SolSec3.1 - 3Chapter Three Solutions Problem and Solutions...

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3- 1 Chapter Three Solutions Problem and Solutions for Section 3.1 (3.1 through 3.11) 3.1 Calculate the solution to ˙˙ ˙ ˙ xxx t xx ++= () () = () = 22 01 00 δπ and plot the response. Solution: Given: ( ) ( ) 0 0 , 1 0 2 2 = = = + + x x t x x x & & & & π δ ωζ ω ζ n d n k m c km == = = = = 1 414 2 0 7071 1 1 2 .. , rad/s, rad/s Total Solution: t x t x t x p h + = Homogeneous: From Equation (1.36) xt A e t A vx x x xt e t h t d n d d d n h t n =+ = + + = + = ζω ωφ φ sin , tan . . sin . 2 0 2 2 1 0 785 1 414 785 rad Particular: From Equation. (3.9) m et e t tt x t p d t d t p t n =− = ()=− ()=− ⇒ −− 11 ωτ τ sin sin sin sin But, sin So, e t t e t e t t t << −> 1 414 0 785 0 1 414 0 785 . sin . . sin . sin This is plotted below using the Heaviside function.
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3- 2 3.2 Calculate the solution to () 1 0 0 0 sin 3 2 = = + = + + x x t t x x x & & & & π δ and plot the response. Solution: Given: ˙˙ ˙ sin , , ˙ xxx t t x x ++= + ( ) = () = 23 00 δπ ωζ ω ζ n d n k m c km == = = = = 1 732 2 0 5774 1 1 414 2 .. , . rad/s, rad/s Total Solution: xt x x t x x x t h p h pp =+ < < + > 1 12 0 Homogeneous: Eq. (1.36) xt A e t A e t h t d t n ζω ωφ φ sin sin . 1 414 Particular: #1 (Chapter 2) xt X t f F m X f t p nn n n p 1 0 0 22 2 2 1 1 11 2 0 3536 2 0 785 0 3536 0 7854 ( ) sin . , tan . . sin . =− = ⇒= + = ωθ ωω θ ζω ω , where rad/s . Note that and rad 0 Particular: #2 Equation 3.9 m et e t p d t d t p t n 2 2 1 1 414 1 414 0 7071 1 414 = ( ) −− ωτ sin . sin . . sin . The total solution for 0< t < π becomes: Ae t t x t Ae t Ae t t xA A A t tt +− ()=− + + + + sin . . sin . ˙ sin( . ) . cos . . cos . sin . . sin ˙ sin . 1 414 0 3536 0 7854 1 414 1 414 1 414 0 3536 0 7854 0 2 5 025 0 1 1 414 φφ cos . . . . . tan +⇒=− () = 075 025 1414025 1 034 075 and A Thus for the first time interval, the response is e t t t t << 0 75 1 414 0 34 0 3536 0 7854 0 . sin . . . sin . Next consider the application of the impulse at t = π : x x x e t t e t t h t t + ()=− + ++ > 0 433 1 414 0 6155 0 3536 0 7854 0 7071 1 414 . sin . . . sin . . sin . ππ The response is plotted in the following (from Mathcad):
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3- 3 3.3 Calculate the impulse response function for a critically damped system. Solution: The change in the velocity from an impulse is m F v ˆ 0 = , while x 0 = 0. So for a critically damped system, we have from Eqs. 1.45 and 1.46 with x 0 = 0: xt vte F m te n n t t () ˆ = ⇒= 0 ω
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3- 4 3.4 Calculate the impulse response of an overdamped system.
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This note was uploaded on 02/11/2010 for the course MECHANICAL ms316 taught by Professor Abduljaba during the Spring '09 term at Kalamazoo.

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SolSec3.1 - 3Chapter Three Solutions Problem and Solutions...

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