By Fowles G.R., Cassiday G.L.

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0032 A n = 0, ±1, ±2, . . n f ( t ) = ∑ cn cos nω t + ∑ cni sin nω t , n = 0, ±1, ±2, . . n T 2 T − 2 T 2 T − 2 n and cn = 1 T∫ f ( t ) e − inωt dt , cn = 1 T∫ f ( t ) cos ( nω t ) dt − n = 0, ±1, ±2, . . i T T 2 T − 2 ∫ f ( t ) sin ( nω t ) dt The first term on cn is the same for n and − n ; the second term changes sign for n vs. − n . The same holds true for the trigonometric terms in f ( t ) . Therefore, when terms that cancel in the summations are discarded: 28 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 1 T2 n = ±1, ±2, .

0032 A n = 0, ±1, ±2, . . n f ( t ) = ∑ cn cos nω t + ∑ cni sin nω t , n = 0, ±1, ±2, . . n T 2 T − 2 T 2 T − 2 n and cn = 1 T∫ f ( t ) e − inωt dt , cn = 1 T∫ f ( t ) cos ( nω t ) dt − n = 0, ±1, ±2, . . i T T 2 T − 2 ∫ f ( t ) sin ( nω t ) dt The first term on cn is the same for n and − n ; the second term changes sign for n vs. − n . The same holds true for the trigonometric terms in f ( t ) . Therefore, when terms that cancel in the summations are discarded: 28 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 1 T2 n = ±1, ±2, .

X = Ae β t −iφ Assuming a solution of the form: ( mβ 2 + cβ + k ) x = FA xeiφ F mα 2 − 2imαω − mω 2 − cα + icω + k = ( cos φ + i sin φ ) A F m (α 2 − ω 2 ) − cα + k = cos φ A F ω ( −2mα + c ) = sin φ A 27 φ = tan −1 ω ( c − 2mα ) m (α 2 − ω 2 ) − cα + k Using sin 2 φ + cos 2 φ = 1 , 2 F2 2 = m (α 2 − ω 2 ) − cα + k + ω 2 ( c − 2mα ) 2 A F A= {m (α − ω ) − cα + k + ω ( c − 2mα ) } 2 2 2 2 1 2 2 and x ( t ) = Ae −α t cos (ω t − φ ) + the transient term. 084 T2 l 4π 2l gives g = 2 , approximately 8% too small.

### Analytical mechanics, 7ed., Solutions manual by Fowles G.R., Cassiday G.L.

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