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Unformatted text preview: k = 40, and the external force F ( t ) = cos 2 t into the equation (23) on page 226 in the text yields 8 y + 3 y + 40 y = cos 2 t. The steadystate (a particular) solution to this equation is given in (6) and (7), page 221, that is, y p ( t ) = F ( k m 2 ) 2 + b 2 2 ( k m 2 ) cos t + b sin t = 1 [40 (8)(2) 2 ] 2 + (3) 2 (2) 2 ( 40 8(2) 2 ) cos 2 t + (3)(2) sin 2 t = 1 100 { 8 cos 2 t + 6 sin 2 t } = 1 10 sin(2 t + ) , where = arctan(8 / 6) . 927 . 244...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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