y ¼ u=‘ ½1Š;
ð1:96Þ
where ‘ ½m] is yet a free constant? Then we have a dimensionless time and a
dimensionless displacement y, and are free to choose the constants x and ‘. They
can be chosen as some characteristic frequency x and some characteristic length ‘
for the system. However, any choice is allowed, and for now, the decision is just
postponed, until we can see which choice makes the resulting dimensionless
equation appear as simple as possible.
From (1.95) It follows that differentiation wrt. physical time t turns into differentiation wrt. dimensionless time s as follows:
_
u ¼
du
dt
¼
du
ds
ds
dt
¼ x
du
ds
;
€ u ¼
d _
u
dt
¼
d _
u
ds
ds
dt
¼ x
d
ds
x
du
ds
¼ x
2 d
2 u
ds 2 :
ð1:97Þ
Insert (1.95)–(1.97) into (1.94) and obtain:
mx
2 d
2
ðy‘Þ
ds
þ c
dðy‘Þ
ds
x þ kðy‘Þ ¼ Q sinðXs=xÞ:
ð1:98Þ
Then redefine overdots to mean differentiation wrt. s, divide by mx
2
‘ to free the
highest-order derivative term (here d
2 y=ds
2 ), and find:
€ y þ
c
xm
_
y þ
k
mx 2 y ¼
Q
mx 2 ‘
sin
X
x
s
1
½ Š:
ð1:99Þ
Now one can choose x and ‘ to give a simple equation, with interpretable
parameters. For example, we can choose:
k
mx 2 ¼ 1 ) x ¼
ffiffiffi ffi
k
m
r
;
ð1:100Þ
Q
mx 2 ‘
¼ 1 ) ‘ ¼
Q
mx 2 ¼
Q
mk=m
¼
Q
k
;
ð1:101Þ
c
xm
¼ 2f ) f ¼
c
2
ffiffiffiffiffiffi
km
p ;
ð1:102Þ
~
X ¼
X
x
:
ð1:103Þ
Note that all of the new parameters introduced by (1.100)–(1.103) have a simple
physical interpretation: x is the undamped natural frequency, ‘ the static deformation of the elastic element having stiffness k when the load is Q, f is the damping
ratio, and ~
X the ratio of excitation frequency to natural frequency. Substituting
(1.100)–(1.103) into (1.99) we find:
30
1 Vibration Basics