Equations of Motion for the Nonlinear Oscillator (2DOF)

Equations of Motion for the Nonlinear Oscillator (2DOF) https://www.youtube.com/watch?v=l4XQtlkyC2o
HK
https://en.wikipedia.org/wiki/Deformation_(physics)
           

初始品質點P的位置為坐标原點，移動後點P坐标 ( x 1 , x 2 ) (x_1,x_2) (x1​,x2​)

彈簧1長為 l 1 l_1 l1​，剛度 k 1 k_1 k1​，彈簧2長為 l 2 l_2 l2​，剛度 k 2 k_2 k2​

彈簧1的固定點P1坐标 ( − l 1 , 0 ) (-l_1,0) (−l1​,0)，彈簧2的固定點P2坐标 0 , − l 2 ) 0,-l_2) 0,−l2​)

彈簧1的變形量 Δ 1 = ∣ P − P 1 ∣ − l 1 \Delta_1 = |P-P_1|-l_1 Δ1​=∣P−P1​∣−l1​

彈簧2的變形量 Δ 2 = ∣ P − P 2 ∣ − l 2 \Delta_2 = |P-P_2|-l_2 Δ2​=∣P−P2​∣−l2​

彈簧1的應變能 V 1 = 1 / 2 K 1 Δ 1 2 V_1 = 1/2K_1\Delta_1^2 V1​=1/2K1​Δ12​

彈簧2的應變能 V 2 = 1 / 2 K 2 Δ 2 2 V_2 = 1/2K_2\Delta_2^2 V2​=1/2K2​Δ22​

品質點的動能 T = 1 / 2 m ( x 1 ′ 2 + x 2 ′ 2 ) T = 1/2m(x_1'^2 + x_2'^2) T=1/2m(x1′2​+x2′2​)

拉格朗日方程

d d t ( ∂ L ∂ q i ′ ) − ∂ L ∂ q i = F \frac{d}{dt}(\frac{\partial L}{\partial q_i'}) - \frac{\partial L}{\partial q_i} = F dtd​(∂qi′​∂L​)−∂qi​∂L​=F

拉格朗日量 L = T − V = T − ( V 1 + V 2 ) L=T-V =T-(V_1 + V_2) L=T−V=T−(V1​+V2​)

此例中無外力，帶入拉格朗日量，得

d d t ( ∂ T ∂ q i ′ ) + ∂ V 1 ∂ q i + ∂ V 2 ∂ q i = 0 \frac{d}{dt}(\frac{\partial T}{\partial q_i'}) + \frac{\partial V_1}{\partial q_i} + \frac{\partial V_2}{\partial q_i} = 0 dtd​(∂qi′​∂T​)+∂qi​∂V1​​+∂qi​∂V2​​=0

列出兩個方程

d d t ( ∂ T ∂ x 1 ) + ∂ V 1 ∂ x 1 + ∂ V 2 ∂ x 1 = 0 \frac{d}{dt}(\frac{\partial T}{\partial x_1}) + \frac{\partial V_1}{\partial x_1} + \frac{\partial V_2}{\partial x_1} = 0 dtd​(∂x1​∂T​)+∂x1​∂V1​​+∂x1​∂V2​​=0

d d t ( ∂ T ∂ x 2 ) + ∂ V 1 ∂ x 2 + ∂ V 2 ∂ x 2 = 0 \frac{d}{dt}(\frac{\partial T}{\partial x_2}) + \frac{\partial V_1}{\partial x_2} + \frac{\partial V_2}{\partial x_2}= 0 dtd​(∂x2​∂T​)+∂x2​∂V1​​+∂x2​∂V2​​=0

得出

m x 1 ′ ′ + α 1 k 1 ( x 1 + l 1 ) + α 2 k 2 x 1 = 0 mx_1'' + \alpha_1k_1(x_1+l_1) + \alpha_2 k_2 x_1 = 0 mx1′′​+α1​k1​(x1​+l1​)+α2​k2​x1​=0

m x 2 ′ ′ + α 1 k 1 x 1 + α 2 k 2 ( x 2 + l 2 ) = 0 mx_2'' + \alpha_1k_1x_1 + \alpha_2 k_2(x_2+l_2) = 0 mx2′′​+α1​k1​x1​+α2​k2​(x2​+l2​)=0

α 1 = 1 − l 1 / ∣ P − P 1 ∣ , α 2 = 1 − l 2 / ∣ P − P 2 ∣ \alpha_1 = 1-l_1/|P - P_1|, \alpha_2 = 1-l_2/|P - P_2| α1​=1−l1​/∣P−P1​∣,α2​=1−l2​/∣P−P2​∣

α 1 = 1 − 1 / ϵ 1 , α 2 = 1 − 1 / ϵ 2 \alpha_1 = 1 - 1/\epsilon_1 , \alpha_2 = 1 - 1/\epsilon_2 α1​=1−1/ϵ1​,α2​=1−1/ϵ2​

α \alpha α：變形後的伸長量與變形後的長度之比

工程應變： ϵ \epsilon ϵ，變形後的伸長量與變形前的長度之比

更簡單的方式得出動力學方程：

F ⃗ 1 = − k 1 Δ 1 e ⃗ 1 , e ⃗ 1 = ( P − P 1 ) / ∣ P − P 1 ∣ \vec{F}_1 = -k_1 \Delta_1 \vec{e}_1 , \vec{e}_1 = (P - P_1)/|P - P_1| F

1​=−k1​Δ1​e

1​,e

1​=(P−P1​)/∣P−P1​∣

F ⃗ 2 = − k 2 Δ 2 e ⃗ 2 , e ⃗ 2 = ( P − P 2 ) / ∣ P − P 2 ∣ \vec{F}_2 = -k_2 \Delta_2 \vec{e}_2 , \vec{e}_2 = (P - P_2)/|P - P_2| F

2​=−k2​Δ2​e

2​,e

2​=(P−P2​)/∣P−P2​∣

m x 1 ′ ′ = ( F ⃗ 1 + F ⃗ 2 ) e ⃗ 1 mx_1'' = (\vec{F}_1 + \vec{F}_2)\vec{e}_1 mx1′′​=(F

1​+F

2​)e

1​

m x 2 ′ ′ = ( F ⃗ 1 + F ⃗ 2 ) e ⃗ 2 mx_2'' = (\vec{F}_1 + \vec{F}_2)\vec{e}_2 mx2′′​=(F

1​+F

2​)e

2​