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# ADS Homework 6 - Homework 6 mv cv kv = p 0 sin t The...

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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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ADS Homework 6 - Homework 6 mv cv kv = p 0 sin t The...

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