Ch3-1(1-2)c - 1 MAE351 Mechanical Vibrations Lecture 11...

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1 1 MAE351 Mechanical Vibrations Lecture 11 (Chap. 3.1-3.2) 2 Types of Input Signals (Forces) (adapted from B&K Technical Note BA767412)
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2 3 Chap. 3. General Forced Response: Linear superposition 12 1 1 2 2 If , are both solutions, then is a solution; xx x ax ax 2 11 2 22 2 2 2 If is a particular sol'n of and is a particular sol'n of , solves n n n x f x f ax x x af b f    : 2 2 dd a+ b c dt dt L Linear operator Nonlinear operator 2 : d bc x dt    N 4 3.1 Impulse Response Function F Impulse excitation ( t ) 0 2 F i.e. area under pulse 0 0 impulse = ( ) 1 F t dt F dt F    + - is a small positive number. 2
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3 5 Properties of impulse function: Dirac delta, δ (t) F Equal Dirac delta ( t ) impulses delta function 0 00 () 0 , Ft F t t F td t F t F         t 6 Recall: Transient Response (Ch. 1) (Free Vibration)     sin n t d xt A e t   m x(t) 0 F A 1 d (v 0 n x 0 ) 2 (x 0 d ) 2 tan 1 x 0 d v 0 n x 0 k c 0 0, 0 : 0. d If x but v v and A  
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4 7 Impulse = Initial Velocity Just after impulse just before impulse Impulse = Momentum change 00 0 () [ ( ) ( ) ] Ft F mv m v t v t   05 1 m x(t) 0 F 0 0 s in n t d d F x te t F h t m  0 10 20 30 40 -1 -0.5 0 0.5 Time h(t) k c 8 0 sin n t d t ht e tt     Impulse Response Function d m 0 10 20 30 40 -1 0 1 h1 0 1
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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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Ch3-1(1-2)c - 1 MAE351 Mechanical Vibrations Lecture 11...

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