Stationary Responses of Seven Classes of Fractional Vibrations Driven by Sinusoidal Force

Abstract

This paper gives the contributions in three folds. First, we propose fractional phasor motion equations of seven classes of fractional vibrators. Second, we put forward fractional phasor responses to seven classes of fractional vibrators. Third, we bring forward the analytical expressions of stationary responses in time to seven classes of fractional vibration systems driven by sinusoidal force using elementary functions. The present results show that there are obvious effects of fractional orders on the sinusoidal stationary responses to fractional vibrations.

1. Introduction

For a vibration system with the primary mass m, primary damping c, and primary stiffness k, its motion equation is a differential equation in the form

$$\{\begin{array}{c}m{\displaystyle \frac{d^{2}x(t)}{d{t}^2}}+c{\displaystyle \frac{dx(t)}{dt}}+kx(t)=f(t),\\ x(0)=x_0,x^{\prime }(0)=v_0,\end{array}$$

where f(t) is driven force, x(t) is response, x₀ and v₀ are initial conditions. To find response x(t) is to find the solution to the above differential equation.

In vibration engineering, a type of specific forced response x(t) when f(t) = F₀sin(ωt + θ) for large t is essential. Such a response is called sinusoidal stationary response or stationary sinusoidal response. In engineering, a sinusoidal stationary response may be solved with an algebraic method called phasor instead of those for a differential equation. The theory of phasor, introduced by Steinmetz originally for its use in electrical engineering (Steinmetz, [1,2,3], Desoer and Kuh [4], Qiu [5]), has been now a widely used tool for dealing with stationary responses to sinusoidal vibrations in vibrations (see e.g., Den Hartog [6], Nakagawa and Ringo [7], Palley et al. [8], Grote and Antonsson [9], Allemang and Avitabile [10], Soong and Grigoriu [11], Rothbart and Brown [12], Jin and Xia [13], Harris [14], and Li [15], just mentioning a few).

Fractional vibrations gain increasing interests of researchers (see e.g., Rossikhin and Shitikova [16,17,18,19], Rossikhin [20], Shitikova [21,22], Uchaikin [23], Pskhu and Rekhviashvili [24], Zelenev et al. [25], Freundlich [26,27], Zurigat [28], Blaszczyk [29], Blaszczyk et al. [30], Blaszczyk and Ciesielski [31], Al-Rabtah et al. [32], Drozdov [33], Stanislavsky [34], Tofighi [35], Ryabov and Puzenko [36], Achar et al. [37,38,39], Xu et al. [40], Shitikova et al. [41], Askarian et al. [42], and Li [43,44], simply citing a few). A motion equation of a fractional vibrator is a fractional differential equation. Usually, responses of a fractional vibrator are dealt with a special function called the Mittag–Leffler function. Reports with respect to responses to a fractional vibrator using elementary functions are rarely seen.

Recently, Li introduced analytical expressions of responses (free, impulse, step) of six classes (classes I–VI in this paper) of fractional vibrators using elementary functions (Li [15]) and analytical expressions of sinusoidal responses of three classes (classes I-III in this paper) of fractional vibrators (Li [43]). However, he did not study sinusoidal stationary responses using phasor, which is an algebraic method. In the field, reports regarding motion phasor equations or sinusoidal stationary responses to fractional vibrations using fractional phasor are seldom seen. This paper aims at presenting analytical expressions of sinusoidal stationary responses of seven classes of fractional vibrators listed in

Table 1. Seven classes of fractional vibration systems.

Classification	Motion Equations	Ranges Fractional Orders
Class I	$m{\displaystyle \frac{d^{\alpha }{x}_1(t)}{d{t}^{\alpha }}}+k{x}_1(t)=f(t)$	0 < α < 3
Class II	$m{\displaystyle \frac{d^2{x}_2(t)}{d{t}^2}}+c{\displaystyle \frac{d^{\beta }{x}_2(t)}{d{t}^{\beta }}}+k{x}_2(t)=f(t)$	0 < β < 2
Class III	$m{\displaystyle \frac{d^{\alpha }{x}_3(t)}{d{t}^{\alpha }}}+c{\displaystyle \frac{d^{\beta }{x}_3(t)}{d{t}^{\beta }}}+k{x}_3(t)=f(t)$	0 < α < 3, 0 < β < 2
Class IV	$m{\displaystyle \frac{d^{\alpha }{x}_4(t)}{d{t}^{\alpha }}}+k{\displaystyle \frac{d^{\lambda }{x}_4(t)}{d{t}^{\lambda }}}=f(t)$	0 < α < 3, 0 ≤ λ < 1
Class V	$m{\displaystyle \frac{d^2{x}_5(t)}{d{t}^2}}+k{\displaystyle \frac{d^{\lambda }{x}_5(t)}{d{t}^{\lambda }}}=f(t)$	0 ≤ λ < 1
Class VI	$m{\displaystyle \frac{d^{\alpha }{x}_6(t)}{d{t}^{\alpha }}}+c{\displaystyle \frac{d^{\beta }{x}_6(t)}{d{t}^{\beta }}}+k{\displaystyle \frac{d^{\lambda }{x}_6(t)}{d{t}^{\lambda }}}=f(t)$	0 < α < 3, 0 < β < 2, 0 ≤ λ < 1
Class VII	$m{\displaystyle \frac{d^2{x}_7(t)}{d{t}^2}}+c{\displaystyle \frac{d^{\beta }{x}_7(t)}{d{t}^{\beta }}}+k{\displaystyle \frac{d^{\lambda }{x}_7(t)}{d{t}^{\lambda }}}=f(t)$	0 < β < 2, 0 ≤ λ < 1

. The novelty and originality of the present results are in three folds. First, we establish the fractional phasor motion equations of seven classes of fractional vibrators. Second, we establish the expressions of fractional phasor responses to seven classes of fractional vibrators. Third, we propose the sinusoidal stationary responses to seven classes of fractional vibrators only using elementary functions.

The rest of the paper is organized as follows. The fractional phasor is described in Section 2. Fractional phasor equations of motions, fractional phasor responses, and stationary sinusoidal responses of classes I–VII fractional vibration systems are proposed in Section 3, Section 4, Section 5, Section 6, Section 7, Section 8 and Section 9, respectively. Discussions are in Section 10. Conclusions are given in Section 11.

2. Fractional Phasor

In this research, we use the fractional derivative of Weyl’s, referring [45,46,47,48,49,50] and Appendix A for the Weyl fractional derivative.

When writing f(t) = F₀sin(ωt + θ) by

f(t) = Im{F₀exp[i(ωt + θ)]},

(1)

we denote the phasor of f(t) by F. It is given by

F = F₀exp(iθ).

(2)

Using F, we write

f(t) = Im[Fexp(iωt)].

(3)

Lemma 1.

Suppose that y(t) = Y₀sin(ωt + θ). Let Y be the phasor of y(t). It is in the form of Y = Y₀exp(iθ). Then, the phasor of y^(v)(t) (v ≥ 0) is given by (iω)^vY.

Proof. 

Since y^(v)(t) = Im{(iω)^vY₀exp[i(ωt + θ)]} = Im[(iω)^vYexp(iωt)], the phasor of y^(v)(t) is (iω)^vY. The proof is finished. □

Theorem 1.

The phasor of y^(v)(t) can be expressed by

$${(i\omega )}^{v}Y=\left({\omega }^v\cos {\displaystyle \frac{v\pi }{2}}+i{\omega }^v\sin {\displaystyle \frac{v\pi }{2}}\right)Y.$$

(4)

Proof. 

Considering the principal branch of i^v, the above holds. This finishes the proof. □

3. Results Regarding Fractional Vibrators of Class I

3.1. Motion Phasor Equation of Class I Fractional Vibrator

Theorem 2.

The motion phasor equation of a class I fractional vibrator is expressed by

m(iω)^αX₁ + kX₁ = F,

(5)

where X₁ is the phasor of response x₁(t) and F is the phasor of excitation f(t) = F₀sin(ωt + θ).

Proof. 

Expressing both sides of $m{\displaystyle \frac{d^{\alpha }{x}_1(t)}{d{t}^{\alpha }}}+k{x}_1(t)=f(t)$ using phasor yields (5). The proof is finished. □

Considering Theorem 1, the left side of (5) can be expressed by

$$m{(i\omega )}^{\alpha }{X}_1+k{X}_1=\left(m{\omega }^{\alpha }\cos {\displaystyle \frac{\alpha \pi }{2}}+im{\omega }^{\alpha }\sin {\displaystyle \frac{\alpha \pi }{2}}+k\right)X_1.$$

(6)

When writing the right side of the above by

$$\begin{array}{l}\left(m{\omega }^{\alpha }\cos {\displaystyle \frac{\alpha \pi }{2}}+im{\omega }^{\alpha }\sin {\displaystyle \frac{\alpha \pi }{2}}+k\right)X_1\\ =\left[{(i\omega )}^2(-1)m{\omega }^{\alpha -2}\cos {\displaystyle \frac{\alpha \pi }{2}}+(i\omega )m{\omega }^{\alpha -1}\sin {\displaystyle \frac{\alpha \pi }{2}}+k\right]X_1,\end{array}$$

(7)

we express the motion phasor equation of a class I fractional vibrator by

$$m{(i\omega )}^{\alpha }{X}_1+k{X}_1=\left[{(i\omega )}^2(-1)m{\omega }^{\alpha -2}\cos {\displaystyle \frac{\alpha \pi }{2}}+(i\omega )m{\omega }^{\alpha -1}\sin {\displaystyle \frac{\alpha \pi }{2}}+k\right]X_1.$$

(8)

Denote by m_eq1 the equivalent mass for a fractional vibrator of class I. Then, from (6),

$$m_{\mathrm{eq}1}=-m{\omega }^{\alpha -2}\cos {\displaystyle \frac{\alpha \pi }{2}},1<\alpha <3.$$

(9)

Let c_eq1 be the equivalent damping for a fractional vibrator of class I. Then, from (6),

$$c_{\mathrm{eq}1}=m{\omega }^{\alpha -1}\sin {\displaystyle \frac{\alpha \pi }{2}},1<\alpha <3.$$

(10)

Therefore, the motion phasor equation in Theorem 2 is expressed by

$$m{(i\omega )}^{\alpha }{X}_1+k{X}_1={(i\omega )}^2{m}_{\mathrm{eq}1}{X}_1+i\omega c_{\mathrm{eq}1}{X}_1+k{X}_1=F.$$

(11)

Denote by ζ_eq1 the equivalent damping ratio for a fractional vibrator of class I. Define it by

$${\varsigma }_{\mathrm{eq}1}={\displaystyle \frac{c_{\mathrm{eq}1}}{2\sqrt{m_{\mathrm{eq}1}k}}}.$$

(12)

Then,

$${\varsigma }_{\mathrm{eq}1}={\displaystyle \frac{{\omega }^{\frac{\alpha }{2}}\sin \frac{\alpha \pi }{2}}{2{\omega }_n\sqrt{-\cos \frac{\alpha \pi }{2}}}},$$

(13)

where ${\omega }_n=\sqrt{{\displaystyle \frac{k}{m}}}$ is the conventional natural angular frequency with damping free.

