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Practice Final Problem 1 solution

# Practice Final Problem 1 solution - 3 2 3 3 cos 2 t Ae t φ...

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(Fall 2000 ) A system is modeled by the differential equations 3 6 dp q dt = - 3 3 dq q p dt = - - Determine the values and the names of whichever of the following parameters that apply: n ω , d ω , ς , τ . First use operator notation to get, 3 6 sp q = - and 3 3 sq q p = - - . We need the characteristic equation. This can be found by solving for either of the state variables ( p , q ). This in turn can be done by numerous methods. I find writing a matrix form of the two expressions, that is 3 6 3 3 0 s p s q - -     =     +     followed by setting the determinant of the left hand of the matrix expression equal to zero, the most expedient. Thus ( ) ( )( ) 3 3 3 0 s s + - - = which simplifies to 2 3 9 0 s s + + = . At this point on could either use the quadratic formula to find 2 3 3 4 9 3 9 36 3 3 3 2 2 2 j s - ± - - ± - - ± = = = i or completing the squares to find 2 2 2 2 2 2 3 3 3 27 3 3 3 3 9 9 2 2 2 4 2 2 s s s s s    + + = + + - = + + = + +       . In either case a
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Unformatted text preview: 3 2 3 3 cos 2 t Ae t φ- + is indicated. As such 2 the characteristic time, 3 = and 3 3 the damped natural frequency, 2 d = are indicated. Alternatively one could realize that 2 3 9 s s + + = is of the form 2 2 2 n n s s ςω + + = . Equating the last terms in the two expressions gives the undamped natural frequency, 3 n = while equating the middle terms gives 2 3 n = . Note that 2 2/ n = and thus, upon substitution 2 the characteristic time, 3 = could be found using this alternative procedure. Rearranging terms and substituting gives, 3 3 2 2 3 n = = i . Thus 1 the percent critical damping is, 2 = . Utilizing this latter information and the expression 2 1 d n =-also leads to 3 3 the damped natural frequency, 2 d = ....
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