Post on 11-Jul-2018
q2
u1
u2
k1
k2 q1
k1
k2
u1
u2
k1
JAWABAN TUGAS XIII DINAMIKA
Diketahui struktur sebagai berikut:
h
h
L
Diketahui data sebagai berikut:
w1 = q1 x L = 4400 kg
w2 = q2 x L = 2900 kg
k1 = 2000 kg/cm
k2 = 1500 kg/cm
Ditanyakan:
a. Tentukan frekuensi alami dan pola dari sistem
b. Tentukan respon pada struktur dengan getaran bebas tak teredam
c. Jika ditambahkan damping (c) sebesar 2c dan c, tentukan Mn, Kn, Cn dan
persamaan geraknya.
c
2c
d. Jika ( ) ( ) ( ) ( ) 00;00;00;10;05,0;100 21211 ====== qqqqkmc ζ , tentukan respon
getaran bebas untuk pola natural pertama.
u1
u2
k1
k2
k1
k2
e. Jika ( ) ( ) ( ) ( ) 00;00;10;00;10,0;100 21212 ====== qqqqkmc ζ , tentukan respon
getaran bebas untuk pola natural kedua.
Penyelesaian:
a. Menentukan frekuensi alami dan pola dari sistem
m2
h ke2
m1
h ke1
L
Jika w1 = 4400 kg à mskg
gwm
2.49,4981440011 ===
w2 = 2900 kg à mskg
gwm
2.96,2981290022 ===
ke1 = k1 + k1 = 2000 + 2000 = 4000 kg/cm
ke2 = k2 + k2 = 1500 + 1500 = 3000 kg/cm
Ø Matriks massa
m = mskg
mm 2
2
1 .96,20049,4
00
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡
Ø Matriks kekakuan
k = cmkg
kkkkk
⎥⎦
⎤⎢⎣
⎡
−
−=⎥
⎦
⎤⎢⎣
⎡
−
−+=⎥
⎦
⎤⎢⎣
⎡
−
−+
3000300030007000
30003000300030004000
22
221
Nilai frekuensi alami (ωn)
det [k - ωn2 m] = 0
[ ] [ ]
⎥⎦
⎤⎢⎣
⎡
−−
−−=
⎥⎦
⎤⎢⎣
⎡−⎥⎦
⎤⎢⎣
⎡
−
−=−
2
2
22
96,230003000300049,47000
96,20049,4
3000300030007000
n
n
nn MK
ω
ω
ωω
[ ] [ ]( )
( )( ) ( )( )
010.2,13419029,13
010.929,13134702072010.1,2
03000300096,2300049,47000
096,230003000
300049,47000det
0det
724
64227
22
2
2
2
=+−
=−+−−
=−−−−−
=⎥⎦
⎤⎢⎣
⎡
−−
−−
=−
nn
nnn
nn
n
n
n MK
ωω
ωωω
ωω
ω
ω
ω
Persamaan penyelesaian:
( ) ( ) ( )( )( )
58,2622,2304434190
29,13210.2,129,1343412034190
10.2,13419029,13
722
7
±=
−−±−−=
=
−=
=
n
cba
ω
40,4628,215358,26
22,2304434190
48,2033,41958,26
22,2304434190
222
121
=⇒=+
=
=⇒=−
=
ωω
ωω
Pola natural untuk sistem I diperoleh dengan mensubstitusikan �n = �1
[ ]{ } 33,4190 221=→=− ωφω nnmk
( )( )
078,17583000
300021,5117
033,41996,230003000
300033,41949,47000
33,419096,230003000
300049,47000
21
11
21
11
2
21
1121
21
1
=⎭⎬⎫
⎩⎨⎧⎥⎦
⎤⎢⎣
⎡
−
−
=⎭⎬⎫
⎩⎨⎧⎥⎦
⎤⎢⎣
⎡
−−
−−
=→=⎭⎬⎫
⎩⎨⎧⎥⎦
⎤⎢⎣
⎡
−−
−−
φ
φ
φ
φ
ωφ
φ
ω
ω
Pola natural ditentukan dengan menentukan satu satuan harga untuk salah satu pola.
