Chapter 8.4, Problem 5E

Interpretation Introduction

Interpretation:

To show that the average equation for the system is $\text{r' = -}\frac{1}{2}\text{(kr + Fsin}\varphi \text{),}$ $\varphi \text{' = - }\frac{1}{8}{\text{(4a - 3br}}^{\text{2}}\text{+ }\frac{\text{4F}}{\text{r}}\text{cos}\varphi \text{)}$ where $\text{x = r cos(t + }\varphi \text{),}$ $\dot{\text{x}}\text{ = -r sin(t + }\varphi \text{)}$, and the prime denotes differentiation with respect to slow time $\text{T = εt}$ as usual.

Concept Introduction:

• ➢ Use average or slow time equation for the weekly nonlinear oscillator.

• ➢ Determine $\text{r' }$ and $\varphi \text{' }$ from the slow time equation.

Answer to Problem 5E

Solution:

It is shown that the average equations for the system are

$\text{r' = -}\frac{1}{2}\text{(kr + Fsin}\varphi \text{),}$ $\varphi \text{' = - }\frac{1}{8}{\text{(4a - 3br}}^{\text{2}}\text{+ }\frac{\text{4F}}{\text{r}}\text{cos}\varphi \text{)}$.

Explanation of Solution

The general equation of the weekly nonlinear oscillator is

$\ddot{\text{x}}\text{ + x + }\epsilon \text{h(x, }\dot{\text{x}}\text{) = 0}$

The average or slow time equation is

$\text{r' =}\langle \text{hsinθ}\rangle \text{ = }\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\text{h(θ)sinθ dθ}}$

$\text{r}\varphi \text{' =}\langle \text{h cos θ}\rangle \text{ = }\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\text{h(θ) cos θ dθ}}$

The equation for the forced Duffing oscillator in the limit where the forcing, detuning, damping is done all week is as below,

$\ddot{\text{x}}\text{ + x + }\epsilon {\text{(bx}}^3\text{+k}\dot{\text{x}}\text{ - ax - Fcos t) = 0}$

Comparing the above equation with the general equation of the weekly nonlinear oscillator and

${\text{h(x, }\dot{\text{x}}\text{) = (bx}}^3\text{+k}\dot{\text{x}}\text{ - ax - Fcos t)}$

By substituting $\text{(θ - }\varphi \text{)}$ for $\text{t}$

$\text{h(x, }\dot{\text{x}}\text{) = }\left({\text{bx}}^3\text{+k}\dot{\text{x}}\text{ - ax - Fcos (}\theta \text{-}\varphi \text{)}\right)$

Consider

$\text{x = r cosθ}$

By differentiating the above equation

$\dot{\text{x}}\text{ = - r sin θ}$

$\text{h(x, }\dot{\text{x}}\text{) = }\left(\text{b}{\left(\text{rcos}\theta \right)}^3\text{+k}\left(\text{- rsin}\theta \right)\text{ - a}\left(\text{rcos}\theta \right)\text{ - Fcos (}\theta \text{-}\varphi \text{)}\right)$

$\text{h(x, }\dot{\text{x}}\text{) = b}{\left(\text{rcos}\theta \right)}^3\text{-krsin}\theta \text{- arcos}\theta \text{ - Fcos (}\theta \text{-}\varphi \text{)}$

The average or slow time equation is

$\text{r' =}\langle \text{hsinθ}\rangle \text{ = }\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\text{h(θ)sinθ dθ}}$

$\text{r' =}\langle \text{h(x, }\dot{\text{x}}\text{)sinθ}\rangle \text{ = }\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\text{h(x, }\dot{\text{x}}\text{) sin}\theta \text{ dθ}}$

$\text{r' = }\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(\text{b}{\left(\text{rcos}\theta \right)}^3\text{-krsin}\theta \text{- arcos}\theta \text{ - Fcos (}\theta \text{-}\varphi \text{) }\right)\text{ sin }\theta \text{ dθ}}$

$\text{r' = }\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left({\text{br}}^3\text{sin}\left(\theta \right){\text{cos}}^3\left(\theta \right){\text{- krsin}}^2\left(\theta \right)\text{- ar sin}\left(\theta \right)\text{cos}\left(\theta \right)\text{ - Fsin}\left(\theta \right)\text{cos (}\theta \text{-}\varphi \text{) }\right)\text{dθ}}$

Using the identity $\text{cos (}\theta \text{-}\varphi \text{) = cos (}\theta \text{)cos (}\varphi \text{)}+\text{sin (}\theta \text{)sin (}\varphi \text{)}$

$\text{r' = }\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left\{\begin{array}{l}{\text{br}}^3\text{sin}\left(\theta \right){\text{cos}}^3\left(\theta \right){\text{- krsin}}^2\left(\theta \right)\text{- ar sin}\left(\theta \right)\text{cos}\left(\theta \right)\\ \text{- Fsin}\left(\theta \right)\text{cos (}\theta \text{)cos (}\varphi {\text{)-Fsin}}^2\left(\theta \right)\text{sin (}\varphi \text{)}\end{array}\right\}\text{dθ}}$