Now, we introduce a term of equivalent damping free natural angular frequency. Let it be ω_eqn1. Define it by

$${\omega }_{\mathrm{eqn}1}=\sqrt{{\displaystyle \frac{k}{m_{\mathrm{eq}1}}}}.$$

(14)

Then,

$${\omega }_{\mathrm{eqn}1}={\displaystyle \frac{{\omega }_n}{\sqrt{-{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}}}},1<\alpha <3.$$

(15)

Without losing the generality, we restrict $\left|{\varsigma }_{\mathrm{eq}1}\right|\le 1$ in what follows. Let ω_eqd1 be the equivalent damped natural angular frequency for a fractional vibrator of class I. Then,

$${\omega }_{\mathrm{eqd}1}={\omega }_{\mathrm{eqn}1}\sqrt{1-{\varsigma }_{\mathrm{eq}1}^2},\left|{\varsigma }_{\mathrm{eq}1}\right|\le 1.$$

(16)

Equation (16) can be written by

$${\omega }_{\mathrm{eqd}1}={\displaystyle \frac{{\omega }_n}{\sqrt{-{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}}}}\sqrt{1-{\displaystyle \frac{{\omega }^{\alpha }{\sin }^2\frac{\alpha \pi }{2}}{4{\omega }_n^2\left|\cos \frac{\alpha \pi }{2}\right|}}}.$$

(17)

Using the notations of m_eq1, c_eq1, ζ_eq1, ω_eqn1, and ω_eqd1, we write the motion phasor equation in Theorem 2 by

$${(i\omega )}^2{X}_1+i2{\zeta }_{\mathrm{eqn}1}{\omega }_{\mathrm{eqn}1}{X}_1+{\omega }_{\mathrm{eqn}1}^2{X}_1={\displaystyle \frac{F}{m_{\mathrm{eq}1}}}.$$

(18)

3.2. Frequency Transfer Function of Class I Fractional Vibrator

Let γ_eq1 be the equivalent frequency ratio of a fractional vibrator of class I. Define it by

$${\gamma }_{\mathrm{eq}1}={\displaystyle \frac{\omega }{{\omega }_{\mathrm{eqn}1}}}.$$

(19)

Then,

$${\gamma }_{\mathrm{eq}1}={\displaystyle \frac{\omega \sqrt{{\omega }^{\alpha -2}\left|\cos \frac{\alpha \pi }{2}\right|}}{{\omega }_n}}.$$

(20)

When writing (7) by

$${(i\omega )}^{\alpha }{X}_1+{\omega }_n^2{X}_1=\left({\omega }^{\alpha }\cos {\displaystyle \frac{\alpha \pi }{2}}+i{\omega }^{\alpha }\sin {\displaystyle \frac{\alpha \pi }{2}}+{\omega }_n^2\right)X_1={\displaystyle \frac{F}{m}},$$

(21)

we have the frequency transfer function in phasor, denoted by H₁(ω), in the form

$$H_1(\omega )={\displaystyle \frac{X_1}{F}}={\displaystyle \frac{1}{k\left(1-\frac{{\omega }^{\alpha }}{{\omega }_n^2}\left|\cos \frac{\alpha \pi }{2}\right|+i\frac{{\omega }^{\alpha }}{{\omega }_n^2}\sin \frac{\alpha \pi }{2}\right)}}.$$

(22)

Using γ_eq1 and ζ_eq1, the above can be rewritten by

$$H_1(\omega )={\displaystyle \frac{1}{k\left(1-{\gamma }_{\mathrm{eq}1}^2+i2{\varsigma }_{\mathrm{eq}1}{\gamma }_{\mathrm{eq}1}\right)}}.$$

(23)

The amplitude of H₁(ω) is given by

$$\left|H_1(\omega )\right|={\displaystyle \frac{1/k}{\sqrt{{\left(1-\frac{{\omega }^{\alpha }}{{\omega }_n^2}\left|\cos \frac{\alpha \pi }{2}\right|\right)}^2+{\left(\frac{{\omega }^{\alpha }}{{\omega }_n^2}\sin \frac{\alpha \pi }{2}\right)}^2}}}$$

(24)

or

$$\left|H_1(\omega )\right|={\displaystyle \frac{1/k}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}1}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}1}{\gamma }_{\mathrm{eq}1}\right)}^2}}}.$$

(25)

The phase of H₁(ω) is in the form

$${\phi }_1(\omega )={\tan }^{-1}{\displaystyle \frac{{\omega }^{\alpha }\sin \frac{\alpha \pi }{2}}{{\omega }_n^2-{\omega }^{\alpha }\left|\cos \frac{\alpha \pi }{2}\right|}}$$

(26)

or

$${\phi }_1={\cos }^{-1}{\displaystyle \frac{1-{\gamma }_{\mathrm{eq}1}^2}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}1}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}1}{\gamma }_{\mathrm{eq}1}\right)}^2}}},$$

(27)

or

$${\phi }_1(\omega )={\cos }^{-1}{\displaystyle \frac{1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}1}}\right)}^2}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}1}}\right)}^2\right]}^2+{\left(2{\varsigma }_{\mathrm{eq}1}\frac{\omega }{{\omega }_{\mathrm{eqn}1}}\right)}^2}}}.$$

(28)

3.3. Stationary Response to Sinusoidal Vibration of Class I Fractional Vibrator

Theorem 3.

The response phasor of a class I fractional vibrator is expressed by

$$X_1={\displaystyle \frac{F}{k\left(1-\frac{{\omega }^{\alpha }}{{\omega }_n^2}\left|\cos \frac{\alpha \pi }{2}\right|+i\frac{{\omega }^{\alpha }}{{\omega }_n^2}\sin \frac{\alpha \pi }{2}\right)}}.$$

(29)

Proof. 

According to (22), we have

$$X_1=H_1(\omega )F={\displaystyle \frac{F}{k\left(1-\frac{{\omega }^{\alpha }}{{\omega }_n^2}\left|\cos \frac{\alpha \pi }{2}\right|+i\frac{{\omega }^{\alpha }}{{\omega }_n^2}\sin \frac{\alpha \pi }{2}\right)}}.$$

The proof is finished. □

Equation (29) can be in the form

$$X_1={\displaystyle \frac{F}{k\left(1-{\gamma }_{\mathrm{eq}1}^2+i2{\varsigma }_{\mathrm{eq}1}{\gamma }_{\mathrm{eq}1}\right)}}.$$

(30)

Theorem 4.

The stationary sinusoidal response to a class I fractional vibrator is expressed by

$$\begin{array}{l}{x}_1(t)={\displaystyle \frac{F_0/k}{\sqrt{{\left(1-\frac{{\omega }^{\alpha }}{{\omega }_n^2}\left|\cos \frac{\alpha \pi }{2}\right|\right)}^2+{\left(\frac{{\omega }^{\alpha }}{{\omega }_n^2}\sin \frac{\alpha \pi }{2}\right)}^2}}}\\ \sin \left\{\omega t+\theta -{\cos }^{-1}{\displaystyle \frac{1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}1}}\right)}^2}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}1}}\right)}^2\right]}^2+{\left(2{\varsigma }_{\mathrm{eq}1}\frac{\omega }{{\omega }_{\mathrm{eqn}1}}\right)}^2}}}\right\}.\end{array}$$

(31)

Proof. 

Note that x₁(t) = Im[X₁exp(iωt)]. As X₁ = H₁(ω)F = H₁(ω)F₀exp(iθ), we have the above. The proof is finished. □

When ω = 2, F₀ = 1, θ = 0, m = 1, k = 1, we use Figure 1 to illustrate x₁(t) for α = 1.2, 2, and 2.5. It exhibits that the effect of fractional order α on x₁(t) is considerable.

4. Results in Fractional Vibrators of Class II

4.1. Motion Phasor Equation of Class II Fractional Vibrator

Theorem 5.

Let X₂ be the phasor of response x₂(t) of a class II fractional vibrator. Let F be the phasor of excitation f(t) = F₀sin(ωt + θ). Then, the motion phasor equation of a class II fractional vibrator is expressed by

m(iω)²X₂ + c(iω)^βX₂ + kX₂ = F.

(32)

Proof. 

Using phasor to express both sides of $m\frac{d^2{x}_2(t)}{d{t}^2}+c\frac{d^{\beta }{x}_2(t)}{d{t}^{\beta }}+k{x}_2(t)=f(t)$ produces (32). This completes the proof. □

Considering the principal branch of i^β, we write the left side of (32) by

$$m{(i\omega )}^2{X}_2+c{(i\omega )}^{\beta }{X}_2+k{X}_2=\left(-m{\omega }^2+c{\omega }^{\beta }\cos {\displaystyle \frac{\beta \pi }{2}}+ic{\omega }^{\beta }\sin {\displaystyle \frac{\beta \pi }{2}}+k\right)X_2.$$

(33)

Writing the right side of the above by

$$\begin{array}{l}\left(-m{\omega }^2+c{\omega }^{\beta }\cos {\displaystyle \frac{\beta \pi }{2}}+ic{\omega }^{\beta }\sin {\displaystyle \frac{\beta \pi }{2}}+k\right)X_2\\ =\left[-m{\omega }^2+c{\omega }^{\beta }\cos {\displaystyle \frac{\beta \pi }{2}}+(i\omega )c{\omega }^{\beta -1}\sin {\displaystyle \frac{\beta \pi }{2}}+k\right]X_2\end{array}$$

(34)

yields the motion phasor equation of a class II fractional vibrator in the form

$$m{(i\omega )}^2{X}_2+c{(i\omega )}^{\beta }{X}_2+k{X}_2=\left[\left(m-c{\omega }^{\beta -2}\cos {\displaystyle \frac{\beta \pi }{2}}\right){(i\omega )}^2+(i\omega )c{\omega }^{\beta -1}\sin {\displaystyle \frac{\beta \pi }{2}}+k\right]X_2.$$

(35)

Let m_eq2 be the equivalent mass for a fractional vibrator of class II. Then, from (35),

$$m_{\mathrm{eq}2}=m-c{\omega }^{\beta -2}\cos {\displaystyle \frac{\beta \pi }{2}},0<\beta <2.$$

(36)

Let c_eq2 be the equivalent damping for a fractional vibrator of class II. Then, from (35),

$$c_{\mathrm{eq}2}=c{\omega }^{\beta -1}\sin {\displaystyle \frac{\beta \pi }{2}},0<\beta <2.$$

(37)

Therefore, we can write the motion phasor equation in Theorem 5 by

$$m{(i\omega )}^2{X}_2+c{(i\omega )}^{\beta }{X}_2+k{X}_2={(i\omega )}^2{m}_{\mathrm{eq}2}{X}_2+i\omega c_{\mathrm{eq}2}{X}_2+k{X}_2=F.$$

(38)

Let ζ_eq2 be the equivalent damping ratio for a fractional vibrator of class II. Define it by

$${\varsigma }_{\mathrm{eq}2}={\displaystyle \frac{c_{\mathrm{eq}2}}{2\sqrt{m_{\mathrm{eq}2}k}}}.$$

(39)

Then,

$${\varsigma }_{\mathrm{eq}2}={\displaystyle \frac{\varsigma {\omega }^{\beta -1}\sin \frac{\beta \pi }{2}}{\sqrt{1-\frac{c}{m}{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}}}},$$

(40)

where $\varsigma ={\displaystyle \frac{c}{2\sqrt{mk}}}.$

Let ω_eqn2 be the equivalent damping free natural angular frequency of a class II fractional vibrator. Define it by

$${\omega }_{\mathrm{eqn}2}=\sqrt{{\displaystyle \frac{k}{m_{\mathrm{eq}2}}}}.$$

(41)

