Chapter03-page15

Chapter03-page15 - Physics 235 Chapter 3 B x (t ) = (ω 0 2...

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Unformatted text preview: Physics 235 Chapter 3 B x (t ) = (ω 0 2 − ω n 2 ) + 4ω n 2 β 2 2 sin (ω n t − δ n ) where ⎛ 2ω β ⎞ δ n = tan −1 ⎜ 2 n 2 ⎟ ⎝ ω0 − ωn ⎠ The solution to the third equation is x= 1 2ω 0 2 The solution of the following differential equation + 2 β x + ω 0 2 x = x ∞ 1 a0 + ∑ ( an cos nω t + bn sin nω t ) 2 n =1 is thus equal to x (t ) = ∞ a0 +∑ 2ω 0 2 n =1 1 (ω 2 0 − ωn ) 22 + 4ω n β 2 2 (a n cos (ω n t − δ n ) + bn sin (ω n t − δ n )) where ⎛ 2ω β ⎞ δ n = tan −1 ⎜ 2 n 2 ⎟ ⎝ ω0 − ωn ⎠ and ω n = nω - 15 - ...
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This note was uploaded on 01/08/2012 for the course PHY 235 taught by Professor Morgan during the Spring '09 term at SUNY Stony Brook.

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