Approximate frequency–amplitude relationship for a singular oscillator

Abstract

In this study, He’s frequency–amplitude formulation is modified for the fast estimation of the frequency–amplitude relationship of a nonlinear singular oscillator. The simple approaches give good approximate solutions with short calculation.

Introduction

It is necessary to get a simple and effective approach to nonlinear singular oscillators on their amplitude–period relationship in engineering^1–5 because the amplitude–period relationship is one of the most important properties of a nonlinear oscillator. There are some useful amplitude–period formulae, e.g. He’s amplitude–frequency formula,^6–11 García’ amplitude–frequency formula, ¹² and Suárez-Antola’s amplitude–frequency formula. ¹³ Tian and Liu ¹⁴ gave a system review on various amplitude–frequency formulae. In this paper, we will also apply two modifications of He’s ¹⁵ amplitude–frequency formula to a singular oscillator.

Consider a nonlinear oscillator in the form

$$\ddot{x}+f\left(x\right)=0$$

(1)

with initial conditions

$$x(0)=A\mathrm{and}\dot{x}\left(0\right)=0$$

(2)

where a dot denotes the differentiation with respect to time $t$ , $A$ is a given constant, and $-f(x)$ is a conservative force.

We will treat the problem formed by equations (1) and (2) when $f\left(x\right)=x^{-1}$ .

Modification of He’s frequency–amplitude formula-I

Consider an oscillator where the force is inversely proportional to the dependent variable^16,17

$$\ddot{x}=-x^{-1},x(0)=A,\dot{x}\left(0\right)=0$$

(3)

Rewrite the first equation in system (3) as

$$\ddot{x}x+1=0$$

(4)

Choose two trial solutions in the form ¹⁸

$$x_1=a\text{cos}{\omega }_{1}t+b\text{cos}3{\omega }_{1}t$$

(5)

$$x_2=a\text{cos}{\omega }_{2}t+b\text{cos}3{\omega }_{2}t$$

(6)

The residuals for two trial solutions are, respectively

$$R_1\left(t\right)=-a^2{\omega }_1^2{\text{cos}}^2{\omega }_{1}t-10ab{\omega }_1^2\cos {\omega }_{1}t\cos 3{\omega }_{1}t+9b^2{\omega }_1^2{\text{cos}}^{2}3{\omega }_{1}t+1$$

(7)

$$R_2\left(t\right)=-a^2{\omega }_2^2{\text{cos}}^2{\omega }_{2}t-10ab{\omega }_2^2\cos {\omega }_{2}t\cos 3{\omega }_{2}t+9b^2{\omega }_2^2{\text{cos}}^{2}3{\omega }_{2}t+1$$

(8)

with constants $a$ and $b$ satisfying the relationship

$$a+b=A$$

(9)

For simplicity, we set $t=0,{\omega }_1=1,\text{\hspace{1em}and\hspace{1em}}{\omega }_2=2$ in equations (7) and (8)

$$R_1\left(0\right)=-a^2-10ab+9b^2+1$$

(10)

$$R_2\left(0\right)=-4a^2-40ab+36b^2+1$$

(11)

By He’s frequency–amplitude formulation,^18–20 we have

$$\begin{array}{c}{\omega }^2=\frac{{\omega }_1^2{R}_2(0)-{\omega }_2^2{R}_1(0)}{R_2(0)-R_1(0)}\\ =\frac{3}{3a^2+30ab-27b^2}\end{array}$$

(12)

To obtain an additional equation, we set

$$R\left(t=\frac{2\pi }{\omega N}\right)=0$$

(13)

with $N \text{being }$ a constant. If we choose $N=12$

$$R\left(t=\frac{\pi }{6\omega }\right)=-a^2{\omega }^2\frac{3}{4}+1=0$$

(14)

That is

$$\begin{array}{c}{a}^2{\omega }^2=\frac{4}{3}\\ {\omega }^2=\frac{4}{3}{a}^{-2}\end{array}$$

(15)

Solving equations (9), (12), and (15) simultaneously, we are able to determine $a$ , $b,\text{and }\omega$ . Therefore, we reach the approximate solution $x=a\text{cos}\omega t+b\text{cos}3\omega t$ . In case $b=0,$ we get the same result as equations (5.37) or (5.47) in He ¹⁵ just from equation (15) with good accuracy of 7.8%.

If we assume the solution can be expressed in a more general form^{18, 21} with constants $a_i$ to be determined

$$x(t)={\displaystyle \sum }_{i=1}^M{a}_i\text{cos}\left(\left(2i-1\right)\omega t\right)$$

(16)

With the same idea as above, we will arrive at any accuracy of the frequency of a nonlinear oscillator.