$\begin{array}{l}\text{r' = }\frac{\text{1}}{\text{2π}}{\displaystyle \underset{\text{0}}{\overset{\text{2π}}{\int }}{\text{br}}^{\text{3}}\text{sin}\left(\text{θ}\right){\text{cos}}^{\text{3}}\left(\text{θ}\right)\text{- }{\displaystyle \underset{\text{0}}{\overset{\text{2π}}{\int }}{\text{krsin}}^{\text{2}}\left(\text{θ}\right)\text{dθ}}}\\ \text{ -}{\displaystyle \underset{\text{0}}{\overset{\text{2π}}{\int }}\text{ar sin}\left(\text{θ}\right)\text{cos}\left(\text{θ}\right)\text{ dθ}}-{\displaystyle \underset{\text{0}}{\overset{\text{2π}}{\int }}\text{ Fsin}\left(\text{θ}\right)\text{cos (θ)cos (}\varphi \text{) dθ}}-{\displaystyle \underset{\text{0}}{\overset{\text{2π}}{\int }}{\text{Fsin}}^{\text{2}}\left(\text{θ}\right)\text{sin (}\varphi \text{)dθ}}\end{array}$

By solving it,

$\text{r' = }\frac{\text{-1}}{\text{2}}\left(\text{kr + F}\sin \varphi \right)$

Similarly,

$\text{r}\varphi \text{' =}\langle \text{hcos θ}\rangle \text{ = }\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\text{h(θ) cos θ dθ}}$

$\text{r}\varphi \text{' =}\langle \text{h(x, }\dot{\text{x}}\text{) cos θ}\rangle \text{ = }\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\text{h(x, }\dot{\text{x}}\text{) cos θ dθ}}$

$\text{r}\varphi \text{' =}\langle \text{h(x, }\dot{\text{x}}\text{) cos θ}\rangle \text{ = }\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left({\text{bx}}^3\text{+k}\dot{\text{x}}\text{ - ax - Fcos (}\theta \text{-}\varphi \text{)}\right)\text{cos θ dθ}}$

By substituting $\text{x = r cosθ}$ and $\dot{\text{x}}\text{ = - r sin θ}$,

$\text{r}\varphi \text{' = }\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(\text{b}{\left(\text{r cosθ}\right)}^3\text{+k}\left(\text{- r sin θ}\right)\text{ - a}\left(\text{r cosθ}\right)\text{ - Fcos (}\theta \text{-}\varphi \text{)}\right)\text{cos θ dθ}}$

$\text{r}\varphi \text{' = }\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left({\text{br}}^3{\text{cos}}^4\left(\text{θ}\right)\text{-krsin }\left(\text{θ}\right)\text{cos}\left(\text{θ}\right){\text{ - arcos}}^2\left(\text{θ}\right){\text{ - Fcos}}^2\left(\text{θ}\right)\text{cos (}\varphi \text{)-Fcos}\left(\text{θ}\right)\text{sin (}\theta \text{)sin (}\varphi \text{)}\right)\text{ dθ}}$

$\text{r}\varphi \text{' = }\frac{1}{2\pi }\left\{\begin{array}{l}{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left({\text{br}}^3{\text{cos}}^4\left(\text{θ}\right)\right)\text{ dθ}}-{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(\text{krsin }\left(\text{θ}\right)\text{cos}\left(\text{θ}\right)\right)\text{ dθ}}-{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left({\text{arcos}}^2\left(\text{θ}\right)\right)\text{ dθ}}\\ \text{ -}{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left({\text{ Fcos}}^2\left(\text{θ}\right)\text{cos (}\varphi \text{)}\right)\text{ dθ}}-{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(\text{Fcos}\left(\text{θ}\right)\text{sin (}\theta \text{)sin (}\varphi \text{)}\right)\text{ dθ}}\end{array}\right\}$

By solving it,

$\text{r}\varphi \text{' = }\frac{3}{8}{\text{br}}^3-\frac{1}{2}\text{ar}-\frac{1}{2}\text{F}\cos \left(\varphi \right)$

$\varphi \text{' = }\frac{3}{8}{\text{br}}^2-\frac{1}{2}\text{a}-\frac{1}{2\text{r}}\text{F}\cos \left(\varphi \right)$

$\varphi \text{' = - }\frac{1}{8}{\text{(4a - 3br}}^{\text{2}}\text{+ }\frac{\text{4F}}{\text{r}}\text{cos}\varphi \text{)}$

Conclusion

It is seen that the average equations for the system are

$\text{r' = -}\frac{1}{2}\text{(kr + Fsin}\varphi \text{),}$ $\varphi \text{' = - }\frac{1}{8}{\text{(4a - 3br}}^{\text{2}}\text{+ }\frac{\text{4F}}{\text{r}}\text{cos}\varphi \text{)}$.