Ch3-2(3-4)c

# Ch3-2(3-4)c - 1 MAE351 Mechanical Vibrations Lecture 12...

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1 1 MAE351 Mechanical Vibrations Lecture 12 (Chap. 3.3-3.4) 2 3.3 Periodic Forcing Functions We know that periodic ) ( ) ( where ) ( 2 2 T t F t F t F x x x n n  We know that periodic functions can be represented by a series of sines and cosines (Fourier) or a series of e j ω t and e -j ω t (complex Fourier) - curve fitting -0.5 0 0.5 1 1.5 2 splacement x(t) T Response is superposition of as many RHS terms as you think that are necessary to represent the forcing function accurately 0 2 4 6 -2 -1.5 -1 Time (s) Dis

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2 3 Fourier Series Expansion Assume F ( t ) a 0 2 a n cos n t b n sin n t  n 1 where a 0 2 T F ( t ) dt 0 T T a n 2 T F ( t ) cos n td t 0 b n 2 T F ( t ) sin n t 0 T 4 Fourier Series: Saw Tooth Function 2 sin 2 sin3 sin 4 [sin ] 234 ttt tt   0 1 1 2s i n ( ) ( 1) N n N nt S Partial Sums: 1 01 -1 -1 -1 -1 0 1 1 1 n n 0 1 0 -1 -1 0 1 -1 -1 0
3 5 Fourier Series: Example F(t) F 0 Step 1: find the F.S. t 1 t 2 =T 1 , 0 t t and determine how many terms you need 0  2 1 1 1 2 0 , ) ( t t t t t t t F t F 6 Fourier Series: Example 1 1.2 F(t) 0.2 0.4 0.6 0.8 Force F(t) 2 coefficients 10 coefficients 100 coefficients 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 -0.2 0 Time (s)

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4 7 Solution as a series of sines & cosines ) ( 2 2 t F x x x n n  The solution can be written as a summation   0 1 0 2 00 0 2 () where ( ) is a solution to
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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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Ch3-2(3-4)c - 1 MAE351 Mechanical Vibrations Lecture 12...

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