Civil Engineering Reference
In-Depth Information
Equation 10.19 represents a sinusoidal oscillation around the equilibrium posi-
tion and is, therefore, the same as the solution given in Equation 10.6. Given the
displacement of node 2, the reaction force at node 1 is obtained via the constraint
equation as
))
(10.20)
Amplitude
C
and phase angle
are determined by application of the initial con-
ditions, as illustrated in the following example.
f
1
=−
ku
2
(
t
)
=−
k
(
st
+
C
sin(
t
+
EXAMPLE 10.1
A simple harmonic oscillator has
k
=
25 lb/in. and
mg
=
20 lb. The mass is displaced
downward a distance of 1.5 in. from the equilibrium position. The mass is released from
that position with zero initial velocity at
t
=
0. Determine (a) the natural circular fre-
quency, (b) the amplitude of the oscillatory motion, and (c) the phase angle of the oscil-
latory motion.
■
Solution
The natural circular frequency is
k
m
=
25
=
4
=
21
.
98
rad/sec
/
.
20
386
where, for consistency of units, the mass is obtained from the weight using
g
=
386.4 in./s
2
.
The given initial conditions are
u
2
(
t
=
0)
=
st
+
1
.
5in
.
u
2
(
t
=
0)
=
0 in./sec
and the static deflection is
st
=
W
/
k
=
20
/
25
=
0
.
8 in.
Therefore, we have
u
2
(0)
=
2
.
3 in.
The motion of node 2 (hence, the mass) is then given by Equation 10.19 as
u
2
(
t
)
=
0
.
8
+
C
sin(21
.
98
t
+
) in.
and the velocity is
d
u
2
d
t
u
2
(
t
)
=
=
21
.
98
C
cos(21
.
98
t
+
) in./sec
Applying the initial conditions results in the equations
u
2
(
t
=
0)
=
2
.
3
=
0
.
8
+
C
sin
u
2
(
t
)
=
0
=
21
.
98
C
cos
The initial velocity equation is satisfied by
C
=
0 or
=
/
2
.
If the former is true, the
initial displacement equation cannot be satisfied, so we conclude that
=
/
2
.
Substi-
tuting into the displacement equation then gives the amplitude
C
as 1.5 in. The complete
motion solution is
u
2
(
t
)
=
0
.
8
+
1
.
5 sin
21
.
98
t
+
2
=
0
.
8
+
1
.
5 cos(21
.
98
t
) in.
