Civil Engineering Reference
In-Depth Information
2
xt
xt T
(
=
)
6
=
10 ,
(2.23)
2
(
=
)
and if further applied to Equation
(2.21)
we have
6
ln (10 )
2.2
T
=
≈
.
(2.24)
ω η
⋅
f
⋅
η
0
0
This expression is applied in measurement methods for determining the loss factor for
materials where the energy losses are relatively small. For materials with high losses the
reverberation time will be too short to obtain reasonable accuracy. A better method is
then to excite the material specimen into resonance and measure the Q factor.
2.4.2
Forced oscillations (vibrations)
Driving our simple mass-spring system using an external force
F
we now obtain
FFFF
x
++=
or
mc
k
2
(2.25)
d
d
x
mc
++⋅ =
t
k
x
F
(),
2
d
t
d
t
where we have, in the last equation, indicated that the external force could be an arbitrary
function of time. There are several available procedures for solving the equation. Our
aim is to determine the transfer function between applied force and displacement and,
later, between force and velocity, which is the driving point impedance. Letting the input
force be a simple harmonic force is the easiest way to solve the differential equation.
Even then the solution will contain two terms. One of these terms will represent a
transient motion as we start when the system is in a stable position. This will be the
solution of the homogeneous Equation
(2.12)
whereas the other term will be the
stationary part, which will be of primary interest. We may solve this term by expressing
the force as
F
(
t
) =
F
0
⋅exp(jω
t
) and then assuming a solution having the form
x
(
t
) =
F
0
x
=
+−
(2.26)
0
2
−
ωω
j
ck
and therefore we may write
j
ω
t
j
ω
t
Fe
⋅
Fe
⋅
0
0
xt
()
=
=
.
(2.27)
⎡
k
⎤
j
ω
Z
⎛
⎞
m
j
ω
c
+
j
ω
m
−
⎜
⎟
⎢
⎥
ω
⎝
⎠
⎣
⎦
When differentiating we get the velocity
v
(
t
) as
