Tugas Komputasi Dasar

Heri suhud K TIKM 2009/2010 HOMEWORK Find the shape s of the x(t) and the current i(t) 1. Mechanical system m d2x + b dx + kx =0 dt2 dt put x(t) = est substitution to differential equation we have : m s2 est + b s est + k est =0 (m s2 + b s + k ) est

=0

Equation m s2 + b s + k =0 Use abc formula : s12 =-b±b2-4mk2m There are 3 shape cases: Case 1 : when b2>4mk ; we have two roots s1 and s2 : s1 = -b+b2-4mk2m s2

=-b-b2-4mk2m

we plot at graph

Case 2 when b2=4mk ; we have same roots s1 = s2 : s1 = s2 = -b2m so; x(t) = e-b.t2m

, we plot at graph

Case 3 when b2<4mk ; we have complex roots s1 = -b+ib2-4mk2m s2

=-b-ib2-4mk2m

2. Current I(t) L d2I + R dI + 1Cx =0 dt2 dt put I(t) = est substitution to differential equation we have : L s2 est + R s est + 1Cest =0 (L s2 + R s + 1C ) est

=0

Equation L s2 + R s + 1C =0 Use abc formula : s12 =-R±R2-4L/C2L There are 3 shape cases: Case 1 : when R2>4L/C ; we have two roots s1 and s2 : s1 = -R+R2-4L/C2L s2 so ; I1(t) =

=-R-R2-4L/C2L e-R+R2-4L/C2L

we plot at graph

I1(t) =

e-R-R2-4L/C2L

We plot at graph:

Case 2 when R2=4L/C ; we have same roots s1 = s2 : s1 = s2 = -R2L so;

x(t) = e-R.t2L

, we plot at graph

Case 3 when R2<4R/C ; we have complex roots s1 = -R+iR2-4L/C2L s2 so ; I1(t) = I2(t) =

=-R-iR2-4L/C2L e-R+iR2-4L/C2L e -R-iR2-4L/C2L