1. Tugas 1 Getaran
Kelompok 20
Reva Risky ( 4112100080 )
Fauzi Rogera ( 4112100091 )
Lendy Hari ( 4112100109 )
Rafli Alfazer ( 4112100117 )
Diketahui :
k = 5x106
m = 2000 lb
X0 = 0.1 inch
V0 = 0.5 inch/sec
ζ = 0.05
Soal :
> Hitung amplitudo ( X )
> Hitung Sudut fase ( α )
> Gambarkan di excel grafik dari f1(t), f2(t), dan x(t)
Jawab :
ωn = √k/m
ωn = √5000000/2000
ωn = 50
d = √1 − 휁2
d = √1-(0.05)2
d = 0.998749
ωnd = 49.9375

2. x(t) = e-ζωnt ( A1 cos ωndt + A2 sin ωndt )
x(0) = e-0.05x50x0 ( A1 cos 49.9375x0 + A2 sin 49.9375x0 )
x(0) = 1 ( A1 x 1 + A2 x 0)
x(0) = A1
A1 = 0.1
ẋ(t) = -ζωne-ζωnt [ A1 cos ωndt + A2 sin ωndt ] + e-ζωnt [ -A1 (ωnd) sin ωndt + A2 (ωnd) cos ωndt ]
ẋ(0) = V0
ẋ(0) = -0.05x50xe-0.05x50x0 [ A1 cos 49.9375x0 + A2 sin 49.9375x0 ] + e-0.05x50x0 [ -A1 (49.9375) sin
49.9375x0 + A2 (49.9375) cos 49.9375x0 ]
ẋ(0) = -2.5 [ A1 cos 0 + A2 sin 0 ] + 1 [ -A1 (49.9375) sin 0 + A2 (49.9375) cos 0 ]
V0 = -2.5 A1 + 49.9375 A2
0.5 = -2.5 A1 + 49.9375 A2
0.5 = -2.5 ( 0.1 ) + 49.9375 A2
A2 = 0.75/49.9375
A2 = 0.0150
maka amplitudonya :
X = √퐴1
2 + A2
2
X = √(0.1)2 + (0.015)2
X = 0.1011
maka sudut fasenya :
α = tg-1 A2/A1
α = tg-1 0.015/0.1
α = 8.5308 deg
α = 0.1489 rad

3. f1(t) = e-ζωnt
f2(t) = X cos ( ωndt-α )
ω = 2휋f
f = ω/2휋
f = 50/2 (3.14)
f = 7.9618 hz
T = 1/f
T = 1/7.9618
T = 0.1256 sec
karena satu periode di bagi menjadi 20 titik maka :
T/20 = 0.00628 sec
kondisi awal :
t = 0.00628 sec
f1(t) = e-ζωnt
maka f1(0.00628) = e-0.05x50x0.00628 = 0.9844
f2(t) = X cos ( ωndt-α )
maka f2(0.00628) = 0.1011 cos ((49.9375x0.00628)-0.1489) = 0.09975
Perhitungan selanjutnya terlampir.