Topics In Linear Algebra and Its Applications - STEM2

T = 1 2 m 1 ˙u 2 1 + 1 2 m 2 ˙u 2 2and the potential energy isThe Lagrangian isV = 1 2 k 1u 2 1 + 1 2 k 2(u 2 − u 1 ) 2 + 1 2 k 3u 2 2.L = T − V.The equations of motion are given by Lagrange’s equationsandWe haveandAlsoandThus Lagrange’s equations ared ∂L− ∂L = 0dt ∂ ˙u 1 ∂u 1d ∂L− ∂L = 0.dt ∂ ˙u 2 ∂u 2∂L∂ ˙u 1= ∂T∂ ˙u 1= m 1 ˙u 1∂L= ∂T = m 2 ˙u 2 .∂ ˙u 2 ∂ ˙u 2− ∂L∂u 1= ∂V∂u 1= k 1 u 1 − k 2 (u 2 − u 1 )− ∂L∂u 2= ∂V∂u 2= k 2 (u 2 − u 1 ) + k 3 u 2 .m 1 ü 1 + k 1 u 1 − k 2 (u 2 − u 1 ) = 0andm 2 ü 2 + k 2 (u 2 − u 1 ) + k 3 u 2 = 0.30