Then,

$${\omega }_{\mathrm{eqn}2}={\displaystyle \frac{{\omega }_n}{\sqrt{1-\frac{c}{m}{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}}}}.$$

(42)

Considering practical vibrations, we adopt $\left|{\varsigma }_{\mathrm{eq}2}\right|\le 1$ in what follows. Let ω_eqd2 be the equivalent damped natural angular frequency for a fractional vibrator of class II. Then,

$${\omega }_{\mathrm{eqd}2}={\omega }_{\mathrm{eqn}2}\sqrt{1-{\varsigma }_{\mathrm{eq}2}^2}.$$

(43)

The above implies

$${\omega }_{\mathrm{eqd}2}={\displaystyle \frac{{\omega }_n}{\sqrt{1-\frac{c}{m}{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}}}}\sqrt{1-{\displaystyle \frac{{\varsigma }^2{\omega }^{2(\beta -1)}si{n}^2\frac{\beta \pi }{2}}{1-\frac{c}{m}{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}}}}.$$

(44)

Adopting the notations of m_eq2, c_eq2, ζ_eq2, and ω_eqn2, we write the motion phasor equation in Theorem 5 by

$${(i\omega )}^2{X}_2+i2{\zeta }_{\mathrm{eqn}2}{\omega }_{\mathrm{eqn}2}{X}_2+{\omega }_{\mathrm{eqn}2}^2{X}_2={\displaystyle \frac{F}{m_{\mathrm{eq}2}}}.$$

(45)

4.2. Frequency Transfer Function of Class II Fractional Vibrator

Denote by γ_eq2 the equivalent frequency ratio of a fractional vibrator of class II. It is defined by

$${\gamma }_{\mathrm{eq}2}={\displaystyle \frac{\omega }{{\omega }_{\mathrm{eqn}2}}}.$$

(46)

Then,

$${\gamma }_{\mathrm{eq}2}=\gamma \sqrt{1-{\displaystyle \frac{c}{m}}{\omega }^{\beta -2}\cos {\displaystyle \frac{\beta \pi }{2}}}.$$

(47)

Rewrite (45) by

$$\left[{(i\omega )}^2+i2{\zeta }_{\mathrm{eqn}2}{\omega }_{\mathrm{eqn}2}+{\omega }_{\mathrm{eqn}2}^2\right]X_2={\displaystyle \frac{F}{m_{\mathrm{eq}2}}}.$$

(48)

Let H₂(ω) be the frequency transfer function in phasor for a class II fractional vibrator. Then,

$$H_2(\omega )={\displaystyle \frac{1/k}{1-{\gamma }^2\left(1-\frac{c}{m}{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)+i\frac{2\varsigma {\omega }^{\beta }\sin \frac{\beta \pi }{2}}{{\omega }_n}}}.$$

(49)

By using γ_eq2 and ζ_eq2, the above can be rewritten by

$$H_2(\omega )={\displaystyle \frac{1}{k\left(1-{\gamma }_{\mathrm{eq}2}^2+i2{\varsigma }_{\mathrm{eq}2}{\gamma }_{\mathrm{eq}2}\right)}}.$$

(50)

The amplitude of H₂(ω) is

$$\left|H_2(\omega )\right|={\displaystyle \frac{1/k}{\sqrt{{\left[1-{\gamma }^2\left(1-\frac{c}{m}{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)\right]}^2+{\left(\frac{1}{{\omega }_n}2\varsigma {\omega }^{\beta }\sin \frac{\beta \pi }{2}\right)}^2}}},$$

(51)

which equals to

$$\left|H_2(\omega )\right|={\displaystyle \frac{1/k}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}2}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}2}{\gamma }_{\mathrm{eq}2}\right)}^2}}}.$$

(52)

The phase of H₂(ω) is given by

$${\phi }_2(\omega )={\cos }^{-1}{\displaystyle \frac{1-{\gamma }_{\mathrm{eq}2}^2}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}2}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}2}{\gamma }_{\mathrm{eq}2}\right)}^2}}}.$$

(53)

Equivalently,

$${\phi }_2(\omega )={\cos }^{-1}{\displaystyle \frac{1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}2}}\right)}^2}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}2}}\right)}^2\right]}^2+{\left(2{\varsigma }_{\mathrm{eq}2}\frac{\omega }{{\omega }_{\mathrm{eqn}2}}\right)}^2}}}.$$

(54)

4.3. Stationary Response to Sinusoidal Vibration of Class II Fractional Vibrator

Theorem 6.

The response phasor of a class II fractional vibrator is given by

$$X_2={\displaystyle \frac{F/k}{1-{\gamma }^2\left(1-\frac{c}{m}{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)+i\frac{2\varsigma {\omega }^{\beta }\sin \frac{\beta \pi }{2}}{{\omega }_n}}}.$$

(55)

Proof. 

Since X₂ = H₂(ω)F, according to (49), we see that the above holds. The proof is finished. □

Equation (55) is equivalently given by

$$X_2={\displaystyle \frac{F}{k\left(1-{\gamma }_{\mathrm{eq}2}^2+i2{\varsigma }_{\mathrm{eq}2}{\gamma }_{\mathrm{eq}2}\right)}}.$$

(56)

Theorem 7.

The stationary sinusoidal response to a class II fractional vibrator is given by

$$\begin{array}{l}{x}_2(t)={\displaystyle \frac{F_0/k}{\sqrt{{\left[1-{\gamma }^2\left(1-\frac{c}{m}{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)\right]}^2+{\left(\frac{1}{{\omega }_n}2\varsigma {\omega }^{\beta }\sin \frac{\beta \pi }{2}\right)}^2}}}\\ \sin \left\{\omega t+\theta -{\cos }^{-1}{\displaystyle \frac{1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}2}}\right)}^2}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}2}}\right)}^2\right]}^2+{\left(2{\varsigma }_{\mathrm{eq}2}\frac{\omega }{{\omega }_{\mathrm{eqn}2}}\right)}^2}}}\right\}.\end{array}$$

(57)

Proof. 

Because x₂(t) = Im[X₂exp(iωt)], X₂ = H₂(ω)F, and F = F₀exp(iθ), we have x₂(t) = Im[H₂(ω)F₀exp(iθ)exp(iωt)] = Im{|H₂(ω)|F₀exp(iωt)exp(iθ)exp[−iφ(ω)]}. Thus, the above holds. The proof is finished. □

Some plots of x₂(t) for β = 0.4, 1, and 1.5 when ω = 2, F₀ = 1, θ = 0, m = 1, and k = 1, are indicated in Figure 2, which illustrates that there is significant effect of fractional order β on x₂(t).

5. Results in Class III Fractional Vibrators

5.1. Motion Phasor Equation of Class III Fractional Vibrator

Theorem 8.

Denote by X₃ the phasor of response x₃(t) of a class III fractional vibrator. Let F be the phasor of excitation f(t) = F₀sin(ωt + θ). Then, the motion phasor equation of a class III fractional vibrator is expressed by

m(iω)^αX₃ + c(iω)^βX₃ + kX₃ = F.

(58)

Proof. 

Using phasor to express both sides of $m{\displaystyle \frac{d^{\alpha }{x}_3(t)}{d{t}^{\alpha }}}+c{\displaystyle \frac{d^{\beta }{x}_3(t)}{d{t}^{\beta }}}+k{x}_3(t)=f(t)$ yields (58). This completes the proof. □

Considering the principal branches of i^α and i^β, we express the left side of (58) by

$$\begin{array}{l}m{(i\omega )}^{\alpha }{X}_3+c{(i\omega )}^{\beta }{X}_3+k{X}_3=\left[m{(i\omega )}^{\alpha }+c{(i\omega )}^{\beta }+k\right]X_3\\ \left(m{\omega }^{\alpha }\cos {\displaystyle \frac{\alpha \pi }{2}}+im{\omega }^{\alpha }\sin {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta }\cos {\displaystyle \frac{\beta \pi }{2}}+ic{\omega }^{\beta }\sin {\displaystyle \frac{\beta \pi }{2}}+k\right)X_3\\ =\left[m{\omega }^{\alpha }\cos {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta }\cos {\displaystyle \frac{\beta \pi }{2}}+i\left(m{\omega }^{\alpha }\sin {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta }\sin {\displaystyle \frac{\beta \pi }{2}}\right)+k\right]X_3.\end{array}$$

(59)

The right side of the above can be written by

$$\begin{array}{l}\left[m{\omega }^{\alpha }\cos {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta }\cos {\displaystyle \frac{\beta \pi }{2}}+i\left(m{\omega }^{\alpha }\sin {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta }\sin {\displaystyle \frac{\beta \pi }{2}}\right)+k\right]X_3\\ =\left[-\left(m{\omega }^{\alpha -2}\cos {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta -2}\cos {\displaystyle \frac{\beta \pi }{2}}\right){(i\omega )}^2+\left(m{\omega }^{\alpha -1}\sin {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta -1}\sin {\displaystyle \frac{\beta \pi }{2}}\right)(i\omega )+k\right]X_3.\end{array}$$

(60)

Therefore,

$$\begin{array}{l}\left[m{(i\omega )}^{\alpha }+c{(i\omega )}^{\beta }+k\right]X_3\\ =\left[-\left(m{\omega }^{\alpha -2}\cos {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta -2}\cos {\displaystyle \frac{\beta \pi }{2}}\right){(i\omega )}^2+\left(m{\omega }^{\alpha -1}\sin {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta -1}\sin {\displaystyle \frac{\beta \pi }{2}}\right)(i\omega )+k\right]X_3.\end{array}$$

(61)

Let m_eq3 be the equivalent mass for a class III fractional vibrator. Then, from (61),

$$m_{\mathrm{eq}3}=-\left(m{\omega }^{\alpha -2}\cos {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta -2}\cos {\displaystyle \frac{\beta \pi }{2}}\right).$$

(62)

Let c_eq3 be the equivalent damping of a class III fractional vibrator. Then, from (61),

$$c_{\mathrm{eq}3}=m{\omega }^{\alpha -1}\sin {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta -1}\sin {\displaystyle \frac{\beta \pi }{2}}.$$

(63)

Therefore, we can write the motion phasor equation in Theorem 8 by

$$\left[m{(i\omega )}^{\alpha }+c{(i\omega )}^{\beta }+k\right]X_3=\left[{(i\omega )}^2{m}_{\mathrm{eq}3}+i\omega c_{\mathrm{eq}3}+k\right]X_3=F.$$

(64)

Let ζ_eq3 be the equivalent damping ratio for a fractional vibrator of class III. It is defined by

$${\varsigma }_{\mathrm{eq}3}={\displaystyle \frac{c_{\mathrm{eq}3}}{2\sqrt{m_{\mathrm{eq}3}k}}}.$$

(65)

Then,

$${\varsigma }_{\mathrm{eq}3}={\displaystyle \frac{{\omega }^{\alpha -1}\sin \frac{\alpha \pi }{2}+2\varsigma {\omega }_n{\omega }^{\beta -1}\sin \frac{\beta \pi }{2}}{2{\omega }_n\sqrt{-\left({\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}+2\varsigma {\omega }_n{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)}}}.$$

(66)

Denote by ω_eqn3 the equivalent damping free natural angular frequency of a class III fractional vibrator. It is defined by

$${\omega }_{\mathrm{eqn}3}=\sqrt{{\displaystyle \frac{k}{m_{\mathrm{eq}3}}}}.$$

(67)

