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5 Vibration of Two Degrees of Freedom System
Example 5.3 A jig contains a screen that reciprocates with a frequency of 600 cpm.
The jig weighs 230 N and has a fundamental frequency of 400 cpm. If an absorber
weighing 60 N is to be installed to eliminate the vibration of the jig frame, determine
the absorber spring stiffness. What will be the resulting two natural frequencies of
the system?
The external frequency ω =
2 × π × 600
60
= 20 π rad/s.
The mass of the jig frame m 1 =
230
9.81
kg.
The mass of the absorber m 2 =
60
9.81
kg.
In order to eliminate the vibration of the jig frame, we get from Eq. (5.26)
ω
2
=
k 2
m 2
where k 2 is the stiffness of the absorber spring. Therefore
k 2 = ω
2 m 2 = 20 π × 20 π ×
60
9.81
= 24,146 N/m
= 24.146 N/mm
If k 1 is the stiffness of the jig frame
ω
2
1 =
k 1
m 1
where
ω 1 =
400 × 2 π
60
=
40 π
3
rad/s
Therefore
k 1 =
40 π
3
×
40 π
3
×
230
9.81
= 41,137 N/m
Natural frequencies of the two degrees of freedom system can be obtained by
Eq. (5.6) as
p
4
−
k 2
m 2
+
k 1 + k 2
m 1
p
2
+
k 1 k 2
m 1 m 2
= 0
or