Question 13.6.1: Suppose a particular nonlinear spring is described by (13.6.......

The first step is to determine the equilibrium positions for each mass. With m = 4 in (13.6.2), the only real root of this cubic equation is y_r = 0.0278 m. With m = 14 in (13.6.2), the only real root is y_r = 0.091 m. Note that f (y_r ) = mg, and that

k_1 y_r+k_2 y_r^3-m g=0          (13.6.2)

The equivalent linear spring constant k is the slope m of the curve f (y) near the specific equilibrium point. Thus, the linearized spring constants for each equilibrium are:
a. For y_r = 0.0278, k = 1430 N/m
b. For y_r = 0.091, k = 1723 N/m
The linearized equations of motion, which are accurate only near their respective equilibrium points, are found from m\ddot{x} + kx = 0. They are
a. For y_r = 0.0278, x = y − 0.0278, 4 \ddot{x} + 1430x = 0, and ω_n = 18.91 rad/s.
b. For y_r = 0.091, x = y − 0.091, 14\ddot{x} + 1723x = 0, and ω_n = 11.09 rad/s.

Note that in each case, the mass oscillates about the specific equilibrium position. The oscillation frequency is different for each equilibrium.