26
2 Free Vibration of Single Degree of Freedom System
identical. For the equivalent spring, the deformation due to unit force is 1/k eq , and
for the two springs, they are 1/k 1 and 1/k 2 . Therefore,
1
k eq
=
1
k 1
+
1
k 2
or
k eq =
k 1 k 2
k 1 + k 2
Substituting numerical values for the problem
I =
1
12
× (100) (150)
3
= 28,125,000 mm
4
k 1 =
48 × 2.1 × 10
5
× 28,125,000
(6000) 3
= 1312.5 N/mm
k eq =
1312.5 × 40
1312.5 + 40
= 38.82 N/mm
The natural frequency of the system is
f =
1
2π
k eq
m
=
1
2π
38.82 × 9810
20
= 21.962 Hz
Example 2.6 A mass m 1 hangs from a spring having stiffness k. A second mass
m 2 drops through a height h and sticks to mass m 1 without rebound (Fig. 2.11).
Determine the subsequent motion.
The velocity of mass m 2 just before impact is given by
v
2
2 = 2gh
Fig. 2.11 Example 2.6
2 Free Vibration of Single Degree of Freedom System
identical. For the equivalent spring, the deformation due to unit force is 1/k eq , and
for the two springs, they are 1/k 1 and 1/k 2 . Therefore,
1
k eq
=
1
k 1
+
1
k 2
or
k eq =
k 1 k 2
k 1 + k 2
Substituting numerical values for the problem
I =
1
12
× (100) (150)
3
= 28,125,000 mm
4
k 1 =
48 × 2.1 × 10
5
× 28,125,000
(6000) 3
= 1312.5 N/mm
k eq =
1312.5 × 40
1312.5 + 40
= 38.82 N/mm
The natural frequency of the system is
f =
1
2π
k eq
m
=
1
2π
38.82 × 9810
20
= 21.962 Hz
Example 2.6 A mass m 1 hangs from a spring having stiffness k. A second mass
m 2 drops through a height h and sticks to mass m 1 without rebound (Fig. 2.11).
Determine the subsequent motion.
The velocity of mass m 2 just before impact is given by
v
2
2 = 2gh
Fig. 2.11 Example 2.6