2014Fall - 132 - Discussion - Week 2 - Tue 1-17
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g x = − d 2 x dt 2 . • • • x (t) = x 0 e rt r g x 0 e rt = −r 2 x 0 e rt ⇒ r 2 = − g ⇒ r = ± g ⇒ x (t) = x 0,1 e ωt + x 0,2 e − ωt = = x 0,1 (cos (ωt) + sin (ωt)) + x 0,2 (cos (ωt) − sin (ωt)) = = (x 0,1 + x 0,2 ) cos (ωt) + (x 0,1 − x 0,2 ) si n (ωt) = = A cos (ωt) + B sin (ωt) = = A 1 2 e ωt + e − ωt + B 1 2 e ωt − e − ωt = • −b dx dt + ω 2 0 x = − d 2 x dt 2 , ω 2 0 = g r x (t) = x 0 e rt r = b 2 ± b 2 4 − ω 2 0 . ∗ b 2 4 > ω 2 0 ⇒ b 2 4 − ω 2 0 , > 0, r = b 2 ± b 2 4 − ω 2 0 ∗ b 2 4 < ω 2 0 ⇒ b 2 4 − ω 2 0 < 0 r = b 2 ± ω 2 0 − b 2 4 .
Transcript of 2014Fall - 132 - Discussion - Week 2 - Tue 1-17
8/12/2019 2014Fall - 132 - Discussion - Week 2 - Tue 1-17
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