2.3 Free Undamped Vibration of the Sdf System
21
Fig. 2.6 Example 2.1
W
g
¨
y + k (y st + y) − W = 0
(2.16)
Substituting y st from Eqs. (2.15)–(2.16), we get
W
g
¨
y + W + ky − W = 0
(2.17)
or
W
g
¨
y + ky = 0
or
¨
y + p
2 y = 0
(2.18)
where p
2
=
kg
W
.
The natural frequency is given by
f =
p
2π
=
1
2π
kg
W
(2.19)
Substituting y st from Eq. (2.15) into Eq. (2.19), we get
f =
1
2π
g
y st
(2.20)
Putting the numerical values of W and k, we get
f =
1
2π
2 × 9810
15
= 5.756 Hz (cycles/s)
21
Fig. 2.6 Example 2.1
W
g
¨
y + k (y st + y) − W = 0
(2.16)
Substituting y st from Eqs. (2.15)–(2.16), we get
W
g
¨
y + W + ky − W = 0
(2.17)
or
W
g
¨
y + ky = 0
or
¨
y + p
2 y = 0
(2.18)
where p
2
=
kg
W
.
The natural frequency is given by
f =
p
2π
=
1
2π
kg
W
(2.19)
Substituting y st from Eq. (2.15) into Eq. (2.19), we get
f =
1
2π
g
y st
(2.20)
Putting the numerical values of W and k, we get
f =
1
2π
2 × 9810
15
= 5.756 Hz (cycles/s)