccos(\omega t)[/LATEX]

 

what would be the solutions to the equations of motion of the three particles ?

 

Would it be (q1+q2+q3)/3, (q1-q2+q3)/3 and (q1-q2-q3)/3 ? What about (q1+q2-q3)/3 ? Don't we have to take all linear combinations, or is it based on the ratios of amplitudes ? (for example, in the 2 bodied one, we got A/B=1 for one normal mode and then A/B=-1 for another normal mode, therefore q1+q2 and q1-q2)

I think I got this figured out.

At the end of the day, you need as many unknowns as there are equations, so for the two body case, we start out with 8 unknowns as follows:

[math]x_a=Acos(\omega_{-}t+\phi_1)+Bcos(\omega_{+}t+\phi_2)[/math]

[math]x_b=Ccos(\omega_{-}t+\phi_3)+Dcos(\omega_{+}t+\phi_4)[/math]

namely, A, B, C, D, ϕ1, ϕ2, ϕ3,and ϕ4.

These being normal modes, dictates that ϕ1= ϕ3,and ϕ2= ϕ4.

It also just so happens that in all our equation grinding, we found the ratio of A:C and B:D as 1 and -1 respectively.

and so, the equations boil down to:

[math]x_a=Acos(\omega_{-}t+\phi_1)+Bcos(\omega_{+}t+\phi_2)[/math]

[math]x_b=Acos(\omega_{-}t+\phi_1)-Bcos(\omega_{+}t+\phi_2)[/math]

4 unknowns, 4 initial conditions, and we're done.

 

If there were 3 masses, then there would be 3 normal modes, each with a frequency and phase associated with it. What I previously mentioned as (q1+q2+q3) actually represent the superposition of normal modes IF the ratio of the amplitudes were 1. But it may be the case that A:B:C is 4:2:1 in which case it would change the general equation.

For example: for normal mode 1, we might get:

[math]x_a=Acos(\omega_{-}t+\phi_1)[/math]

[math]x_b=0.71Acos(\omega_{-}t+\phi_1)[/math]

[math]x_c=-Acos(\omega_{-}t+\phi_1)[/math]

i.e: the ratio of A:B:C (in normal mode 1) is 1:0.71:-1

 

and in normal mode 2, A:B:C=-2:1:5, and in normal mode 3, A:B:C=4:1:-3

so the general equation for the body is:

[math]x_a=Acos(\omega_{-}t+\phi_1)-2Bcos(\omega_{+}t+\phi_2)+4Ccos(\omega_{++}t+\phi_3)[/math]

etc.

6 unknowns, 3 initial positions, 3 initial velocities, and we're set for life.

 

Please correct me if I am wrong.