Hooke's law assumes the restoring force exerted by a spring...

Hooke's law assumes the restoring force exerted by a spring under tension or compression is proportional to the displacement (the distance stretched or foreshortened). This assumption cannot be valid for large displacements since there are limits to the amount a spring can be stretched or compressed. Suppose we assume that the restoring force $F_{R}(x)$ is related to spring displacement $x$ by $$ F_{R}(x)=-\frac{2 k \delta}{\pi} \tan \left(\frac{\pi x}{2 \delta}\right) $$ In this model, the restoring force has vertical asymptotes at $x=\pm \delta$; the value $\delta$ represents the maximum amount the spring can be stretched or compressed. Consider the figure, illustrating a mass $m$ attached to such a spring. Assume that the mass moves on a frictionless horizontal surface and that the spring has unstretched length $l$. Newton's second law of motion leads to the nonlinear differential equation $$ m x^{\prime \prime}+\frac{2 k \delta}{\pi} \tan \left(\frac{\pi x}{2 \delta}\right)=0 . $$ (a) Consider $\tan (\pi x / 2 \delta)$ as a function of $x$ defined on $-\delta<x<\delta$. Expand this function in a Maclaurin series. Show that if we assume $|\pi x / 2 \delta|$ is small and approximate $\tan (\pi x / 2 \delta)$ by the first nonvanishing term in this series, we obtain the linear differential equation found previously when we assumed Hooke's law to be valid. (b) Show that if the first two nonvanishing terms of the Maclaurin expansion are retained, we obtain the differential equation $$ m x^{\prime \prime}+k\left[x+\frac{1}{3}\left(\frac{\pi}{2 \delta}\right)^{2} x^{3}\right]=0 . $$ Equation (14) is often used to model the onset of nonlinear effects and is referred to as modeling a spring-mass system with cubic nonlinearity. (c) Rewrite differential equations (13) and (14) as equivalent first order systems. (d) For each nonlinear system obtained in part (c), determine the points, if any, where the hypotheses of Theorem $6.1$ are not satisfied.

Hooke's law assumes the restoring force exerted by a spring under tension or compression is proportional to the displacement (the distance stretched or foreshortened). This assumption cannot be valid for large displacements since there are limits to the amount a spring can be stretched or compressed. Suppose we assume that the restoring force $F_{R}(x)$ is related to spring displacement $x$ by
$$
F_{R}(x)=-\frac{2 k \delta}{\pi} \tan \left(\frac{\pi x}{2 \delta}\right)
$$
In this model, the restoring force has vertical asymptotes at $x=\pm \delta$; the value $\delta$ represents the maximum amount the spring can be stretched or compressed. Consider the figure, illustrating a mass $m$ attached to such a spring. Assume that the
mass moves on a frictionless horizontal surface and that the spring has unstretched length $l$. Newton's second law of motion leads to the nonlinear differential equation
$$
m x^{\prime \prime}+\frac{2 k \delta}{\pi} \tan \left(\frac{\pi x}{2 \delta}\right)=0 .
$$
(a) Consider $\tan (\pi x / 2 \delta)$ as a function of $x$ defined on $-\delta<x<\delta$. Expand this function in a Maclaurin series. Show that if we assume $|\pi x / 2 \delta|$ is small and approximate $\tan (\pi x / 2 \delta)$ by the first nonvanishing term in this series, we obtain the linear differential equation found previously when we assumed Hooke's law to be valid.
(b) Show that if the first two nonvanishing terms of the Maclaurin expansion are retained, we obtain the differential equation
$$
m x^{\prime \prime}+k\left[x+\frac{1}{3}\left(\frac{\pi}{2 \delta}\right)^{2} x^{3}\right]=0 .
$$
Equation (14) is often used to model the onset of nonlinear effects and is referred to as modeling a spring-mass system with cubic nonlinearity.
(c) Rewrite differential equations (13) and (14) as equivalent first order systems.
(d) For each nonlinear system obtained in part (c), determine the points, if any, where the hypotheses of Theorem $6.1$ are not satisfied.

