m=1.2x103
k=1.25x106
b=1.4x102;
g=9.8
yo=pertambahan
panjang mula mula pegas = 1
t1=Masukkan akar
waktu (t1) pertama = 1
t2=Masukkan akar
waktu (t2) kedua = 2
Fo=gaya mula
mula = 50
M =Masukkan
jumlah iterasi = 12
wo=sqrt(k/m)
wo=sqrt(1.25x106/1.2x103)
wo=32.2749
w =sqrt((wo)^2-(b^2)/(2*((m)^2)))
w
=sqrt((32.27-(1.4x102)^2/(2*(1.2x103)))
w =32.2748
1.
Menghitung pada
interval awal, t1=1 dan t2=2 sehingga diperoleh
f1= Yt1
f2= Yt2
f1= (yo*cos(wo*t1))+(yo*sin(wo*t1))+((Fo/(wo^2-w^2))*(cos(w*t1)))+
((m*g)/(wo^2))
f1= (1*cos(32.2749*1))+(1*sin(32.2749*1))+((50/32.2749^2-32.278))*
cos(32.2748*1)))+(( 1.2x103*9.8)/( 32.2749^2))
= 4.8126x103
F2= (yo*cos(wo*t2))+(yo*sin(wo*t2))+((Fo/(wo^2-w^2))*(cos(w*t2)))+
((m*g)/(wo^2))
f2= (1*cos(32.2749*2))+(1*sin(32.2749*2))+((50/32.2749^2-32.278))*
cos(32.2748*2)))+(( 1.2x103*9.8)/( 32.2749^2))
= -1.0630x103
2.
Menghitung
estimasi sub interval pertama;
t3= (t1+t2)/2
= (1+2)/2
= 1.5
ft1=(1*cos(32.2749*1.5))+(1*sin(32.2749*1.5))+((50/32.2749^2-32.278))*
cos(32.2748*1.5)))+(( 1.2x103*9.8)/( 32.2749^2))
= -2.0383x103
3. Menghitung interval t4 dari t3 dan t2;
ft2=Y(t)
t4= (t3+t2)/2
= (1.5+2)/2
= 1.75
ft2=(1*cos(32.2749*1.75))+(1*sin(32.2749*1.75))+((50/32.2749^2-32.278))*
cos(32.2748*1.75)))+(( 1.2x103*9.8)/( 32.2749^2))
= 7.3310x103
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