522
14 Nonlinear Vibration
Fig. 14.3 Example 14.2
x 2 = 0 + 0 +
1
6
× (0.1)
2
× [2 × 18 + 18] = 0.09
Substituting the above values in Eq. (a), we get
¨
x 2 =
1.5
0.3
−
40 × 0.09
0.3
= −7
The assumed value of ¨
x 2 is not correct. Assume a new value ¨
x 3 = 3.
14 Nonlinear Vibration
Fig. 14.3 Example 14.2
x 2 = 0 + 0 +
1
6
× (0.1)
2
× [2 × 18 + 18] = 0.09
Substituting the above values in Eq. (a), we get
¨
x 2 =
1.5
0.3
−
40 × 0.09
0.3
= −7
The assumed value of ¨
x 2 is not correct. Assume a new value ¨
x 3 = 3.