6.6 Determination of Absolute Displacement of Free Vibration …
195
¨
ξ 1 + p
2
1 ξ 1 = 0
¨
ξ 2 + p
2
2 ξ 2 = 0
(a)
Solutions of Eq. (a) are
ξ 1 = A 1 cos p 1 t + B 1 sin p 1 t
ξ 2 = A 2 cos p 2 t + B 2 sin p 2 t
(b)
Now,
x 1
x 2
=
1.000 1.000
1.618 − 0.618
A 1 cos p 1 t + B 1 sin p 1 t
A 2 cos p 2 t + B 2 sin p 2 t
or
x 1 = (A 1 cos p 1 t + B 1 sin p 1 t) + (A 2 cos p 2 t + B 2 sin p 2 t)
x 2 = 1.618 (A 1 cos p 1 t + B 1 sin p 1 t)
− 0.618 (A 2 cos p 2 t + B 2 sin p 2 t)
⎫
⎬
⎭
(c)
Substituting the given initial values into Eq. (c) and their derivatives
A 1 + A 2 = x 0
1.618 A 1 − 0.618 A 2 = − x 0
B 1 p 1 + B 2 p 2 = ν
1.618 B 1 p 1 − 0.618 B 2 p 2 = 0
⎫
⎪ ⎪ ⎬
⎪ ⎪ ⎭
(d)
Solving Eq. (d)
A 1 = − 0.1771 x 0 , A 2 = 1.171 x 0 , B 1 =
0.276v
p 1
and B 2 =
0.724v
p 2
Hence, the equations of motion can be written as
x 1 =
− 0.1771 x 0 cos p 1 t +
0.276 v
p 1
sin p 1 t
+
1.171 x 0 cos p 2 t +
0.72 4v
p 2
sin p 2 t
x 2 = 1.618
− 0.171 x 0 cos p 1 t +
0.27 6v
p 1
sin p 1 t
− 0.618
1.171 x 0 cos p 2 t +
0.724 v
p 2
sin p 2 t