0kE=-(k+m02)A22mmAverage<E, >=<E, >=Ik'A(conservative system)4How to judge?O f =-kx Or v(x) is parabola. f = -kx+CE=Asin wt coswtF+WE=01Conservative system. not only3= = Acos(wt +p)Example:Finding , A And @
2 2 4 1 E p = E k = k A 2 2 ( ) 4 1 E = k + m A 2 2 1 = kA m k = 2 (conservative system) Average: How to judge? f = −kx Or v(x) is parabola. f = −kx +C 0 2 + = •• w = Asin wt coswt Conservative system. not only cos( ) = A wt +0 Example: Finding , A And .
Mkkm0M+mVimyX=0M+mymVm'v?(m+M)A三X.+k/k(M +m)W2(M+m)"x= Acosβ. = 0t=0V= -OAsin Po = Vo <0元元D十P二22Kvm元cOs(X:t+二2M+mk(M+m)
k m v1 M X=0 M m k + = M m mv v + = 1 0 k m M M m m v w v A x ( ) ( ) 2 2 1 2 2 2 0 + + = + = ( ) 1 k M m v m + = t = 0 x = Acos0 = 0 sin 0 v = −A 0 = v0 ) 2 cos( ( ) 0 + + + = t M m k k M m v m x 2 0 = 2 0 =
1-3 The damped harmonic oscillatorFriction harmonicf=-W+05=0mx=-kx1resistance coefficientmmx=(-w-kx)M#+二X+00x=0mx+2βx+0gx=02β=damping constantmSolution:x = Ae-βt cos(o't + po)(1). 00 > β
1-3 The damped harmonic oscillator f = −v 0 2 + 0 = •• m x = −kx •• 0 2 0 . + + = •• x x m x 2 0 2 0 . + + = •• x x x m x = (− v − kx) •• Friction harmonic resistance coefficient m 2 = damping constant. Solution: (1).0 cos( ) 0 ' = + − x Ae t t m
A , ?, are determined by initial2元@' = 0o -β2 slow down. Quasi-period:SAD店Lightly damped:三<1 00 ~ 0000Heavily damped00 >>0
A ' 2 0 ' = − ' 2 T = 1 0 0 0 slow down. Lightly damped: Quasi-period: , are determined by initial. ' 0 Heavily damped: S 0 t S t 0
+C2e-(β+/p?-0)0x=Cle-(β-B?-)(2). β>00 SA*C , C, initial condition0十(3). β =Wo x=(C +C,t)e-*fasterSA* critical dampingSummary :ODynamical e.q0.<β0.-βQuasi-period のo>t0.>β0@ Solution behavior
0 t t x C e C e ( ) 2 ( ) 1 2 0 2 2 0 2 − − − − + − = + C 1 C 2 , initial condition. (2). S t 0 (3). =0 t x C C t e − = ( + ) 1 2 faster critical damping Summary : Dynamical e.q. Quasi-period ' 0 = S t Solution behavior