Then,

$${\omega }_{\mathrm{eqn}3}={\displaystyle \frac{{\omega }_n}{\sqrt{-\left({\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}+\frac{c}{m}{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)}}}.$$

(68)

From a practice view, we adopt $\left|{\varsigma }_{\mathrm{eq}3}\right|\le 1$ in what follows. Denote by ω_eqd3 the equivalent damped natural angular frequency for a fractional vibrator of class III. Then,

$${\omega }_{\mathrm{eqd}3}={\omega }_{\mathrm{eqn}3}\sqrt{1-{\varsigma }_{\mathrm{eq}3}^2}.$$

(69)

Thus,

$${\omega }_{\mathrm{eqd}3}={\displaystyle \frac{{\omega }_n\sqrt{1-{\left[\frac{{\left({\omega }^{\alpha -1}\sin \frac{\alpha \pi }{2}+2\varsigma {\omega }_n{\omega }^{\beta -1}sin\frac{\beta \pi }{2}\right)}^2}{4{\omega }_n^2\left[-\left({\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}+2\varsigma {\omega }_n{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)\right]}\right]}^2}}{\sqrt{-\left({\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}+\frac{c}{m}{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)}}}.$$

(70)

When using the notations of m_eq3, c_eq3, ζ_eq3, ω_eqn3, and ω_eqd3, we write the motion phasor equation of a class III fractional vibrator by

$${(i\omega )}^2{X}_3+i2{\zeta }_{\mathrm{eqn}3}{\omega }_{\mathrm{eqn}3}{X}_3+{\omega }_{\mathrm{eqn}3}^2{X}_3={\displaystyle \frac{F}{m_{\mathrm{eq}3}}}.$$

(71)

5.2. Frequency Transfer Function of Class III Fractional Vibrator

Let γ_eq3 be the equivalent frequency ratio of a fractional vibrator of class III. Define it by

$${\gamma }_{\mathrm{eq}3}={\displaystyle \frac{\omega }{{\omega }_{\mathrm{eqn}3}}}.$$

(72)

Then,

$${\gamma }_{\mathrm{eq}3}=\gamma \sqrt{-\left({\omega }^{\alpha -2}\cos {\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{c}{m}}{\omega }^{\beta -2}\cos {\displaystyle \frac{\beta \pi }{2}}\right)}.$$

(73)

Let H₃(ω) be the frequency transfer function in phasor for a class III fractional vibrator. Then,

$$\begin{array}{l}{H}_3(\omega )\\ ={\displaystyle \frac{1/k}{1-{\gamma }^2\left({\omega }^{\alpha -2}\left|\cos \frac{\alpha \pi }{2}\right|-2\varsigma {\omega }_n{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)+i\frac{\gamma \left({\omega }^{\alpha -1}\sin \frac{\alpha \pi }{2}+2\varsigma {\omega }_n{\omega }^{\beta -1}\sin \frac{\beta \pi }{2}\right)}{{\omega }_n\left({\omega }^{\alpha -2}\left|\cos \frac{\alpha \pi }{2}\right|-2\varsigma {\omega }_n{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)}}}.\end{array}$$

(74)

Using γ_eq3 and ζ_eq3, the above can be rewritten by

$$H_3(\omega )={\displaystyle \frac{1}{k\left(1-{\gamma }_{\mathrm{eq}3}^2+i2{\varsigma }_{\mathrm{eq}3}{\gamma }_{\mathrm{eq}3}\right)}}.$$

(75)

The amplitude of H₃(ω) is in the form

$$\left|H_3(\omega )\right|={\displaystyle \frac{1/k}{\sqrt{\left\{\begin{array}{l}{\left[1-{\gamma }^2\left({\omega }^{\alpha -2}\left|\cos \frac{\alpha \pi }{2}\right|-2\varsigma {\omega }_n{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)\right]}^2\\ +{\left[\frac{\gamma \left({\omega }^{\alpha -1}\sin \frac{\alpha \pi }{2}+2\varsigma {\omega }_n{\omega }^{\beta -1}sin\frac{\beta \pi }{2}\right)}{{\omega }_n\left({\omega }^{\alpha -2}\left|\cos \frac{\alpha \pi }{2}\right|-2\varsigma {\omega }_n{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)}\right]}^2\end{array}\right\}}}}.$$

(76)

which is equivalent to

$$\left|H_3(\omega )\right|={\displaystyle \frac{1/k}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}3}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}3}{\gamma }_{\mathrm{eq}3}\right)}^2}}}.$$

(77)

The phase of H₃(ω) is given by

$${\phi }_3(\omega )={\cos }^{-1}{\displaystyle \frac{1-{\gamma }_{\mathrm{eq}3}^2}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}3}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}3}{\gamma }_{\mathrm{eq}3}\right)}^2}}}.$$

(78)

Equivalently,

$${\phi }_3(\omega )={\cos }^{-1}{\displaystyle \frac{1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}3}}\right)}^2}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}3}}\right)}^2\right]}^2+{\left(2{\varsigma }_{\mathrm{eq}3}\frac{\omega }{{\omega }_{\mathrm{eqn}3}}\right)}^2}}}.$$

(79)

5.3. Stationary Response to Sinusoidal Vibration of Class III Fractional Vibrator

Theorem 9.

The response phasor of a class III fractional vibrator is expressed by

$$X_3={\displaystyle \frac{F/k}{1-{\gamma }^2\left({\omega }^{\alpha -2}\left|\cos \frac{\alpha \pi }{2}\right|-2\varsigma {\omega }_n{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)+i\frac{\gamma \left({\omega }^{\alpha -1}\sin \frac{\alpha \pi }{2}+2\varsigma {\omega }_n{\omega }^{\beta -1}\sin \frac{\beta \pi }{2}\right)}{{\omega }_n\left({\omega }^{\alpha -2}\left|\cos \frac{\alpha \pi }{2}\right|-2\varsigma {\omega }_n{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)}}}.$$

(80)

Proof. 

As X₃ = H₃(ω)F, according to (74), we have the above. The proof is finished. □

Equation (80) is equivalently to

$$X_3={\displaystyle \frac{F}{k\left(1-{\gamma }_{\mathrm{eq}3}^2+i2{\varsigma }_{\mathrm{eq}3}{\gamma }_{\mathrm{eq}3}\right)}}.$$

(81)

Theorem 10.

The stationary sinusoidal response to a class III fractional vibrator is expressed by

$$\begin{array}{l}{x}_3(t)={\displaystyle \frac{F_0/k}{\sqrt{\left\{\begin{array}{l}{\left[1-{\gamma }^2\left({\omega }^{\alpha -2}\left|\cos \frac{\alpha \pi }{2}\right|-2\varsigma {\omega }_n{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)\right]}^2\\ +{\left[\frac{\gamma \left({\omega }^{\alpha -1}\sin \frac{\alpha \pi }{2}+2\varsigma {\omega }_n{\omega }^{\beta -1}\sin \frac{\beta \pi }{2}\right)}{{\omega }_n\left({\omega }^{\alpha -2}\left|\cos \frac{\alpha \pi }{2}\right|-2\varsigma {\omega }_n{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)}\right]}^2\end{array}\right\}}}}\\ \sin \left\{\omega t+\theta -{\cos }^{-1}{\displaystyle \frac{1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}3}}\right)}^2}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}3}}\right)}^2\right]}^2+{\left(2{\varsigma }_{\mathrm{eq}3}\frac{\omega }{{\omega }_{\mathrm{eqn}3}}\right)}^2}}}\right\}.\end{array}$$

(82)

Proof. 

Since x₃(t) = Im[X₃exp(iωt)], X₃ = H₃(ω)F, and F = F₀exp(iθ), we have x₃(t) = Im{|H₃(ω)|F₀exp(iωt)exp(iθ)exp[−iφ(ω)]}. Therefore, the above is valid. The proof is finished. □

Figure 3 indicates some plots of x₃(t) for different values of α and β when ω = 2, F₀ = 1, θ = 0, m = 1, and k = 1. That shows the considerable effect of α and β on the response x₃(t).

6. Results about Class IV Fractional Vibrators

6.1. Motion Phasor Equation of Class IV Fractional Vibrator

Theorem 11.

Let X₄ be the phasor of response x₄(t) of a class IV fractional vibrator. Denote by F the phasor of excitation f(t) = F₀sin(ωt + θ). Then, the motion phasor equation of a class IV fractional vibrator is in the form

m(iω)^αX₄ + k(iω)^λX₄ = F.

(83)

Proof. 

Using phasor to express both sides of $m{\displaystyle \frac{d^{\alpha }{x}_4(t)}{d{t}^{\alpha }}}+k{\displaystyle \frac{d^{\lambda }{x}_4(t)}{d{t}^{\lambda }}}=f(t)$ yields (83). The proof is finished. □

Taking into account the principal branches of i^α and i^λ, we write the left side of the above equation by

$$\begin{array}{l}m{(i\omega )}^{\alpha }{X}_4+k{(i\omega )}^{\lambda }{X}_4=\left(m{\omega }^{\alpha }\cos \frac{\alpha \pi }{2}+im{\omega }^{\alpha }\sin \frac{\alpha \pi }{2}+k{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}+ik{\omega }^{\lambda }\sin \frac{\lambda \pi }{2}\right)X_4\\ =\left[m{\omega }^{\alpha }\cos \frac{\alpha \pi }{2}+i\left(m{\omega }^{\alpha }\sin \frac{\alpha \pi }{2}+k{\omega }^{\lambda }\sin \frac{\lambda \pi }{2}\right)+k{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}\right]X_4.\end{array}$$

(84)

The right side of the above can be written by

$$\begin{array}{l}\left[m{\omega }^{\alpha }\cos {\displaystyle \frac{\alpha \pi }{2}}+i\left(m{\omega }^{\alpha }\sin {\displaystyle \frac{\alpha \pi }{2}}+k{\omega }^{\lambda }\sin {\displaystyle \frac{\lambda \pi }{2}}\right)+k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}\right]X_4\\ =\left[-\left(m{\omega }^{\alpha -2}\cos {\displaystyle \frac{\alpha \pi }{2}}\right){(i\omega )}^2+\left(m{\omega }^{\alpha -1}\sin {\displaystyle \frac{\alpha \pi }{2}}+k{\omega }^{\lambda -1}\sin {\displaystyle \frac{\lambda \pi }{2}}\right)(i\omega )+k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}\right]X_4.\end{array}$$

(85)

Therefore,

$$\begin{array}{l}m{(i\omega )}^{\alpha }{X}_4+k{(i\omega )}^{\lambda }{X}_4\\ =\left[-\left(m{\omega }^{\alpha -2}\cos {\displaystyle \frac{\alpha \pi }{2}}\right){(i\omega )}^2+\left(m{\omega }^{\alpha -1}\sin {\displaystyle \frac{\alpha \pi }{2}}+k{\omega }^{\lambda -1}\sin {\displaystyle \frac{\lambda \pi }{2}}\right)(i\omega )+k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}\right]X_4.\end{array}$$

(86)

Denote by m_eq4 the equivalent mass for a fractional vibrator of class IV. Then, from (86),

$$m_{\mathrm{eq}4}=m_{\mathrm{eq}1}=-m{\omega }^{\alpha -2}\cos {\displaystyle \frac{\alpha \pi }{2}}.$$

(87)