Key Concepts

Hooke's Law

Hooke's law is a fundamental principle in mechanics stating that the restoring force exerted by an elastic material, such as a spring, is proportional to its displacement from equilibrium. This linear relationship underpins many basic models of oscillatory systems and serves as the starting point for analyzing more complex behaviors when deviations from linearity occur.

Nonlinear Restoring Forces

When displacements become large, the relationship between force and displacement deviates from linearity, leading to nonlinear restoring forces. These nonlinearities capture more realistic behavior in mechanical systems, such as the finite stretch limits of springs, and require alternative modeling strategies that extend beyond simple proportional approximations.

Maclaurin Series Expansion

The Maclaurin series is a method of expanding functions into an infinite sum of terms calculated from the derivatives of the function at zero. This expansion is particularly useful for approximating nonlinear functions near the equilibrium point, allowing complex expressions to be replaced by simpler polynomials, which facilitate analysis and solution of differential equations.

Linearization Techniques

Linearization involves approximating a nonlinear system by a linear one near the equilibrium point, often by retaining only the lowest-order term(s) in a series expansion. This technique simplifies analysis and solution by converting a problem with potentially complex dynamics into a more tractable linear system that retains the essential features of the behavior in the vicinity of equilibrium.

Cubic Nonlinearity

Cubic nonlinearity refers to the inclusion of terms involving the cube of the variable in the model, which captures the first significant correction to linear behavior in systems experiencing moderate deviations from equilibrium. In mechanical systems, incorporating a cubic term helps model the onset of nonlinear effects that influence the system's dynamics, such as changes in natural frequency or amplitude-dependent behaviors.

Conversion to First Order Systems

Many differential equations, particularly second order ones common in mechanics, can be rewritten as equivalent systems of first order equations. This transformation simplifies the application of theoretical tools such as phase plane analysis, numerical integration, and existence and uniqueness theorems, by decomposing the problem into a set of first order problems with standard properties.

Existence and Uniqueness Theorems

Existence and uniqueness theorems, such as the Picard-Lindelöf theorem, provide conditions under which differential equations have a unique solution that depends continuously on initial conditions. These theorems are pivotal in understanding the behavior of dynamical systems, including identifying where the standard assumptions may fail due to singularities or discontinuities, which is crucial when extending linear models to incorporate nonlinear effects.

Transcript

00:01 Okay, so for part a here, we need to show that we can write this system as a second, or as a system of two first order differential equations.

00:11 So we're going to make the substitution that q is going to equal dp, dt.

00:18 So if we let, or if we try to find the derivative of dqdt, dqdt, is going to be d squared p, dt.

00:30 So.

00:32 Now we're going to make this substitution for dqdt.

00:40 So dqdt, we're substituting this in here.

00:47 So we have m times the qdt is equal to negative kp or dqdt is equal to negative k over m times p.

01:01 So now we have our system is going to consist of this equation here and this equation here.

01:09 So we have dp, well, let me have do that in a different color, we have d .p.

01:15 D .t.

01:17 Is equal to q.

01:19 And then we have dq, d .t is equal to negative k over m times p.

01:29 Now for part b, let's draw the phase plane.

01:34 Okay.

01:35 So b, b, so first the null clients, they're just going to be, so when d pdt equals zero, so q equals zero, and then we're going to have negative k over m p equals zero.

01:51 So if we graph the null clients, so q equals zero, well first off, we're going to let p be on the horizontal axis and q is on the vertical axis.

02:08 So q equals 0 is going to be the horizontal line here.

02:12 And then p, well, this just becomes p equals 0, which is the vertical line here.

02:20 So the equilibrium is going to be at 0 here.

02:26 Let me make that actually such that it's going to be like this, should be like this.

02:39 And then we have q and then p.

02:42 So q equals 0, again, that's going to be horizontal line, and then p equals 0 is going to be vertical line.

02:50 So our equilibrium is just going to be at the point 0.

02:56 Okay, so at the origin...