This preview shows page 1. Sign up to view the full content.
Unformatted text preview: − B cos 0 = 24 − B ⇒ B = 8 . Therefore, we see that M ( t ) = 24 − 8 cos ωt, where ω = π/ 12. As in Example 2, using the fact that B = M + H /K = M + 0 /K = M , we see that T ( t ) = 24 − 8 F ( t ) + Ce − Kt , where F ( t ) = cos ωt + ( ω/K ) sin ωt 1 + ( ω/K ) 2 = ± 1 + ² ω K ³ 2 ´ − 1 / 2 cos( ωt − α ) . In the last expression, α is chosen such that tan α = ω/K . By assuming that the exponential term dies o±, we obtain T ( t ) = 24 − 8 ± 1 + ² ω K ³ 2 ´ − 1 / 2 cos( ωt − α ) . This function will reach a minimum when cos( ωt − α ) = 1 and it will reach a maximum when cos( ωt − α ) = − 1. 120...
View
Full
Document
This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

Click to edit the document details