Question

Use the mass-spring analogy to explain the qualitative nature of the solutions to the Rayleigh equation (22) $\mathrm{y}-\left[1-{\left(\mathrm{y}\text{'}\right)}^2\right]\mathrm{y}\text{'}+\mathrm{y}=0$depicted in Figures 4.24 and 4.25.

Short answer

Therefore, the solution to the Rayleigh equation is we can expect a limit cycle.

Step-by-step solution

General form 

The Energy Integral Lemma:

Let y(t) be a solution to the differential equation $\mathrm{y}=\mathrm{f}\left(\mathrm{y}\right)$, where f(y) is a continuous function that does not depend on y’ or the independent variable t. Let F(y) is an indefinite integral of $\mathrm{f}\left(\mathrm{y}\right)$, that is, $\mathrm{f}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{F}\left(\mathrm{y}\right)$. Then the quantity is $\mathrm{E}\left(\mathrm{t}\right):=\frac{1}{2}\mathrm{y}\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{F}\left(\mathrm{y}\left(\mathrm{t}\right)\right)$constant; i.e., $\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{E}\left(\mathrm{t}\right)=0$.

Change of angular momentum:

$\mathrm{m}{\ell }^2\mathrm{\theta }=-\ell \mathrm{mgsin\theta }$ …… (1)

Newton’s rotational law: The rate of change of angular momentum is equal to torque.

The mass–spring oscillator equation:

$$\begin{gathered} {\mathrm{F}}_{\mathrm{ext}}=\left[\mathrm{inertia}\right]\mathrm{y}+\left[\mathrm{damping}\right]\mathrm{y}\text{'}+\left[\mathrm{stiffness}\right]\mathrm{y} \\ =\mathrm{my}+\mathrm{by}\text{'}+\mathrm{ky} \end{gathered}$$…… (2)

Prove the given equation.

Given that, $\mathrm{y}-\left[1-{\left(\mathrm{y}\text{'}\right)}^2\right]\mathrm{y}\text{'}+\mathrm{y}=0$…… (3)

Compare equation (3) with equation (2).

Then, $\mathrm{m}=1,\mathrm{b}={\left(\mathrm{y}\text{'}\right)}^2-1$and $\mathrm{k}=1$.

And the damping coefficient is ${\left(\mathrm{y}\text{'}\right)}^2-1$. It indicates that low velocities are boosted and high velocities are slowed.

So, we can expect a limited cycle.