Coupled differential equations, free oscillations of carbon and oxygen atoms

Question

$x''(t) = -(A/M_C)*(x-y)$ and $y''(t)=-(A/M_O)*(y-x)$

how would i go about solving the set of coupled differential equations (see snippet) i have made use of the substitution so that i obtained a fourth order differential equation of the form $x'''' + x''(A/M_c + A/M_0)=0$ from this i have formed the characteristic polynomial by letting $x(t)=Kexp(L*T)$ which after substituting into the equation, gave me the equation of the format $Kexp(LT)(L^2(L^2 + (A/M_o+A/M_c))=0$ then from this i obtained the solution set $L={0,sqrt(-(A/M_c +A/M_0)), - sqrt(-(A/M_c +A/M_0))}$ but since $M_c$ and $M_o$ are masses, theyre clearly $>=0$ and so to the variable A which is the spring constant which is also >=0 thus the final solutions would have complex conjugate roots.

my overall solution for $x(t)$ is as follows:

$x(t) = \alpha$ $+$ $\beta *t$ + $\gamma*cos(t(A/M_o + A/M_c))$ + $\delta*sin(t(A/M_o + A/M_c))$

$y(t) = (-M_C /A)(A/M_o +A/M_c)^2$[$\gamma*cos(t(A/M_o + A/M_c$)) + $\delta$sin(t(A/M_o + A/M_c))] + $\alpha$ + $\beta*t$ + $\gamma*cos(t(A/M_o + A/M_c$)+$\delta*sin(t(A/M_o + A/M_c$)

then making use of the conversion equation $y(t)=(M_C/A)*x''(t)+x(t)$ to get the corresponding equation for y(t).

where would i go from this point?

Also how is the sum $M_cX + M_oY$ Dependant on time? And i'd verify this by adding the corresponding solutions.

Answer 1

Your method is correct, but not the most efficient. From the point you’ve gotten to, you would write down the general solution for $x(t)$ in terms of the complex exponentials you’ve found. (Note that the $L=0$ case corresponds to the solutions $x=At +B$ for constants $A$ and $B$.). Plug the general solution into the first of your equations, and you will get an equation for $y(t)$.

However, it would be more efficient to do the following: write down two new equations. One is the difference between your two equations. The other is $M_c$ times the first equation plus $M_o$ times the second equation. The resulting equations are written in terms of combinations of $x$ and $y$ that are known as normal coordinates. Importantly, the resulting equations are decoupled when written in terms of these coordinates, and so they can be solved separately.

Answer 2

Just to expand on @MichaelSeifert's answer:

Let $w:=M_Cx+M_Oy$ so $w^{\prime\prime}=0$, i.e. $w$ is linear in time. Let $z:=x-y,\,\omega:=\sqrt{\frac{A(M_C+M_O)}{M_CM_O}}$ so $z^{\prime\prime}=-\omega^2z$. Solving for $w$ and $z$ (up to degrees of freedom from initial conditions), you can get $x,\,y$ from$$x=\frac{w+M_Oz}{M_C+M_O},\,y=\frac{w-M_Cz}{M_C+M_O}.$$