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Unformatted text preview: Homework 6: mv + cv + kv = p 0 sin t The solution of the diferantial equation for the given system has two parts. These are homogeneous solution and particular solution. v c (t ) = e wt ( A. sin D t + B. cos D t )
v p (t ) = E. sin t + F . cost : : Homogeneous Solution Particular Solution Determine the constants (E and F) of the particular solution. Solution: v p (t ) = E. sin t + F . cost v p (t ) = E. cost  F . sin t v p (t ) =  2 E. sin t  2 F . cost mv + cv + kv = p 0 sin t v + p p c k v + v = 0 sin t v + 2v + 2 v = 0 sin t m m m m ( 2 E. sin t  2 F . cost ) + 2(E. cost  F . sin t ) + 2 ( E. sin t + F . cost ) = p0 sin t m p0 sin t m
(1) ( 2 E  2 + 2 E ) sin t + ( 2 F + 2 + 2 F ) cost = F E
From equation 1 we can get;  2 E  2 + 2 E = F p0 m (2) (3)  2 F + 2 + 2 F = 0 E
By using equation 2 we can write F in terms of E. F= 2 E 2 2 (4) 1 If we put F in equation 1 we get;  2 E  2 p 2 E + 2E = 0 2 2 m  (5) We know that m = k / 2 . If we put this expression in equation 5, we can get; 2 (2 ) 2 p 2 E 2  2  2 + 1 = 0 k 2 We know that = / . If we put this expression in equation 6, we can get; (6) (2 ) 2 p 0 E  2  + 1 = 1 2 k
and also we can obtain the expression of E. E= p0 1 2 k (1  2 ) 2 + (2 ) 2 (7) We know that;
F= 2 E 2 2 2 2 2 2 F= 2 E F = 2 2 E F =  E 2  1 2  2 2 If we put equation 7 into the expression of F that is written above; F = p 2 p 0 1 2 2 F = 0 2 2 2 2 2 2 k (1  ) + (2 ) 2 1  k (1  ) + (2 ) we can obtain the expression of F. So vp(t) can be written as;
p0 p 1 2 2 v p (t ) = . sin t  0 . cost 2 2 2 2 2 k (1  ) + (2 ) k (1  ) + ( 2 ) 2
v p (t ) = p k 1  2 sin t  ( 2 ) cost 2 (1  ) + (2 )
0 2 2 [( ) 2 ...
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This note was uploaded on 11/23/2010 for the course DEP 503E 10852 taught by Professor Prof.zekaicelep during the Spring '10 term at Istanbul Technical University.
 Spring '10
 Prof.ZekaiCELEP

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