MIT18_03S10_c12 - 18.03 Class 12 March 1 2010 Good...

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Good Vibrations [0] Summary table [on the blackboard] [1] Overdamping [2] Underdamping [3] Real solution theorem [4] Natural and damped circular frequencies [5] Critical damping [6] Transience [7] Root diagram [0] Summary table of unforced system responses. Name* m,b,k relation Char. roots Exp. sol's Basic real solns Overdamped b^2/4m > k Two diff. real e^{r1 t}, e^{r2 t} same roots, r1, r2 Critically b^2/4m = k Repeated root e^{rt} e^{rt}, te^{rt} damped r = - b/2m Underdamped b^2/4m < k -b/2m +- iw e^{r1 t}, e^{r2 t} e^{-bt/2m}cos(w t) e^{-bt/2m}sin(w t) ( w = omega_d = sqrt{ (k/m) - (b/2m)^2 } ) * The names here are appropriate under the assumption that b and k are both non-negative. The rest of the table makes sense in general, but it doesn't have a good interpretation in terms of a mechanical system. We are studying equations of the form mx" + bx' + kx = 0 (*) which model a mass, dashpot, spring system without external forcing term. Homogeneous constant coefficient linear second order. We found that (*) has an exponential solution e^{rt} exactly when r is a root of the "characteristic polynomial" p(s) = ms^2 + bs + k . These are called the MODES of the system. [1] Example. [Overdamped: Distinct real roots] x" + 5x' + 4x = 0 . We did this on Friday: The characteristic polynomial s^2 + 5s + 4 factors as (s + 1)(s + 4) so the roots are r = -1 and r = -4 . The corresponding exponential solutions are e^{-t} and e^{-4t} . These are called *modes* of the system. They represent pure states. The general solution is mixture of the two states, x = c_1 e^{-t} + c_2 e^{-4t} . But where's the vibration?
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This note was uploaded on 10/21/2011 for the course MATH 18.03 taught by Professor Vogan during the Spring '09 term at MIT.

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MIT18_03S10_c12 - 18.03 Class 12 March 1 2010 Good...

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