3.1.3 Damped Free Vibration 3.1.3 Damped Free Vibration When the radical is zero,we have Cer m2+c2+k=0 c2-4mt-0 ce-plmt Define: Its roots: 1--ctve-imk 2m We obtain:c=25m Its solution: And the roots become: x(t)=Ae+Be 大01: 3.1.3 Damped Free Vibration 3.1.3 Damped Free Vibration When: When: 5=>1 Over damped The solution is: Ce The solution is: Ce x(1)=Ae-+Bte- )= No mare vibration ,、No mare vibration t t 大生 3.1.3 Damped Free Vibration 3.1.3 Damped Free Vibration (<1 When: =C<1 damped Solving for the initial conditions: The imaginaryso )9 -.-cu o.--5o Using Euler transformation: x()-eeod-giot)Dsim-go 。 2π 2015/101 移际大林Ξ 01610w14 Characteristic equation is: Its roots: Its solution: 常系数齐次线性微分方程的特 征方程 t t x t Ae Be 1 2 m c c mk 2 4 2 0 2 m c k 3.1.3 Damped Free Vibration 2015/10/13 熊海贝,同济大学土木工程学院 xionghaibei@tongji.edu.cn 19 When the radical is zero, we have Cc: Define: We obtain: And the roots become: 4 0 2 cc mk c c c c 2m 1 2 Cc 2 mk 3.1.3 Damped Free Vibration 2015/10/13 熊海贝,同济大学土木工程学院 xionghaibei@tongji.edu.cn 20 The solution is : x t x0 v0 t t x t Ae Bte 2015/10/13 熊海贝,同济大学土木工程学院 xionghaibei@tongji.edu.cn 21 3.1.3 Damped Free Vibration 1 cc c When: No mare vibration critical damped x t x0 v0 t t t x t e Ae t Be 1 1 2 2 The solution is : Over damped 2015/10/13 熊海贝,同济大学土木工程学院 xionghaibei@tongji.edu.cn 22 3.1.3 Damped Free Vibration When: 1 c c c No mare vibration The imaginary solution is: Using Euler transformation: t i t i t x t e Ae Be 2 2 1 1 xt e Ccos t D sin t t 2 2 1 1 2015/10/13 熊海贝,同济大学土木工程学院 xionghaibei@tongji.edu.cn 23 3.1.3 Damped Free Vibration 1 c c c When: damped Solving for the initial conditions: x t x0 v0 T a 1 sin t v x x t e x cos t a a a t 0 0 0 2 a 1 2 1 2 2 a Ta Damped frequency 2015/10/13 熊海贝,同济大学土木工程学院 xionghaibei@tongji.edu.cn 24 3.1.3 Damped Free Vibration