Misalkan iniberikutanalisisdari nilaidiperoleh sehingga ,1 1121 φφ =
58625,022,124121,2117
022,124121,2117
078,175830000300021,5117
00178,17583000
300021,5117
11
11
11
11
11
11
=
=
=−
+=+−
=−
==⎭⎬⎫
⎩⎨⎧⎥⎦
⎤⎢⎣
⎡
−
−
φ
φ
φ
φ
φ
φ
Pola natural untuk sistem II diperoleh dengan mensubstitusikan �n = �2
( )( )
071,33733000
300023,2668
028,215396,230003000
300028,215349,47000
28,2153096,230003000
300049,47000
22
12
22
12
22
22
1222
22
=⎭⎬⎫
⎩⎨⎧⎥⎦
⎤⎢⎣
⎡
−−
−−
=⎭⎬⎫
⎩⎨⎧⎥⎦
⎤⎢⎣
⎡
−−
−−
=→=⎭⎬⎫
⎩⎨⎧⎥⎦
⎤⎢⎣
⎡
−−
−−
φ
φ
φ
φ
ωφ
φ
ω
ω
Pola natural ditentukan dengan menentukan satu satuan harga untuk salah satu pola.
Misalkan iniberikutanalisisdari nilaidiperoleh sehingga ,1 1222 φφ =
12641,171,37377,331
071,37377,331
071,337330000300023,2668
0171,33733000
300023,2668
12
12
12
12
12
12
−=
−=
=+
−=−−
=−−
=⎭⎬⎫
⎩⎨⎧⎥⎦
⎤⎢⎣
⎡
−−
−−
φ
φ
φ
φ
φ
φ
Sehingga, pola natural dari sistem tersebut adalah:
{ }
{ }⎭⎬⎫
⎩⎨⎧−
=⎭⎬⎫
⎩⎨⎧
=
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
=
112641,1
158625,0
22
122
21
111
φ
φφ
φ
φφ
u1
u2
k k
k k 2m
21φ
11φ
12φ
22φ
m
h
h
Pola Natural I
Pola Natural II
Kontrol kondisi orthogonal
{ }[ ]{ }
{ }
{ }[ ]{ }
{ } OK
M
OK
K
T
T
0112641,1
96,20049,4
158625,0
0
0112641,1
3000300030007000
158625,0
0
21
21
≈⎭⎬⎫
⎩⎨⎧−⎥⎦
⎤⎢⎣
⎡
=
≈⎭⎬⎫
⎩⎨⎧−⎥⎦
⎤⎢⎣
⎡
−
−
=
φφ
φφ
b. Penentuan Respons Getaran Bebas jika diketahui:
!! 0 = 1, !! 0 = 1, !! 0 = 0, !! 0 = 0.
Persamaan gerak:
u (t) = ( ) ( ) ( )∑∑==
=⎥⎦
⎤⎢⎣
⎡+
n
nnn
n
nn
n
nnnn tqtSinqtCosq
11.00 φω
ωωφ
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )tCostSinqtCosqtq
tCostSinqtCosqtq
40,46100
48,20100
22
2122
11
1111
=+=
=+=
ωω
ω
ωω
ω
Persamaan Getaran à u (t) = ( )∑=
n
nnn tq
1
.φ
( )( )
( ) ( )
( )( )
( ) ( )
( ) ( ) ( )( ) ( ) ( )tCostCostu
tCostCostu
tCostCostutu
tqtqtutu
40,4648,2040,4612641,148,2058625,0
40,46112641,1
48,20158625,0
..
