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August 13th, 2014, 03:23 PM   #1
Joined: Apr 2012

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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!

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August 30th, 2014, 12:41 PM   #2
Joined: Jun 2008

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If you have linear system of equations (and in this case you have)

$\displaystyle \begin{equation*}
\frac{dx}{dt} = ax + by\\
\frac{dy}{dt} = cx + dy\\
\end{array} \right.

the you must find eigenvalues

$\displaystyle \begin{equation*}
a-\lambda & b\\
c & d-\lambda\\
\end{array} \right| = 0


$\displaystyle \begin{equation*}
\frac{dx}{dt} = \frac{1}{m}\cdot p\\
\frac{dp}{dt} = -kx - \frac{c}{m}\cdot p\\
\end{array} \right.

$\displaystyle \begin{equation*}
-\lambda & \frac{1}{m}\\
-k & -\frac{c}{m}-\lambda\\
\end{array} \right| = 0.

$\displaystyle (-\lambda)\cdot (-\frac{c}{m}-\lambda) - (-k)\cdot \frac{1}{m}=0$

$\displaystyle \lambda^2 + \frac{c}{m}\lambda + \frac{k}{m}=0$

$\displaystyle \lambda_{1;2}=\dfrac{-\frac{c}{m}\pm\sqrt{\big(\frac{c}{m}\big)^2-4\frac{k}{m}}}{2}$

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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