Let c_eq4 be the equivalent damping for a fractional vibrator of class IV. Then, from (86),

$$c_{\mathrm{eq}4}=m{\omega }^{\alpha -1}\sin {\displaystyle \frac{\alpha \pi }{2}}+k{\omega }^{\lambda -1}\sin {\displaystyle \frac{\lambda \pi }{2}}.$$

(88)

Denote by k_eq4 the equivalent stiffness of a fractional vibrator of class IV. Then, from (86),

$$k_{\mathrm{eq}4}=k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}.$$

(89)

Using m_eq4, c_eq4, and k_eq4, we write the motion phasor equation in Theorem 11 by

$$\left[m{(i\omega )}^{\alpha }+k{(i\omega )}^{\lambda }\right]X_4=\left[{(i\omega )}^2{m}_{\mathrm{eq}4}+i\omega c_{\mathrm{eq}4}+k_{\mathrm{eq}4}\right]X_4=F.$$

(90)

Denote by ζ_eq4 the equivalent damping ratio for a fractional vibrator of class IV. Define it by

$${\varsigma }_{\mathrm{eq}4}={\displaystyle \frac{c_{\mathrm{eq}4}}{2\sqrt{m_{\mathrm{eq}4}{k}_{\mathrm{eq}4}}}}.$$

(91)

Then,

$${\varsigma }_{\mathrm{eq}4}={\displaystyle \frac{m{\omega }^{\alpha -1}\sin \frac{\alpha \pi }{2}+k{\omega }^{\lambda -1}\sin \frac{\lambda \pi }{2}}{2\sqrt{mk{\omega }^{\alpha +\lambda -2}\left|\cos \frac{\alpha \pi }{2}\right|\cos \frac{\lambda \pi }{2}}}}.$$

(92)

Let ω_eqn4 be the equivalent damping free natural angular frequency of a class IV fractional vibrator. Define it by

$${\omega }_{\mathrm{eqn}4}=\sqrt{{\displaystyle \frac{k_{\mathrm{eq}4}}{m_{\mathrm{eq}4}}}}.$$

(93)

Then,

$${\omega }_{\mathrm{eqn}4}={\omega }_n\sqrt{{\displaystyle \frac{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}{-{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}}}}.$$

(94)

From the point of view of practice, we adopt |ζ_eq4| ≤ 1 in what follows. Let ω_eqd4 be the equivalent damped natural angular frequency for a fractional vibrator of class IV. Then,

$${\omega }_{\mathrm{eqd}4}={\omega }_{\mathrm{eqn}4}\sqrt{1-{\varsigma }_{\mathrm{eq}4}^2}.$$

(95)

Thus,

$${\omega }_{\mathrm{eqd}4}={\omega }_n\sqrt{{\displaystyle \frac{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}{-{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}}}}\sqrt{1-{\left({\displaystyle \frac{m{\omega }^{\alpha -1}\sin \frac{\alpha \pi }{2}+k{\omega }^{\lambda -1}\sin \frac{\lambda \pi }{2}}{2\sqrt{mk{\omega }^{\alpha +\lambda -2}\left|\cos \frac{\alpha \pi }{2}\right|\cos \frac{\lambda \pi }{2}}}}\right)}^2}.$$

(96)

Considering the notations of m_eq4, c_eq4, ζ_eq4, ω_eqn4, and ω_eqd4, we write the motion phasor equation of a class IV fractional vibrator by

$${(i\omega )}^2{X}_4+i2{\zeta }_{\mathrm{eqn}4}{\omega }_{\mathrm{eqn}4}{X}_4+{\omega }_{\mathrm{eqn}4}^2{X}_4={\displaystyle \frac{F}{m_{\mathrm{eq}4}}}.$$

(97)

6.2. Frequency Transfer Function of Class IV Fractional Vibrator

Denote by γ_eq4 the equivalent frequency ratio of a fractional vibrator of class IV. Define it by

$${\gamma }_{\mathrm{eq}4}={\displaystyle \frac{\omega }{{\omega }_{\mathrm{eqn}4}}}.$$

(98)

Then,

$${\gamma }_{\mathrm{eq}4}=\gamma \sqrt{{\displaystyle \frac{-{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}}.$$

(99)

Denote by H₄(ω) the frequency transfer function in phasor for a class IV fractional vibrator. Then,

$$H_4(\omega )={\displaystyle \frac{1/k_{\mathrm{eq}4}}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}\left(\begin{array}{l}1-{\gamma }^2\frac{-{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}\\ +i2\gamma \frac{m{\omega }^{\alpha -1}\sin \frac{\alpha \pi }{2}+k{\omega }^{\lambda -1}\sin \frac{\lambda \pi }{2}}{2\sqrt{mk{\omega }^{\alpha +\lambda -2}\left|\cos \frac{\alpha \pi }{2}\right|\cos \frac{\lambda \pi }{2}}}\sqrt{\frac{-{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}\end{array}\right)}}.$$

(100)

With γ_eq4 and ζ_eq4, we write the above by

$$H_4(\omega )={\displaystyle \frac{1}{k_{\mathrm{eq}4}\left(1-{\gamma }_{\mathrm{eq}4}^2+i2{\varsigma }_{\mathrm{eq}4}{\gamma }_{\mathrm{eq}4}\right)}}.$$

(101)

The above can be written by

$$H_4(\omega )={\displaystyle \frac{1}{k{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}\left(\begin{array}{l}1-{\gamma }^2\frac{-{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}\\ +i2\gamma \frac{m{\omega }^{\alpha -1}\sin \frac{\alpha \pi }{2}+k{\omega }^{\lambda -1}\sin \frac{\lambda \pi }{2}}{2\sqrt{mk{\omega }^{\alpha +\lambda -2}\left|\cos \frac{\alpha \pi }{2}\right|\cos \frac{\lambda \pi }{2}}}\sqrt{\frac{-{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}\end{array}\right)}}.$$

(102)

The amplitude of H₄(ω) is

$$\left|H_4(\omega )\right|={\displaystyle \frac{1}{k_{\mathrm{eq}4}}}{\displaystyle \frac{1}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}4}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}4}{\gamma }_{\mathrm{eq}4}\right)}^2}}}.$$

(103)

The phase of H₄(ω) is in the form

$${\phi }_4(\omega )={\cos }^{-1}{\displaystyle \frac{1-{\gamma }_{\mathrm{eq}4}^2}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}4}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}4}{\gamma }_{\mathrm{eq}4}\right)}^2}}}.$$

(104)

6.3. Stationary Response to Sinusoidal Vibration of Class IV Fractional Vibrator

Theorem 12.

The response phasor of a class IV fractional vibrator is expressed by

$$X_4={\displaystyle \frac{F}{k{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}\left(\begin{array}{l}1-{\gamma }^2\frac{-{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}\\ +i2\gamma \frac{m{\omega }^{\alpha -1}\sin \frac{\alpha \pi }{2}+k{\omega }^{\lambda -1}\sin \frac{\lambda \pi }{2}}{2\sqrt{mk{\omega }^{\alpha +\lambda -2}\left|\cos \frac{\alpha \pi }{2}\right|\cos \frac{\lambda \pi }{2}}}\sqrt{\frac{-{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}\end{array}\right)}}.$$

(105)

Proof. 

Since X₄ = H₄(ω)F and (100), we see that the above is true. The proof is finished. □

The above equation can be written by

$$X_4={\displaystyle \frac{F}{k_{\mathrm{eq}4}\left(1-{\gamma }_{\mathrm{eq}4}^2+i2{\varsigma }_{\mathrm{eq}4}{\gamma }_{\mathrm{eq}4}\right)}}.$$

(106)

Theorem 13.

The stationary sinusoidal response to a class IV fractional vibrator is given by

$$\begin{array}{l}{x}_4(t)={\displaystyle \frac{F_0/k_{\mathrm{eq}4}}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}\left|\left(\begin{array}{l}1-{\gamma }^2\frac{-{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}\\ +i2\gamma \frac{m{\omega }^{\alpha -1}sin\frac{\alpha \pi }{2}+k{\omega }^{\lambda -1}sin\frac{\lambda \pi }{2}}{2\sqrt{mk{\omega }^{\alpha +\lambda -2}\left|\cos \frac{\alpha \pi }{2}\right|\cos \frac{\lambda \pi }{2}}}\sqrt{\frac{-{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}\end{array}\right)\right|}}\\ \sin \left\{\omega t+\theta -{\cos }^{-1}{\displaystyle \frac{1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}4}}\right)}^2}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}4}}\right)}^2\right]}^2+{\left(2{\varsigma }_{\mathrm{eq}4}\frac{\omega }{{\omega }_{\mathrm{eqn}4}}\right)}^2}}}\right\}.\end{array}$$

(107)

Proof. 

Since x₄(t) = Im[X₄exp(iωt)], X₄ = H₄(ω)F, F = F₀exp(iθ), and

$${\phi }_4(\omega )={\cos }^{-1}{\displaystyle \frac{1-{\gamma }_{\mathrm{eq}4}^2}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}4}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}4}{\gamma }_{\mathrm{eq}4}\right)}^2}}}={\cos }^{-1}{\displaystyle \frac{1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}4}}\right)}^2}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}4}}\right)}^2\right]}^2+{\left(2{\varsigma }_{\mathrm{eq}4}\frac{\omega }{{\omega }_{\mathrm{eqn}4}}\right)}^2}}},$$

we have x₄(t) = Im{|H₄(ω)|F₀exp(iωt)exp(iθ)exp[−iφ(ω)]}, which equals to (107). The proof is finished. □

Figure 4 shows some plots of x₄(t) for different values of α and λ when ω = 2, F₀ = 1, θ = 0, m = 1, and k = 1. The significant effect of fractional orders α and λ on the response x₄(t) can be easily seen from Figure 4.

7. Results for Class V Fractional Vibrators

7.1. Motion Phasor Equation of Class V Fractional Vibrator

Theorem 14.

Denote by X₅ the phasor of response x₅(t) of a class V fractional vibrator. Let F be the phasor of excitation f(t) = F₀sin(ωt + θ). The motion phasor equation of a class V fractional vibrator is expressed by

m(iω)²X₅ + k(iω)^λX₅ = F.

(108)

Proof. 

When using phasor to express both sides of $m{\displaystyle \frac{d^2{x}_5(t)}{d{t}^2}}+k{\displaystyle \frac{d^{\lambda }{x}_5(t)}{d{t}^{\lambda }}}=f(t),$ we see that the above is true. The proof is finished. □

Using the principal branch of i^λ, we express the left side of the above equation by

$$\begin{array}{l}m{(i\omega )}^2{X}_5+k{(i\omega )}^{\lambda }{X}_5=\left(m{(i\omega )}^2+k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}+ik{\omega }^{\lambda }\sin {\displaystyle \frac{\lambda \pi }{2}}\right)X_5\\ =\left[m{(i\omega )}^2+ik{\omega }^{\lambda }\sin {\displaystyle \frac{\lambda \pi }{2}}+k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}\right]X_5.\end{array}$$

(109)

Rewriting the right side of the above by

$$\begin{array}{l}\left[m{(i\omega )}^2+ik{\omega }^{\lambda }\sin {\displaystyle \frac{\lambda \pi }{2}}+k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}\right]X_5\\ =\left[m{(i\omega )}^2+k{\omega }^{\lambda -1}\sin {\displaystyle \frac{\lambda \pi }{2}}(i\omega )+k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}\right]X_5,\end{array}$$

(110)

we have

$$m{(i\omega )}^2{X}_5+k{(i\omega )}^{\lambda }{X}_5=\left[m{(i\omega )}^2+k{\omega }^{\lambda -1}\sin {\displaystyle \frac{\lambda \pi }{2}}(i\omega )+k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}\right]X_5.$$

(111)

Let m_eq5 be the equivalent mass for a fractional vibrator of class V. Then, from (110),

m_eq5 = m.