2
1
2
1
22112
1
+=
−=⎭⎬⎫
⎩⎨⎧−
+⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
+=⎭⎬⎫
⎩⎨⎧
φφ
Data: ω1 = 20.480 rad/dt
ω2 = 46.400 rad/dt
No
Time Cos ω 1t Cos ω 2t u1 (t) u2 (t) t (s) (q1 (t)) (q2 (t))
1 0 1.000 1.000 -‐0.540 2.000 2 1.5 0.767 0.885 -‐0.547 1.652 3 3 0.178 0.565 -‐0.532 0.743 4 4.5 -‐0.494 0.116 -‐0.420 -‐0.379 5 6 -‐0.937 -‐0.361 -‐0.143 -‐1.297 6 7.5 -‐0.943 -‐0.754 0.296 -‐1.697 7 9 -‐0.511 -‐0.973 0.796 -‐1.485 8 10.5 0.158 -‐0.968 1.183 -‐0.810 9 12 0.755 -‐0.740 1.276 0.015 10 13.5 1.000 -‐0.341 0.970 0.659 11 15 0.780 0.137 0.303 0.917 12 16.5 0.197 0.583 -‐0.541 0.780 13 18 -‐0.477 0.894 -‐1.287 0.418 14 19.5 -‐0.930 1.000 -‐1.671 0.070 15 21 -‐0.950 0.875 -‐1.542 -‐0.075 16 22.5 -‐0.528 0.548 -‐0.927 0.019 17 24 0.139 0.094 -‐0.025 0.233 18 25.5 0.741 -‐0.381 0.863 0.361 19 27 0.999 -‐0.768 1.451 0.231
No Time Cos ω 1t Cos ω 2t u1 (t) u2 (t) t (s) (q1 (t)) (q2 (t))
20 28.5 0.792 -‐0.978 1.566 -‐0.186 21 30 0.217 -‐0.963 1.211 -‐0.746 22 31.5 -‐0.459 -‐0.725 0.548 -‐1.185 23 33 -‐0.922 -‐0.321 -‐0.179 -‐1.243 24 34.5 -‐0.956 0.158 -‐0.738 -‐0.798 25 36 -‐0.545 0.600 -‐0.995 0.055 26 37.5 0.119 0.904 -‐0.948 1.023 27 39 0.728 0.999 -‐0.699 1.727 28 40.5 0.998 0.864 -‐0.388 1.862 29 42 0.804 0.530 -‐0.125 1.334 30 43.5 0.236 0.073 0.056 0.309 31 45 -‐0.442 -‐0.400 0.192 -‐0.842
Respon getaran bebas yang terjadi:
a. Hubungan antara qn(t) dan waktu (t)
b. Hubungan antara u1(t) dan waktu (t)
-‐1.500
-‐1.000
-‐0.500
0.000
0.500
1.000
1.500
0 5 10 15 20 25 30 35 40 45 50 qn (t)
t (sec)
q1(t)
q2(t)
-‐2.000
-‐1.500
-‐1.000
-‐0.500
0.000
0.500
1.000
1.500
2.000
0 5 10 15 20 25 30 35 40 45 50 u1 (t)
t (sec)
c. Hubungan antara u2(t) dan waktu (t)
c. Jika ditambahkan damping (c) sebesar 2c dan c, tentukan Mn, Kn, Cn dan
persamaan geraknya.
{ }
{ }⎭⎬⎫
⎩⎨⎧−
=⎭⎬⎫
⎩⎨⎧
=
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
=
112641,1
158625,0
22
122
21
111
φ
φφ
φ
φφ
m = ⎥⎦
⎤⎢⎣
⎡
96,20049,4
k = ⎥⎦
⎤⎢⎣
⎡
−
−
3000300030007000
c = ⎥⎦
⎤⎢⎣
⎡
−
−=⎥
⎦
⎤⎢⎣
⎡
−
−+
1113
11112
cc
Menyusun matriks Mn, Kn, Cn
Kn = φnT K φn
[ ]
{ } [ ]{ }
{ }
32,1888158625,0
