Question 13.6.1: Linearizing a Nonlinear Spring Model Suppose a particular no...

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^{3}_{r}  −  mg = 0        (13.6.2)
k = \left(\frac{d  f}{d y} \right)_{r} = k_{1} + 3k_{2} y^{2}_{r} = 1400 + 39 000 y^{2}_{r}
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