example_IV_2_03

# example_IV_2_03 - Example V.2.3 Find the particular...

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Example V.2.3 Find the particular solution of the EOM for the system shown below. Let ˙ ˙ y t ( ) = g t ( ) where g(t) is a general T-periodic function of time. SOLUTION EOMs': m 0 0 m ˙ ˙ x 1 ˙ ˙ x 1 + 2 k k k k x 1 x 1 = ky t ( ) 0 = k 0 g t ( ) = f g t ( ) Particular solution Using equation (10): x p t ( ) = g 0 K [ ] 1 f + H j Ω ( ) [ ] g cj cos j Ω t + g sj sin j Ω t ( ) j = 1 f where H j Ω ( ) [ ] = j Ω ( ) 2 M [ ] + K [ ] [ ] 1 = j Ω ( ) 2 m + 2 k k k j Ω ( ) 2 m + k 1 = 1 Δ j j Ω ( ) 2 m + k k k j Ω ( ) 2 m + 2 k y(t) x 1 (t) x 2 (t)

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Δ j = j Ω ( ) 2 m + k ( ) j Ω ( ) 2 m + 2 k ( ) k 2 K [ ] 1 = 2 k k k k 1 = 1 k 1 1 1 2 f = k 1 0 Therefore: x p t ( ) = g 0 1 1 1 2 1 0 k Δ
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## This document was uploaded on 12/23/2011.

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example_IV_2_03 - Example V.2.3 Find the particular...

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