3000300030007000
158625,0
1111
2221
1211
=⎭⎬⎫
⎩⎨⎧⎥⎦
⎤⎢⎣
⎡
−
−=
=
⎥⎦
⎤⎢⎣
⎡=
φφ Kk
kkkk
K
T
n
-‐2.000 -‐1.500 -‐1.000 -‐0.500 0.000 0.500 1.000 1.500 2.000 2.500
0 5 10 15 20 25 30 35 40 45 50
u2 (t)
t (sec)
{ } [ ]{ }
{ }
0112641,1
3000300030007000
158625,0
2112
≈⎭⎬⎫
⎩⎨⎧−⎥⎦
⎤⎢⎣
⎡
−
−=
= φφ Kk T
{ } [ ]{ }
{ }
0158625,0
3000300030007000
112641,1
1221
≈⎭⎬⎫
⎩⎨⎧⎥⎦
⎤⎢⎣
⎡
−
−−=
= φφ Kk T
{ } [ ]{ }
{ }
06,18640112641,1
3000300030007000
112641,1
2222
=⎭⎬⎫
⎩⎨⎧−⎥⎦
⎤⎢⎣
⎡
−
−−=
= φφ Kk T
[ ] ⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡=
06.186400032,1888
2221
1211
kkkk
Kn
Mn = φnT M φn
[ ]
{ } [ ]{ }
{ }
50,4158625,0
96,20049,4
158625,0
1111
2221
1211
=⎭⎬⎫
⎩⎨⎧⎥⎦
⎤⎢⎣
⎡=
=
⎥⎦
⎤⎢⎣
⎡=
φφ Mm
mmmm
M
T
n
{ } [ ]{ }
{ }
0112641,1
96,20049,4
158625,0
2112
=⎭⎬⎫
⎩⎨⎧−⎥⎦
⎤⎢⎣
⎡=
= φφ Mm T
{ } [ ]{ }
{ }
0158625,0
96,20049,4
112641,1
1221
=⎭⎬⎫
⎩⎨⎧⎥⎦
⎤⎢⎣
⎡−=
= φφ Mm T
{ } [ ]{ }
{ }
66,8112641,1
96,20049,4
112641,1
2222
=⎭⎬⎫
⎩⎨⎧−⎥⎦
⎤⎢⎣
⎡−=
= φφ Mm T
[ ] ⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡=
66,80050,4
2221
1211
mmmm
Mn
Cn = φnT C φn
[ ]
{ } [ ]{ }
{ }
86,0158625,0
1113
158625,0
1111
2221
1211
=⎭⎬⎫
⎩⎨⎧⎥⎦
⎤⎢⎣
⎡
−
−=
=
⎥⎦
⎤⎢⎣
⎡=
φφ Cc
cccc
C
T
n
{ } [ ]{ }
{ }
44,0112641,1
1113
158625,0
2112
−=⎭⎬⎫
⎩⎨⎧−⎥⎦
⎤⎢⎣
⎡
−
−=
= φφ Cc T
{ } [ ]{ }
{ }
44,0158625,0
1113
112641,1
1221
−=⎭⎬⎫
⎩⎨⎧⎥⎦
⎤⎢⎣
⎡
−
−−=
= φφ Cc T
{ } [ ]{ }
{ }
06,7112641,1
1113
112641,1
2222
=⎭⎬⎫
⎩⎨⎧−⎥⎦
⎤⎢⎣
⎡
−
−−=
= φφ Mc T
[ ] ⎥⎦
⎤⎢⎣
⎡
−
−=⎥
⎦
⎤⎢⎣
⎡=
06,744,044,086,0
2221
1211
cccc
Cn
d. Jika ( ) ( ) ( ) ( ) 00;00;00;10;05,0;100 21211 ====== qqqqkmc ζ , tentukan respon
getaran bebas untuk pola natural pertama.
{ }⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
=158625,0
21
111 φ
φφ
∑=
−
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛ ++=
n
ndn
dn
onnodno
tn tqqtqet nn
1sincos)u( ω
ωωζ
ωφ ωζ
( )
{ } ( ) ( ) ( )
( )( )
( ) ( )
( ) ( )
( ) ( )[ ]tSintCose
tSintCose
tSintCosetutu
tqqtqetu
rad
t
t
t
dd
dt
d
45,2005,045,20158625,0
45,2045,20
1.48,20.05,0045,20158625,0
45,2045,20
1.48,20.05,0045,20158625,0
21
sin00cos0)(
sec/45,2005,0148,201
48,20.05,0
48,20.05,0
48,20.05,0
11
1111111
2211