(112)

Denote by c_eq5 the equivalent damping for a fractional vibrator of class V. Then, from (110),

$$c_{\mathrm{eq}5}=k{\omega }^{\lambda -1}\sin {\displaystyle \frac{\lambda \pi }{2}}.$$

(113)

Let k_eq5 be the equivalent stiffness of a fractional vibrator of class V. Then, from (110),

$$k_{\mathrm{eq}5}=k_{\mathrm{eq}4}=k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}.$$

(114)

Thus, we can write the motion phasor equation in Theorem 14 by

$$\left[m{(i\omega )}^2+k{(i\omega )}^{\lambda }\right]X_5=\left[{(i\omega )}^{2}m+i\omega c_{\mathrm{eq}5}+k_{\mathrm{eq}5}\right]X_5=F.$$

(115)

Let ζ_eq5 be the equivalent damping ratio for a fractional vibrator of class V. It is defined by

$${\varsigma }_{\mathrm{eq}5}={\displaystyle \frac{c_{\mathrm{eq}5}}{2\sqrt{m{k}_{\mathrm{eq}5}}}}.$$

(116)

Then,

$${\varsigma }_{\mathrm{eq}5}={\displaystyle \frac{{\omega }_n{\omega }^{\frac{\lambda }{2}-1}\sin \frac{\lambda \pi }{2}}{2\sqrt{\cos \frac{\lambda \pi }{2}}}}.$$

(117)

Let ω_eqn5 be the equivalent damping free natural angular frequency of a class V fractional vibrator. Define it by

$${\omega }_{\mathrm{eqn}5}=\sqrt{{\displaystyle \frac{k_{\mathrm{eq}5}}{m}}}.$$

(118)

Then,

$${\omega }_{\mathrm{eqn}5}={\omega }_n\sqrt{{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}}.$$

(119)

From a view of practical vibrations, we use |ζ_eq5| ≤ 1 in what follows. Denote by ω_eqd5 the equivalent damped natural frequency for a fractional vibrator of class V. Then,

$${\omega }_{\mathrm{eqd}5}={\omega }_{\mathrm{eqn}5}\sqrt{1-{\varsigma }_{\mathrm{eq}5}^2}.$$

(120)

Thus,

$${\omega }_{\mathrm{eqd}5}={\omega }_n\sqrt{{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}}\sqrt{1-{\left({\displaystyle \frac{k{\omega }^{\lambda -1}\sin \frac{\lambda \pi }{2}}{2\sqrt{mk{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}}\right)}^2}.$$

(121)

Using the notations of m_eq5, c_eq5, ζ_eq5, ω_eqn5, and ω_eqd5, we write the motion phasor equation of a class V fractional vibrator by

$${(i\omega )}^2{X}_5+i2{\zeta }_{\mathrm{eqn}5}{\omega }_{\mathrm{eqn}5}{X}_5+{\omega }_{\mathrm{eqn}5}^2{X}_5={\displaystyle \frac{F}{m}}.$$

7.2. Frequency Transfer Function of Class V Fractional Vibrator

Let γ_eq5 be the equivalent frequency ratio of a fractional vibrator of class V. Define it by

$${\gamma }_{\mathrm{eq}5}={\displaystyle \frac{\omega }{{\omega }_{\mathrm{eqn}5}}}.$$

(122)

Then,

$${\gamma }_{\mathrm{eq}5}=\gamma \sqrt{{\displaystyle \frac{1}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}}.$$

(123)

Let H₅(ω) be the frequency transfer function in phasor for a class V fractional vibrator. Then,

$$H_5(\omega )={\displaystyle \frac{1/k_{\mathrm{eq}5}}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}\left(1-\frac{{\gamma }^2}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}+i2\gamma \frac{k{\omega }^{\lambda -1}\sin \frac{\lambda \pi }{2}}{2\sqrt{mk{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}\sqrt{\frac{1}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}\right)}}.$$

(124)

Using γ_eq5 and ζ_eq5, we have

$$H_5(\omega )={\displaystyle \frac{1}{k_{\mathrm{eq}5}\left(1-{\gamma }_{\mathrm{eq}5}^2+i2{\varsigma }_{\mathrm{eq}5}{\gamma }_{\mathrm{eq}5}\right)}}.$$

(125)

The amplitude of H₅(ω) is expressed by

$$\left|H_5(\omega )\right|={\displaystyle \frac{1}{k_{\mathrm{eq}5}}}{\displaystyle \frac{1}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}5}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}5}{\gamma }_{\mathrm{eq}5}\right)}^2}}}.$$

(126)

The phase of H₅(ω) is

$${\phi }_5(\omega )={\cos }^{-1}{\displaystyle \frac{1-{\gamma }_{\mathrm{eq}5}^2}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}5}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}5}{\gamma }_{\mathrm{eq}5}\right)}^2}}}.$$

(127)

7.3. Stationary Response to Sinusoidal Vibration of Class V Fractional Vibrator

Theorem 15.

The response phasor of a class V fractional vibrator is given by

$$X_5={\displaystyle \frac{F/k}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}\left(1-\frac{{\gamma }^2}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}+i2\gamma \frac{k{\omega }^{\lambda -1}\sin \frac{\lambda \pi }{2}}{2\sqrt{mk{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}\sqrt{\frac{1}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}\right)}}.$$

(128)

Proof. 

Because X₅ = H₅(ω)F and (124), we see that the above is true. The proof is finished. □

The above equation can be written by

$$X_5={\displaystyle \frac{F}{k_{\mathrm{eq}5}\left(1-{\gamma }_{\mathrm{eq}5}^2+i2{\varsigma }_{\mathrm{eq}5}{\gamma }_{\mathrm{eq}5}\right)}}.$$

(129)

Theorem 16.

The stationary sinusoidal response to a class V fractional vibrator is given by

$$\begin{array}{l}{x}_5(t)={\displaystyle \frac{F_0/k_{\mathrm{eq}5}}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}\left|\left(1-\frac{{\gamma }^2}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}+i2\gamma \frac{k{\omega }^{\lambda -1}\sin \frac{\lambda \pi }{2}}{2\sqrt{mk{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}\sqrt{\frac{1}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}\right)\right|}}\\ \sin \left\{\omega t+\theta -{\cos }^{-1}{\displaystyle \frac{1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}5}}\right)}^2}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}5}}\right)}^2\right]}^2+{\left(2{\varsigma }_{\mathrm{eq}5}\frac{\omega }{{\omega }_{\mathrm{eqn}5}}\right)}^2}}}\right\}.\end{array}$$

(130)

Proof. 

As x₅(t) = Im[X₅exp(iωt)], X₅ = H₅(ω)F, F = F₀exp(iθ), and

$${\phi }_5(\omega )={\cos }^{-1}{\displaystyle \frac{1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}5}}\right)}^2}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}5}}\right)}^2\right]}^2+{\left(2{\varsigma }_{\mathrm{eq}5}\frac{\omega }{{\omega }_{\mathrm{eqn}5}}\right)}^2}}},$$

we have x₅(t) = Im{|H₅(ω)|F₀exp(iωt)exp(iθ)exp[−iφ(ω)]}, which equals to (130). The proof is finished. □

Figure 5 indicates some plots of x₅(t) for different values of λ when ω = 2, F₀ = 1, θ = 0, m = 1, and k = 1. It exhibits the considerable effect of fractional order λ on the response x₅(t).

8. Results Regarding Class VI Fractional Vibrators

8.1. Motion Phasor Equation of Class VI Fractional Vibrator

Theorem 17.

Let X₆ be the phasor of response x₆(t) to a class VI fractional vibrator. Let F be the phasor of excitation f(t) = F₀sin(ωt + θ). The motion phasor equation of a class VI fractional vibrator is given by

m(iω)^αX₆ + c(iω)^βX₆ + k(iω)^λX₆ = F.

(131)

Proof. 

Using phasor to express both sides of $m{\displaystyle \frac{d^{\alpha }{x}_6(t)}{d{t}^{\alpha }}}+c{\displaystyle \frac{d^{\beta }{x}_6(t)}{d{t}^{\beta }}}+k{\displaystyle \frac{d^{\lambda }{x}_6(t)}{d{t}^{\lambda }}}=f(t)$ yields the above. The proof is finished. □

Using the principal branches of i^α, i^β, and i^λ, we write the left side of the above equation by

$$\begin{array}{l}m{(i\omega )}^{\alpha }{X}_6+c{(i\omega )}^{\beta }{X}_6+k{(i\omega )}^{\lambda }{X}_6=\left(m{\omega }^{\alpha }\cos {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta }\cos {\displaystyle \frac{\beta \pi }{2}}\right)X_6\\ +i\left(m{\omega }^{\alpha }\sin {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta }\sin {\displaystyle \frac{\beta \pi }{2}}+k{\omega }^{\lambda }\sin {\displaystyle \frac{\lambda \pi }{2}}\right)X_6+k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}{X}_6.\end{array}$$

(132)

Writing the right side of the above by

$$\begin{array}{l}\left[\begin{array}{l}\left(m{\omega }^{\alpha }\cos {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta }\cos {\displaystyle \frac{\beta \pi }{2}}\right)\\ +i\left(m{\omega }^{\alpha }\sin {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta }\sin {\displaystyle \frac{\beta \pi }{2}}+k{\omega }^{\lambda }\sin {\displaystyle \frac{\lambda \pi }{2}}\right)+k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}\end{array}\right]X_6\\ =\left[\begin{array}{l}-\left(m{\omega }^{\alpha -2}\cos {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta -2}\cos {\displaystyle \frac{\beta \pi }{2}}\right){(i\omega )}^2\\ +i\omega \left(m{\omega }^{\alpha -1}\sin {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta -1}\sin {\displaystyle \frac{\beta \pi }{2}}+k{\omega }^{\lambda -1}\sin {\displaystyle \frac{\lambda \pi }{2}}\right)+k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}\end{array}\right]X_6,\end{array}$$

(133)

we have

$$\begin{array}{l}m{(i\omega )}^{\alpha }{X}_6+c{(i\omega )}^{\beta }{X}_6+k{(i\omega )}^{\lambda }{X}_6\\ =\left[\begin{array}{l}-\left(m{\omega }^{\alpha -2}\cos {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta -2}\cos {\displaystyle \frac{\beta \pi }{2}}\right){(i\omega )}^2\\ +i\omega \left(m{\omega }^{\alpha -1}\sin {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta -1}\sin {\displaystyle \frac{\beta \pi }{2}}+k{\omega }^{\lambda -1}\sin {\displaystyle \frac{\lambda \pi }{2}}\right)+k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}\end{array}\right]X_6.\end{array}$$

(134)

Denote by m_eq6 the equivalent mass for a fractional vibrator of class VI. Then, from the above, we have

$$m_{\mathrm{eq}6}=m_{\mathrm{eq}3}=-\left(m{\omega }^{\alpha -2}\cos {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta -2}\cos {\displaystyle \frac{\beta \pi }{2}}\right).$$

(135)