Modification of He’s frequency–amplitude formula-II

Filobello-Nino et al. ²² proposed the basic idea for improving the order of the original equation in their enhanced perturbation method, and this idea was extended to the homotopy perturbation method. ²³ We will extend the idea to He’s frequency–amplitude formula.

Reconsider the problem (3) by improving the order of the equation.

Get equation (4) in this form

$$\ddot{x}x=-1.$$

Differentiating the equation twice with respect to time, we obtain the following equations

$$\begin{array}{c}\ddot{x}\dot{x}+x\stackrel{⃛}{x}=0\\ x^{(4)}x+2\stackrel{⃛}{x}\dot{x}+{\ddot{x}}^2=0\\ x^{(4)}+\frac{1+2\stackrel{⃛}{x}{\stackrel{x}{\hspace{0.17em}}}^2\dot{x}}{x^3}=0\end{array}$$

(17)

Denote

$$F\left(x\right)=\frac{1+2\stackrel{⃛}{x}{x}^2\dot{x}}{x^3}$$

(18)

Suppose the approximate solution of problem (3) be the form

$$x_1=A\text{cos}{\omega }_{1}t$$

(19)

Then, inserting equation (19) into equation (17) leads to the residual

$$\begin{array}{l}{R}_1=F\left(x_1\right)+{\omega }_1^4{x}_1\\ \hspace{1em} =\frac{1-2{\omega }_1^4{A}^4{\text{sin}}^2{\omega }_{1}t{\text{ cos}}^2{\omega }_{1}t}{A^3{\text{cos}}^3{\omega }_{1}t}+{\omega }_1^{4}A\text{ cos}{\omega }_{1}t\\ \hspace{1em} =\frac{1-2{\omega }_1^4{A}^4{\text{cos}}^2{\omega }_{1}t+3{\omega }_1^4{A}^4{\text{ cos}}^4{\omega }_{1}t}{A^3{\text{cos}}^3{\omega }_{1}t}\end{array}$$

(20)

Let

$$x_2=A\text{cos}{\omega }_{2}t$$

(21)

with ${\omega }_1\ne {\omega }_2$ , by the same operations, there comes the residual

$$\begin{array}{l}{R}_2=F\left(x_2\right)+{\omega }_2^4{x}_2\\ \hspace{1em} =\frac{1-2{\omega }_2^4{A}^4{\text{cos}}^2{\omega }_{2}t+3{\omega }_2^4{A}^4{\text{ cos}}^4{\omega }_{2}t}{A^3{\text{cos}}^3{\omega }_{2}t}\end{array}$$

(22)

Specially, assume the location point be $\frac{\text{π}}{4}$ , we arrive respectively at

$$R_1=\frac{4-{\omega }_1^4{A}^4}{\sqrt{2}{A}^3}$$

(23)

$$R_2=\frac{4-{\omega }_2^4{A}^4}{\sqrt{2}{A}^3}$$

(24)

Get along with the modified frequency–amplitude formulation^18–20

$${\omega }^4=\frac{{\omega }_1^4{R}_2-{\omega }_2^4{R}_1}{R_2-R_1}$$

(25)

Taking equations (23) and (24) into equation (25), it is not difficult to get

$$\text{ω}=\frac{\sqrt{2}}{A}$$

(26)

This is the same as equation (5.19) in He. ¹⁵

If the location point happened to be $\frac{\text{π}}{3}$ , then

$$R_1=\frac{16-5{\omega }_1^4{A}^4}{2A^3}$$

(27)

$$R_2=\frac{16-5{\omega }_2^4{A}^4}{2A^3}$$

(28)

From equations (25), (27), and (28), it is easy to have

$$\text{ω}=\frac{2}{\sqrt[4]{5}A}$$

(29)

This result is between equation (5.19) and equation (5.37) in He. ¹⁵

There exists flexibility in the location points. When necessary, different frequency–amplitude relationship results, such as equations (26) and (29) can be utilized to get an average one. Furthermore, combined with equation (19) or (21) we get the acceptable period solution.

$$x=A\text{cos}\omega t$$

(30)

Conclusions

Fast estimation is necessary in engineering. This is especially useful before starting a new project. Sometimes, the simpler the better. Two modifications of He’s frequency–amplitude formula are used here to treat a nonlinear singular oscillator with good solutions. Moreover, more use of estimation^23,24 is available today. We are convinced that the method used here have wider range of application.