11
+⎭⎬⎫
⎩⎨⎧
=
⎥⎦
⎤⎢⎣
⎡⎟⎠
⎞⎜⎝
⎛ ++
⎭⎬⎫
⎩⎨⎧
=
⎥⎦
⎤⎢⎣
⎡⎟⎠
⎞⎜⎝
⎛ ++
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛ ++=
=−=−=
−
−
−
− ωωωζ
ωφ
ζωω
ωζ
Tabel perhitungan respon getaran bebas untuk pola natural pertama
No Time Cos
20,45t Sin 20,45t u1 (t) u2 (t) q1 (t) t (s)
1 0 1.000 0.000 0.586 1.000 1.000 2 0.25 0.390 -‐0.921 0.156 0.266 0.343 3 0.5 -‐0.697 -‐0.718 -‐0.257 -‐0.439 -‐0.732 4 0.75 -‐0.932 0.362 -‐0.249 -‐0.424 -‐0.914 5 1 -‐0.030 1.000 0.004 0.007 0.020 6 1.25 0.909 0.417 0.152 0.259 0.930 7 1.5 0.738 -‐0.675 0.089 0.152 0.704 8 1.75 -‐0.334 -‐0.942 -‐0.037 -‐0.064 -‐0.381 9 2 -‐0.998 -‐0.059 -‐0.076 -‐0.129 -‐1.001 10 2.25 -‐0.443 0.896 -‐0.023 -‐0.040 -‐0.399 11 2.5 0.653 0.758 0.031 0.053 0.691 12 2.75 0.952 -‐0.306 0.033 0.056 0.937 13 3 0.089 -‐0.996 0.001 0.002 0.039 14 3.25 -‐0.883 -‐0.470 -‐0.019 -‐0.033 -‐0.906 15 3.5 -‐0.777 0.630 -‐0.012 -‐0.021 -‐0.745 16 3.75 0.278 0.961 0.004 0.007 0.326 17 4 0.993 0.118 0.010 0.017 0.999 18 4.25 0.496 -‐0.868 0.003 0.006 0.452 19 4.5 -‐0.607 -‐0.795 -‐0.004 -‐0.006 -‐0.647 20 4.75 -‐0.968 0.249 -‐0.004 -‐0.007 -‐0.956 21 5 -‐0.148 0.989 0.000 -‐0.001 -‐0.098 22 5.25 0.853 0.521 0.002 0.004 0.879 23 5.5 0.813 -‐0.583 0.002 0.003 0.783 24 5.75 -‐0.220 -‐0.975 0.000 -‐0.001 -‐0.269 25 6 -‐0.984 -‐0.177 -‐0.001 -‐0.002 -‐0.993 26 6.25 -‐0.546 0.838 0.000 -‐0.001 -‐0.504 27 6.5 0.559 0.829 0.000 0.001 0.600 28 6.75 0.982 -‐0.191 0.001 0.001 0.972 29 7 0.206 -‐0.979 0.000 0.000 0.157 30 7.25 -‐0.821 -‐0.571 0.000 -‐0.001 -‐0.850 31 7.5 -‐0.846 0.534 0.000 0.000 -‐0.819
Respon getaran bebas yang terjadi:
a. Hubungan antara u1(t) dan waktu (t)
b. Hubungan antara u2(t) dan waktu (t)
c. Hubungan antara q1(t) dan waktu (t)
-‐0.400
-‐0.200
0.000
0.200
0.400
0.600
0.800
0 1 2 3 4 5 6 7 8
u1 (t)
t (sec)
-‐0.600 -‐0.400 -‐0.200 0.000 0.200 0.400 0.600 0.800 1.000 1.200
0 1 2 3 4 5 6 7 8
u2 (t)
t (sec)
-‐1.500
-‐1.000
-‐0.500
0.000
0.500
1.000
1.500
0 10 20 30 40 50 60 70 q1 (t)
t (sec)
a. Jika ( ) ( ) ( ) ( ) 00;00;00;10;1,0;100 21122 ====== qqqqkmc ζ , tentukan respon
getaran bebas untuk pola natural kedua.
