Civil Engineering Reference
In-Depth Information
k
(
st
x
)
k
L
L
k
m
x
x
st
m
x
m
mg
(a)
(b)
(c)
Figure 10.1
(a) Simple harmonic oscillator. (b) Static equilibrium
position. (c) Free-body diagram for arbitrary
position
x
.
gravitational force is in equilibrium with the spring force so
F
x
st
(10.1)
where
st
is the equilibrium elongation of the spring and
x
is measured positive
downward from the equilibrium position; that is, when
x
=
0
=
mg
−
k
=
0
, the system is at its
equilibrium position.
If, by some action, the mass is displaced from its equilibrium position,
the force system becomes unbalanced, as shown by the free-body diagram of
Figure 10.1c. We must apply Newton's second law to obtain
F
x
m
d
2
x
d
t
2
`
=
ma
x
=
=
mg
−
k
(
st
+
x
)
(10.2)
Incorporating the equilibrium condition expressed by Equation 10.1, Equation 10.2
becomes
m
d
2
x
d
t
2
+
kx
=
0
(10.3)
Equation 10.3 is a second-order, linear, ordinary differential equation with con-
stant coefficients. (And physically, we assume that the coefficients
m
and
k
are
positive.) Equation 10.3 is most-often expressed in the form
d
2
x
d
t
2
d
2
x
d
t
2
k
m
x
2
x
+
=
+
=
0
(10.4)
The general solution for Equation 10.4 is
x
(
t
)
t
(10.5)
where
A
and
B
are the constants of integration. Recall that the solution of a
second-order differential equation requires the specification of two constants to
determine the solution to a specific problem. When the differential equation de-
scribes the time response of a mechanical system, the constants of integration are
most-often called the
initial conditions.
=
A
sin
t
+
B
cos
