Linearizing a Nonlinear Spring Model

Suppose a particular nonlinear spring is described by (13.6.1) with $k_{1} = 1400 N/m$ and $k_{2} = 13000 N/m^{3}$ . Determine its equivalent linear spring constant k for the two equilibrium positions corresponding to the following masses:

$m \ddot{y} = − f (y) + mg = −(k_{1} y + k_{2} y^{3}) + mg$ (13.6.1)

a. m = 4 kg

b. m = 14 kg

In addition, find the linearized equations for motion for each equilibrium.

Question

Linearizing a Nonlinear Spring Model

Suppose a particular nonlinear spring is described by (13.6.1) with [latex]k_{1} = 1400 N/m[/latex] and [latex]k_{2} = 13000 N/m^{3}[/latex]. Determine its equivalent linear spring constant k for the two equilibrium positions corresponding to the following masses:

[latex]m \ddot{y} = − f (y) + mg = −(k_{1} y + k_{2} y^{3}) + mg [/latex] (13.6.1)

a. m = 4 kg

b. m = 14 kg

In addition, find the linearized equations for motion for each equilibrium.

Accepted Answer

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 [latex]y_{r} = 0.0278 m[/latex]. With m = 14 in (13.6.2), the only real root is [latex]y_{r} = 0.091 m[/latex]. Note that [latex]f (y_{r}) = mg[/latex], and that

[latex]k_{1} y_{r} + k_{2} y^{3}_{r} − mg = 0 [/latex] (13.6.2)
[latex]k = \left(\frac{d f}{d y} \right)_{r} = k_{1} + 3k_{2} y^{2}_{r} = 1400 + 39 000 y^{2}_{r} [/latex]
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 [latex]y_{r} = 0.0278, k = 1430 N/m[/latex]
b. For [latex]y_{r} = 0.091, k = 1723 N/m[/latex]
The linearized equations of motion, which are accurate only near their respective equilibrium points, are found from [latex]m \ddot{x} + kx = 0[/latex]. They are
a. For [latex]y_{r} = 0.0278, x = y − 0.0278, 4\ddot{x} + 1430x = 0[/latex], and [latex]ω_{n} = 18.91 rad/s[/latex].
b. For [latex]y_{r} = 0.091, x = y − 0.091, 14\ddot{x} + 1723x = 0[/latex], and [latex]ω_{n} = 11.09 rad/s[/latex].

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

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

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