{ }⎭⎬⎫
⎩⎨⎧−
=⎭⎬⎫
⎩⎨⎧
=112641,1
22
122 φ
φφ
∑=
−
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛ ++=
n
ndn
dn
onnodno
tn tqqtqet nn
1sincos)u( ω
ωωζ
ωφ ωζ
( )
{ } ( ) ( ) ( )
( )( )
( ) ( )
( ) ( )
( ) ( )[ ]tSintCose
tSintCose
tSintCosetutu
tqqtqetu
rad
t
t
t
dd
dt
d
17,461,017,46112641,1
17,4645,20
1.48,20.05,0017,46112641,1
17,4617,46
1.40,46.1,0017,46112641,1
21
sin00cos0)(
sec/17,461,0140,461
64,4
64,4
40,46.1,0
22
2222222
2222
22
+⎭⎬⎫
⎩⎨⎧−
=
⎥⎦
⎤⎢⎣
⎡⎟⎠
⎞⎜⎝
⎛ ++
⎭⎬⎫
⎩⎨⎧−
=
⎥⎦
⎤⎢⎣
⎡⎟⎠
⎞⎜⎝
⎛ ++
⎭⎬⎫
⎩⎨⎧−
=⎭⎬⎫
⎩⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛ ++=
=−=−=
−
−
−
− ωω
ωζωφ
ζωω
ωζ
Tabel perhitungan respon getaran bebas untuk pola natural kedua
No Time Cos ω 1t Sin ω 1t u1 (t) u2 (t) q2 (t) t (s)
1 0 1.000 0.000 -‐1.126 1.000 1.000 2 0.100 -‐0.095 -‐0.995 0.138 -‐0.122 -‐0.195 3 0.200 -‐0.982 0.190 0.429 -‐0.381 -‐0.963 4 0.300 0.282 0.959 -‐0.106 0.094 0.378 5 0.400 0.928 -‐0.372 -‐0.157 0.139 0.891 6 0.500 -‐0.459 -‐0.888 0.061 -‐0.054 -‐0.548 7 0.600 -‐0.841 0.542 0.055 -‐0.049 -‐0.786 8 0.700 0.619 0.785 -‐0.031 0.027 0.698 9 0.800 0.723 -‐0.691 -‐0.018 0.016 0.654 10 0.900 -‐0.757 -‐0.654 0.014 -‐0.013 -‐0.822 11 1.000 -‐0.579 0.816 0.005 -‐0.005 -‐0.497 12 1.100 0.867 0.498 -‐0.006 0.006 0.917 13 1.200 0.413 -‐0.911 -‐0.001 0.001 0.322 14 1.300 -‐0.946 -‐0.325 0.003 -‐0.002 -‐0.978 15 1.400 -‐0.233 0.972 0.000 0.000 -‐0.136 16 1.500 0.990 0.140 -‐0.001 0.001 1.004 17 1.600 0.045 -‐0.999 0.000 0.000 -‐0.055 18 1.700 -‐0.999 0.051 0.000 0.000 -‐0.994 19 1.800 0.146 0.989 0.000 0.000 0.245 20 1.900 0.971 -‐0.239 0.000 0.000 0.947 21 2.000 -‐0.331 -‐0.944 0.000 0.000 -‐0.425 22 2.100 -‐0.908 0.419 0.000 0.000 -‐0.866
No Time Cos ω 1t Sin ω 1t u1 (t) u2 (t) q2 (t) t (s)
23 2.200 0.504 0.864 0.000 0.000 0.590 24 2.300 0.812 -‐0.584 0.000 0.000 0.754 25 2.400 -‐0.658 -‐0.753 0.000 0.000 -‐0.734 26 2.500 -‐0.687 0.727 0.000 0.000 -‐0.614 27 2.600 0.789 0.614 0.000 0.000 0.851 28 2.700 0.536 -‐0.844 0.000 0.000 0.452 29 2.800 -‐0.891 -‐0.454 0.000 0.000 -‐0.937 30 2.900 -‐0.367 0.930 0.000 0.000 -‐0.274 31 3.000 0.961 0.276 0.000 0.000 0.989
Respon getaran bebas yang terjadi:
a. Hubungan antara u1(t) dan waktu (t)
b. Hubungan antara u2(t) dan waktu (t)
-‐1.400 -‐1.200 -‐1.000 -‐0.800 -‐0.600 -‐0.400 -‐0.200 0.000 0.200 0.400 0.600
0 0.5 1 1.5 2 2.5 3
u1 (t)
t (sec)
-‐0.600 -‐0.400 -‐0.200 0.000 0.200 0.400 0.600 0.800 1.000 1.200
0 0.5 1 1.5 2 2.5 3
u2 (t)
t (sec)
c. Hubungan antara q2(t) dan waktu (t)
-‐1.500
-‐1.000
-‐0.500
0.000
0.500
1.000
1.500
0 20 40 60 80 100 120 q2 (t)
t (sec)