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−ζ 2

d = √1-(0.05)2

d = 0.998749

ωnd = 49.9375

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 = √ A12 + 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

f1(t) = e-ζωnt

f2(t) = X cos ( ωndt-α )

ω = 2πf

f = ω/2π

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.