Exercise 5.7

Consider the differential equation of a harmonic oscillator given by a damped mass-spring-damper system (eq. 4, Chapter 3):

$ m\ddot{y}(t) + c\dot{y}(t) + ky(t) = 0$

Suppose we have a spring that stretches 20 cm in length caused by a weight of 98 Newton. Thus $ k = 490$ N/m and $ m = 10$ kg. The damping $ c$ is chosen such that $ c^{2} < 4mk$ holds, that is sub-critically damped.

$ 10\ddot{y}(t) + 50\dot{y}(t) + 490y(t) = 0$

This differential equation is preferably written as set of first order differential equations.
Suppose $ x_{1}(t) = y(t)$ and $ x_{2}(t) = \dot{y}(t)$ , then we get the following equations:

$ \dot{x}_{1} = x_{2}$
$ \dot{x}_{2} = -5x_{2} - 49x_{1}$
a)
Plot the solution $ y(t)$ with initial conditions $ y(t) = 0.2$ and $ \dot{y}(t) = 0$ on a time-interval that is large enough and plot this solution in the phase plane as well.
What happens if you choose other initial conditions?
b)
Plot for the super-critically damped system, $ c > 140$ , the solution. Use the same initial conditions.
c)
See if you can find initial conditions in the case of critical damping, $ c = 140$ , for which the solution at least possesses one zero (at least a crossing of the equilibrium position).



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Esteur 2010-03-22