Let c_eq6 be the equivalent damping for a fractional vibrator of class VI. Then, from (134),

$$c_{\mathrm{eq}6}=c_{\mathrm{eq}6}(\omega ,\alpha ,\beta ,\lambda )=m{\omega }^{\alpha -1}\sin {\displaystyle \frac{\alpha \pi }{2}}+c{\omega }^{\beta -1}\sin {\displaystyle \frac{\beta \pi }{2}}+k{\omega }^{\lambda -1}\sin {\displaystyle \frac{\lambda \pi }{2}}.$$

(136)

Let k_eq6 be the equivalent stiffness of a fractional vibrator of class VI. According to (134), we have

$$k_{\mathrm{eq}6}=k_{\mathrm{eq}5}=k_{\mathrm{eq}4}=k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}.$$

(137)

Now, we write the motion phasor equation in Theorem 17 by

$$m{(i\omega )}^{\alpha }{X}_6+c{(i\omega )}^{\beta }{X}_6+k{(i\omega )}^{\lambda }{X}_6=\left[{(i\omega )}^2{m}_{\mathrm{eq}6}+i\omega c_{\mathrm{eq}6}+k_{\mathrm{eq}5}\right]X_6=F.$$

(138)

Denote by ζ_eq6 the equivalent damping ratio for a fractional vibrator of class VI. Define it by

$${\varsigma }_{\mathrm{eq}6}={\displaystyle \frac{c_{\mathrm{eq}6}}{2\sqrt{m_{\mathrm{eq}6}{k}_{\mathrm{eq}6}}}}.$$

(139)

Then,

$${\varsigma }_{\mathrm{eq}6}={\displaystyle \frac{m{\omega }^{\alpha -1}\sin \frac{\alpha \pi }{2}+c{\omega }^{\beta -1}\sin \frac{\beta \pi }{2}+k{\omega }^{\lambda -1}\sin \frac{\lambda \pi }{2}}{2\sqrt{-\left(m{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}+c{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)k{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}}.$$

(140)

Denote by ω_eqn6 the equivalent damping free natural frequency of a class VI fractional vibrator. It is defined by

$${\omega }_{\mathrm{eqn}6}=\sqrt{{\displaystyle \frac{k_{\mathrm{eq}6}}{m_{\mathrm{eq}6}}}}.$$

(141)

Then,

$${\omega }_{\mathrm{eqn}6}=\sqrt{{\displaystyle \frac{k{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}{-\left(m{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}+c{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)}}}.$$

(142)

From a practice view of vibrations, |ζ_eq6| ≤ 1 is used in what follows. Let ω_eqd6 be the equivalent damped natural angular frequency for a fractional vibrator of class VI. Then,

$${\omega }_{\mathrm{eqd}6}={\omega }_{\mathrm{eqn}6}\sqrt{1-{\varsigma }_{\mathrm{eq}6}^2}.$$

(143)

Thus,

$$\begin{array}{l}{\omega }_{\mathrm{eqd}6}=\sqrt{{\displaystyle \frac{k{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}{-\left(m{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}+c{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)}}}\\ \sqrt{1-{\left({\displaystyle \frac{m{\omega }^{\alpha -1}\sin \frac{\alpha \pi }{2}+c{\omega }^{\beta -1}\sin \frac{\beta \pi }{2}+k{\omega }^{\lambda -1}\sin \frac{\lambda \pi }{2}}{2\sqrt{-\left(m{\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}+c{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)k{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}}\right)}^2}.\end{array}$$

(144)

Taking into account the notations of m_eq6, c_eq6, ζ_eq6, ω_eqn6, and ω_eqd6, we have the motion phasor equation of a class VI fractional vibrator in the form

$${(i\omega )}^2{X}_6+i2{\zeta }_{\mathrm{eqn}6}{\omega }_{\mathrm{eqn}6}{X}_6+{\omega }_{\mathrm{eqn}6}^2{X}_6={\displaystyle \frac{F}{m_{\mathrm{eq}6}}}.$$

(145)

8.2. Frequency Transfer Function of Class VI Fractional Vibrator

Denote by γ_eq6 the equivalent frequency ratio of a fractional vibrator of class VI. It is defined by

$${\gamma }_{\mathrm{eq}6}={\displaystyle \frac{\omega }{{\omega }_{\mathrm{eqn}6}}}.$$

(146)

Then,

$${\gamma }_{\mathrm{eq}6}=\gamma \sqrt{{\displaystyle \frac{-\left({\omega }^{\alpha -2}\cos \frac{\alpha \pi }{2}+2\varsigma {\omega }_n{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}}.$$

(147)

Denote by H₆(ω) the frequency transfer function in phasor for a class VI fractional vibrator. Then,

$$H_6(\omega )={\displaystyle \frac{X_6}{F}}={\displaystyle \frac{1}{k_{\mathrm{eq}6}\left(1-{\gamma }_{\mathrm{eq}6}^2+i2{\varsigma }_{\mathrm{eq}6}{\gamma }_{\mathrm{eq}6}\right)}}.$$

(148)

|H₆(ω)| is given by

$$\left|H_6(\omega )\right|={\displaystyle \frac{1}{k_{\mathrm{eq}6}}}{\displaystyle \frac{1}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}6}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}6}{\gamma }_{\mathrm{eq}6}\right)}^2}}}.$$

(149)

The phase of H₆(ω) is in the form

$${\phi }_6(\omega )={\cos }^{-1}{\displaystyle \frac{1-{\gamma }_{\mathrm{eq}6}^2}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}6}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}6}{\gamma }_{\mathrm{eq}6}\right)}^2}}}.$$

(150)

8.3. Stationary Response to Sinusoidal Vibration of Class VI Fractional Vibrator

Theorem 18.

The response phasor of a class VI fractional vibrator is in the form

$$X_6={\displaystyle \frac{F}{k_{\mathrm{eq}6}\left(1-{\gamma }_{\mathrm{eq}6}^2+i2{\varsigma }_{\mathrm{eq}6}{\gamma }_{\mathrm{eq}6}\right)}}.$$

(151)

Proof. 

Because X₆ = H₆(ω)F and (148), the above is true. The proof is finished. □

Theorem 19.

The stationary sinusoidal response to a class VI fractional vibrator is in the form

$$\begin{array}{l}{x}_6(t)={\displaystyle \frac{1}{k_{\mathrm{eq}6}}}{\displaystyle \frac{F}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}6}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}6}{\gamma }_{\mathrm{eq}6}\right)}^2}}}\\ \sin \left\{\omega t+\theta -{\cos }^{-1}{\displaystyle \frac{1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}6}}\right)}^2}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}6}}\right)}^2\right]}^2+{\left(2{\varsigma }_{\mathrm{eq}6}\frac{\omega }{{\omega }_{\mathrm{eqn}6}}\right)}^2}}}\right\}.\end{array}$$

(152)

Proof. 

Note that x₆(t) = Im[X₆exp(iωt)], X₆ = H₆(ω)F, F = F₀exp(iθ), and

$${\phi }_6(\omega )={\cos }^{-1}{\displaystyle \frac{1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}6}}\right)}^2}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}6}}\right)}^2\right]}^2+{\left(2{\varsigma }_{\mathrm{eq}6}\frac{\omega }{{\omega }_{\mathrm{eqn}6}}\right)}^2}}},$$

(153)

we have x₆(t) = Im{|H₆(ω)|F₀exp(iωt)exp(iθ)exp[−iφ(ω)]}, which equals to (152). The proof is finished. □

Figure 6 illustrates some plots of x₆(t) for different values of α, β, and λ when ω = 2, F₀ = 1, θ = 0, m = 1, and k = 1. The significant effect of fractional orders α, β, and λ on the response x₆(t) can be seen from Figure 6.

9. Results Regarding Class VII Fractional Vibrators

9.1. Motion Phasor Equation of Class VII Fractional Vibrator

Theorem 20.

Let X₇ be the phasor of response x₇(t) to a class VII fractional vibrator. Let F be the phasor of excitation f(t) = F₀sin(ωt + θ). The motion phasor equation of a class VII fractional vibrator is given by

m(iω)²X₇ + c(iω)^βX₇ + k(iω)^λX₇ = F.

(154)

Proof. 

Using phasor to express both sides of $m{\displaystyle \frac{d^2{x}_7(t)}{d{t}^2}}+c{\displaystyle \frac{d^{\beta }{x}_7(t)}{d{t}^{\beta }}}+k{\displaystyle \frac{d^{\lambda }{x}_7(t)}{d{t}^{\lambda }}}=f(t)$ yields the above. The proof is finished. □

Using the principal branches of i^β and i^λ, the left side of the above equation is written by

$$\begin{array}{l}-m{\omega }^2{X}_7+c{(i\omega )}^{\beta }{X}_7+k{(i\omega )}^{\lambda }{X}_7=\left(-m{\omega }^2+c{\omega }^{\beta }\cos {\displaystyle \frac{\beta \pi }{2}}\right)X_7\\ +i\left(c{\omega }^{\beta }\sin {\displaystyle \frac{\beta \pi }{2}}+k{\omega }^{\lambda }\sin {\displaystyle \frac{\lambda \pi }{2}}\right)X_7+k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}{X}_7.\end{array}$$

(155)

We write the right side of the above by

$$\begin{array}{l}\left[\left(-m{\omega }^2+c{\omega }^{\beta }\cos {\displaystyle \frac{\beta \pi }{2}}\right)+i\left(+c{\omega }^{\beta }\sin {\displaystyle \frac{\beta \pi }{2}}+k{\omega }^{\lambda }\sin {\displaystyle \frac{\lambda \pi }{2}}\right)+k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}\right]X_7\\ =\left[\begin{array}{l}-\left(-m+c{\omega }^{\beta -2}\cos {\displaystyle \frac{\beta \pi }{2}}\right){(i\omega )}^2\\ +i\omega \left(+c{\omega }^{\beta -1}\sin {\displaystyle \frac{\beta \pi }{2}}+k{\omega }^{\lambda -1}\sin {\displaystyle \frac{\lambda \pi }{2}}\right)+k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}\end{array}\right]X_7.\end{array}$$

(156)

Therefore,

$$\begin{array}{l}m{(i\omega )}^2{X}_7+c{(i\omega )}^{\beta }{X}_7+k{(i\omega )}^{\lambda }{X}_7\\ =\left[\begin{array}{l}-\left(-m+c{\omega }^{\beta -2}\cos {\displaystyle \frac{\beta \pi }{2}}\right){(i\omega )}^2\\ +i\omega \left(c{\omega }^{\beta -1}\sin {\displaystyle \frac{\beta \pi }{2}}+k{\omega }^{\lambda -1}\sin {\displaystyle \frac{\lambda \pi }{2}}\right)+k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}\end{array}\right]X_7.\end{array}$$

(157)

Denote by m_eq7 the equivalent mass for a fractional vibrator of class VII. Then, we have

$$m_{\mathrm{eq}7}=m_{\mathrm{eq}2}=m-c{\omega }^{\beta -2}\cos {\displaystyle \frac{\beta \pi }{2}}.$$

(158)

Let c_eq7 be the equivalent damping for a fractional vibrator of class VII. Then,

$$c_{\mathrm{eq}7}=c{\omega }^{\beta -1}\sin {\displaystyle \frac{\beta \pi }{2}}+k{\omega }^{\lambda -1}\sin {\displaystyle \frac{\lambda \pi }{2}}.$$

(159)

