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 August 13th, 2014, 03:23 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, 12:41 PM #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 Tags critical, finding, points, types Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post allylee Calculus 1 October 21st, 2012 11:26 PM Timk Calculus 3 November 29th, 2011 11:59 AM maximus101 Calculus 1 March 11th, 2011 08:26 AM Stud778 Calculus 4 June 17th, 2010 11:13 PM stainsoftime Calculus 3 November 24th, 2008 05:24 AM

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