MAT244-2013S > Ch 3

Problem of the week 4b

(1/1)

Victor Ivrii:
Consider two identical and connected harmonic oscillators:
\begin{equation}
\left\{\begin{aligned}
&y''+K y + L (y-z)=0\\
&z''+Kz + L(z-y)=0
\end{aligned}\right.
\label{eq-1}
\end{equation}
with $K>0$, $L>0$.

Even if this is a system one can add or subtract equations getting system of two equations describing $y+z$ and $y-z$ separately.

1) Find $y+z$ and $y-z$ and then $y$ and $z$  (so, find the general solution of (\ef{eq-1}).

2) What frequencies has the described system?

Changyu Li:
2)
guess $y = A e^{rt}$, $z = B e^{rt}$
$$A r^2 + K A + L\left(A-B\right) = 0 \\ B r^2 + K B + L\left(B-A\right) = 0 \\ \left( \begin{array}{cc} r^2 + K + L & - L \\ -L & r^2 + K + L \\ \end{array} \right) \left( \begin{array}{c} A \\ B \end{array} \right) = 0$$

nontrivial solution exists if there is no inverse to the ugly matrix, therefore its determinant is 0

$$\left( r^2 + K + L \right)^2 + L^2 = 0$$

$$r^2 = K - L \pm L$$

therefore the frequencies are $\pm K$, $\pm\left(K-2L\right)$

Brian Bi:
Add and subtract the first and second equations to obtain:
\begin{align}
(y+z)'' + K(y+z) &= 0 \label{added} \\
(y-z)'' + (K+2L)(y-z) &= 0 \label{subtracted}
\end{align}
Since $K, L > 0$, both equations are of the form $u'' + \omega^2 u = 0$, with general solution $u = A \cos (\omega t) + B \sin (\omega t)$. So the general solution to $(\ref{added})$ is
\begin{equation}
y+z = A \cos (\sqrt{K} t) + B \sin (\sqrt{K} t)
\end{equation}
and the general solution to $(\ref{subtracted})$ is
\begin{equation}
y - z = C \cos (\sqrt{K+2L} t) + D \sin(\sqrt{K+2L}t)
\end{equation}
Using the identities $y = \frac{1}{2}((y+z)+(y-z))$ and $z = \frac{1}{2}((y+z)-(y-z))$ we obtain the general solution to $(\ref{eq-1})$:
\begin{align}
y &= A' \cos(\omega_1 t) + B' \sin(\omega_1 t) + C' \cos(\omega_2 t) + D' \sin(\omega_2 t) \\
z &= A' \cos(\omega_1 t) + B' \sin(\omega_1 t) - C' \cos(\omega_2 t) - D' \sin(\omega_2 t)
\end{align}
where $A' = A/2$ and so on, and the frequencies are $\omega_1 = \sqrt{K}$, and $\omega_2 = \sqrt{K+2L}$.