Analytical Mech Homework Solutions 29

Analytical Mech Homework Solutions 29 - 1 T f ( t ) = c +...

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Unformatted text preview: 1 T f ( t ) = c + ∑ ∫ 2T f ( t ) cos ( nω t ) dt cos nω t − n T 2 T 1 + ∑ ∫ 2T f ( t ) sin ( nω t ) dt sin nω t , − n T 2 1T n = ±1, ±2, . . ., and c = ∫ 2T f ( t ) dt T −2 Now, due to the equality of terms in ± n : 2 T f ( t ) = c + ∑ ∫ 2T f ( t ) cos ( nω t ) dt cos nω t − n T 2 T 2 + ∑ ∫ 2T f ( t ) sin ( nω t ) dt sin nω t , − n T 2 n = 1, 2, 3, . . . Equations 3.9.9 and 3.9.10 follow directly. 3.21 f ( t ) = ∑ cn e inω t n T= 2π ω so , 1T cn = ∫ 2T f ( t ) e− inωt dt , T −2 ω cn = 2π = π ω π − ω ∫ and n = 0, ±1, ±2, … f ( t ) e− inω t dt π ω0 − inω t ) dt + ∫0ω e−inωt dt π ( −e ∫− 2π ω ω 1 −inω t = e 2π inω 0 − π ω π 1 − inω t ω − e inω 0 1 1 − e + inπ − e− inπ + 1 2π in For n even, einπ = e − inπ = 1 and the term in brackets is zero. For n odd, einπ = e − inπ = −1 4 cn = , n = ±1, ±3, . . . 2π in 4 inω t f (t ) = ∑ e , n = ±1, ±3, . . . n 2π in 4 1 1 inω t − inω t =∑ ( e − e ) , n = 1, 3, 5, . . . n π n 2i 41 =∑ sin ( nω t ) , n = 1, 3, 5, . . . nπn 4 1 1 f ( t ) = sin ω t + sin 3ω t + sin 5ω t + … π 3 5 = 8 ...
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

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