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 August 13th, 2014, 02:23 PM #1 Member   Joined: Apr 2012 Posts: 46 Thanks: 0 Finding different types of critical points Hi all, Really struggling with this critical point question. The oscillations of a mass on a spring are shown by the system: m.(dx(t)/dt) = p(t) (dp(t)/dt) = -kx(t) - (c/m).p(t) where m is mass, k is spring constant and c is damping constant and are all non negative (with m >0 ). In the case where c>0, show that there are three possible types of critical points at the origin and then find the values of c in terms of k and m relating to each of these cases. Anything I do just ends up making no sense at all. Help please! Thanks!
 August 30th, 2014, 11:41 AM #2 Member   Joined: Jun 2008 Posts: 45 Thanks: 0 If you have linear system of equations (and in this case you have) $\displaystyle \begin{equation*} \left\{ \begin{array}{l} \frac{dx}{dt} = ax + by\\ \frac{dy}{dt} = cx + dy\\ \end{array} \right. \end{equation*}$ the you must find eigenvalues $\displaystyle \begin{equation*} \left| \begin{array}{cc} a-\lambda & b\\ c & d-\lambda\\ \end{array} \right| = 0 \end{equation*}$ So $\displaystyle \begin{equation*} \left\{ \begin{array}{l} \frac{dx}{dt} = \frac{1}{m}\cdot p\\ \frac{dp}{dt} = -kx - \frac{c}{m}\cdot p\\ \end{array} \right. \end{equation*}$ $\displaystyle \begin{equation*} \left| \begin{array}{cc} -\lambda & \frac{1}{m}\\ -k & -\frac{c}{m}-\lambda\\ \end{array} \right| = 0. \end{equation*}$ $\displaystyle (-\lambda)\cdot (-\frac{c}{m}-\lambda) - (-k)\cdot \frac{1}{m}=0$ $\displaystyle \lambda^2 + \frac{c}{m}\lambda + \frac{k}{m}=0$ $\displaystyle \lambda_{1;2}=\dfrac{-\frac{c}{m}\pm\sqrt{\big(\frac{c}{m}\big)^2-4\frac{k}{m}}}{2}$ Now you are given that m > 0 and c > 0. If (c / m)^2 - 4k / m > 0 or c^2 > 4km then both $\displaystyle \lambda_1$ and $\displaystyle \lambda_2$ are negative and you have node. If (c / m)^2 - 4k / m = 0 or c^2 = 4km then $\displaystyle \lambda_1=\lambda_2\neq 0$ and you have improper node. If (c / m)^2 - 4k / m < 0 or c^2 < 4km then both $\displaystyle \lambda_1$ and $\displaystyle \lambda_2$ are non-real and you have spiral. See this for good explanation: Differential Equations: Qualitative Methods

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