Author Topic: Problem of the week 4b  (Read 8593 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2607
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Problem of the week 4b
« on: January 31, 2013, 05:50:11 PM »
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

  • Full Member
  • ***
  • Posts: 16
  • Karma: 10
    • View Profile
Re: Problem of the week 4b
« Reply #1 on: January 31, 2013, 10:45:31 PM »
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

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

expand and use quadratic equation
$$
r^2 = K - L \pm L
$$

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

Brian Bi

  • Moderator
  • Full Member
  • *****
  • Posts: 31
  • Karma: 13
    • View Profile
Re: Problem of the week 4b
« Reply #2 on: February 01, 2013, 12:13:33 AM »
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 $\omega_2 = \sqrt{K+2L}$.