Let k_eq7 be the equivalent stiffness of a fractional vibrator of class VII. Then,

$$k_{\mathrm{eq}7}=k_{\mathrm{eq}6}=k_{\mathrm{eq}5}=k_{\mathrm{eq}4}=k{\omega }^{\lambda }\cos {\displaystyle \frac{\lambda \pi }{2}}.$$

(160)

Now, write the motion phasor equation of a class VII fractional vibrator by

$$\left[m{(i\omega )}^2+c{(i\omega )}^{\beta }+k{(i\omega )}^{\lambda }\right]X_7=\left[{(i\omega )}^2{m}_{\mathrm{eq}7}+i\omega c_{\mathrm{eq}7}+k_{\mathrm{eq}7}\right]X_7=F.$$

(161)

Let ζ_eq7 be the equivalent damping ratio for a fractional vibrator of class VII. Define it by

$${\varsigma }_{\mathrm{eq}7}={\displaystyle \frac{c_{\mathrm{eq}7}}{2\sqrt{m_{\mathrm{eq}7}{k}_{\mathrm{eq}7}}}}.$$

(162)

Then,

$${\varsigma }_{\mathrm{eq}7}={\displaystyle \frac{c{\omega }^{\beta -1}\sin \frac{\beta \pi }{2}+k{\omega }^{\lambda -1}\sin \frac{\lambda \pi }{2}}{2\sqrt{\left(m-c{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)k{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}}.$$

(163)

Let ω_eqn7 be the equivalent damping free natural angular frequency of a class VII fractional vibrator. Define it by

$${\omega }_{\mathrm{eqn}7}=\sqrt{{\displaystyle \frac{k_{\mathrm{eq}7}}{m_{\mathrm{eq}7}}}}.$$

(164)

Then,

$${\omega }_{\mathrm{eqn}7}=\sqrt{{\displaystyle \frac{k{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}{m-c{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}}}}.$$

(165)

From the point of view of practice vibrations, |ζ_eq7| ≤ 1 is used in what follows. Let ω_eqd7 be the equivalent damped natural angular frequency for a fractional vibrator of class VII. Then,

$${\omega }_{\mathrm{eqd}7}={\omega }_{\mathrm{eqn}7}\sqrt{1-{\varsigma }_{\mathrm{eq}7}^2}.$$

(166)

Therefore,

$$\begin{array}{l}{\omega }_{\mathrm{eqd}7}=\sqrt{{\displaystyle \frac{k{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}{m-c{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}}}}\\ \sqrt{1-{\left({\displaystyle \frac{c{\omega }^{\beta -1}\sin \frac{\beta \pi }{2}+k{\omega }^{\lambda -1}\sin \frac{\lambda \pi }{2}}{2\sqrt{\left(m-c{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}\right)k{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}}\right)}^2}.\end{array}$$

(167)

Considering m_eq7, c_eq7, ζ_eq7, ω_eqn7, and ω_eqd7, we express the motion phasor equation of a class VII fractional vibrator in the form

$${(i\omega )}^2{X}_7+i2{\zeta }_{\mathrm{eqn}7}{\omega }_{\mathrm{eqn}7}{X}_7+{\omega }_{\mathrm{eqn}7}^2{X}_7={\displaystyle \frac{F}{m_{\mathrm{eq}7}}}.$$

(168)

9.2. Frequency Transfer Function of Class VII Fractional Vibrator

Let γ_eq7 be the equivalent frequency ratio of a fractional vibrator of class VII. Define it by

$${\gamma }_{\mathrm{eq}7}={\displaystyle \frac{\omega }{{\omega }_{\mathrm{eqn}7}}}.$$

(169)

Then,

$${\gamma }_{\mathrm{eq}7}=\gamma \sqrt{{\displaystyle \frac{1-2\varsigma {\omega }_n{\omega }^{\beta -2}\cos \frac{\beta \pi }{2}}{{\omega }^{\lambda }\cos \frac{\lambda \pi }{2}}}}.$$

(170)

Let H₇(ω) be the frequency transfer function in phasor for a class VII fractional vibrator. Then,

$$H_7(\omega )={\displaystyle \frac{X_7}{F}}={\displaystyle \frac{1}{k_{\mathrm{eq}7}\left(1-{\gamma }_{\mathrm{eq}7}^2+i2{\varsigma }_{\mathrm{eq}7}{\gamma }_{\mathrm{eq}7}\right)}}.$$

(171)

Consequently,

$$\left|H_7(\omega )\right|={\displaystyle \frac{1}{k_{\mathrm{eq}7}}}{\displaystyle \frac{1}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}7}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}7}{\gamma }_{\mathrm{eq}7}\right)}^2}}}.$$

(172)

The phase of H₇(ω) is in the form

$${\phi }_7(\omega )={\cos }^{-1}{\displaystyle \frac{1-{\gamma }_{\mathrm{eq}7}^2}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}7}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}7}{\gamma }_{\mathrm{eq}7}\right)}^2}}}.$$

(173)

9.3. Stationary Response to Sinusoidal Vibration of Class VII Fractional Vibrator

Theorem 21.

The response phasor of a class VII fractional vibrator is in the form

$$X_7={\displaystyle \frac{F}{k_{\mathrm{eq}7}\left(1-{\gamma }_{\mathrm{eq}7}^2+i2{\varsigma }_{\mathrm{eq}7}{\gamma }_{\mathrm{eq}7}\right)}}.$$

(174)

Proof. 

Because X₇ = H₇(ω)F, the above is valid. This finishes the proof. □

Theorem 22.

The stationary sinusoidal response to a class VII fractional vibrator is expressed by

$$\begin{array}{l}{x}_7(t)={\displaystyle \frac{1}{k_{\mathrm{eq}7}}}{\displaystyle \frac{F}{\sqrt{{\left(1-{\gamma }_{\mathrm{eq}7}^2\right)}^2+{\left(2{\varsigma }_{\mathrm{eq}7}{\gamma }_{\mathrm{eq}7}\right)}^2}}}\\ \sin \left\{\omega t+\theta -{\cos }^{-1}{\displaystyle \frac{1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}7}}\right)}^2}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}7}}\right)}^2\right]}^2+{\left(2{\varsigma }_{\mathrm{eq}7}\frac{\omega }{{\omega }_{\mathrm{eqn}7}}\right)}^2}}}\right\}.\end{array}$$

(175)

Proof. 

Due to x₇(t) = Im[X₇exp(iωt)], X₇ = H₇(ω)F, F = F₀exp(iθ), and

$${\phi }_7(\omega )={\cos }^{-1}{\displaystyle \frac{1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}7}}\right)}^2}{\sqrt{{\left[1-{\left(\frac{\omega }{{\omega }_{\mathrm{eqn}7}}\right)}^2\right]}^2+{\left(2{\varsigma }_{\mathrm{eq}7}\frac{\omega }{{\omega }_{\mathrm{eqn}7}}\right)}^2}}},$$

we have x₇(t) = Im{|H₇(ω)|F₀exp(iωt)exp(iθ)exp[−iφ(ω)]}, which equals to (175). The proof is finished. □

Figure 7 shows some plots of x₇(t) for different values of β and λ when ω = 2, F₀ = 1, θ = 0, m = 1, and k = 1. We can easily see that the effect of fractional orders β and λ on x₇(t) is considerable.

10. Discussions

This paper addresses theoretic research for bringing forward the analytical expressions of sinusoidal stationary responses to seven classes of fractional vibrator with elementary functions by using the method of fractional phasor. The theory of conventional vibrations is mature in applications. However, that of fractional vibrations is quite academic at present. In Li [15], Chap. 14 gives a possible explanation of the Rayleigh damping assumption, which may be taken as an application of fractional vibrations.

With respect to the present results regarding sinusoidal stationary responses, fractional oscillators and fractional resonant circuits in electronic engineering may be an application area, where sinusoidal stationary responses are essential as those in structural engineering. We discuss a possible application with respect to the fatigue test below.

Structural fatigue is a subject that is strongly associated with structural vibration (Palley et al. [8], Lalanne [51,52,53]). Regarding the structural fatigue test, a main form of loading is cyclic load, which is usually taken as a sinusoidal function (Palley et al. [8], Lalanne [51,52,53], Sandor [54], Swanson [55], Daggan and Bryne [56]). Let S(t) be a stress function of cyclic load. Let S_max and S_min be the maximum and minimum of S(t), respectively. Denote by S_m = (S_max + S_min)/2, which is the average of S(t). Denote by S_r = (S_max − S_min). It is called stress range. The quantity R = S_min/S_max is called stress ratio. High cycle fatigue test implies that the cycle of a load is large. Low cycle fatigue test means that the cycle of a load is small.

Suppose that a structure follows the class VI fractional vibrator. Then, a cyclic load used in a fatigue test may take the form of sinusoidal stationary response of a class VI fractional vibrator. The function form of that load, therefore, is x₆(t). Without losing the generality, let S(t) = x₆(t). Different from conventional cyclic load of sinusoidal function, in fatigue test of a structure, S_r and load cycle (or load period) of x₆(t) are considerably affected by fractional order triple (α, β, λ) and vibration frequency ω. In order to make it clear, we show some plots of x₆(t) in the case of R = −1 and S_m = 5 in Figure 8. From Figure 8, we conclude that there are significant effects of (α, β, λ) and ω on S_r of cyclic load x₆(t). Because stress range is crucial to test and evaluation of fatigue performances of a structure [51,52,53,54,55,56], that has to be paid attention to in fatigue test for cyclic loading from the point of view of fractional vibrations.

From Table 1, it is seen that

$${x_6(t)|}_{c=0,\lambda =0}=x_1(t),$$

(176)

$${x_6(t)|}_{\alpha =2,\lambda =0}=x_2(t),$$

(177)

$${x_6(t)|}_{\lambda =0}=x_3(t),$$

(178)

$${x_6(t)|}_{c=0}=x_4(t),$$

(179)

$${x_6(t)|}_{\alpha =2,c=0}=x_5(t),$$

(180)

and

$${x_6(t)|}_{\alpha =2}=x_7(t).$$

(181)

Therefore, the conclusion regarding cyclic load x₆(t) mentioned above is also true, respectively, for x₁(t) (class I fractional vibrators), x₂(t) (class II fractional vibrators), x₃(t) (class III fractional vibrators), x₄(t) (class IV fractional vibrators), x₅(t) (class V fractional vibrators), and x₇(t) (class V fractional vibrators). To distinguish x_j(t) (j = 1, …, 7) from conventional cyclic loads, we call x_j(t) fractional cyclic loads. To be precise, if a structure follows class I fractional vibrators, its cyclic load in fatigue test is affected by fractional order α and vibration frequency ω. When a structure follows class II fractional vibrators, its cyclic load for fatigue test is affected by fractional order β and ω. Suppose that a structure follows class III fractional vibrators. Then, its cyclic load in fatigue test is affected by fractional order (α, β) and ω. If a structure follows class IV fractional vibrators, its cyclic load in fatigue test is affected by (α, λ) and ω. When a structure follows class V fractional vibrators, its cyclic load in fatigue test is affected by λ and ω. Suppose that a structure follows class VII fractional vibrators. Then, its cyclic load in fatigue test is affected by (β, λ) and ω. We shall take fractional fatigue loading as one of our future works.

In passing, we note that random fatigue loading is a crucial issue in fatigue test of structures or materials ([51,52,53,54,55,56], Sunder [57], Shang et al. [58], Swanson [59], and our previous work [60,61,62,63]). Fractionally random fatigue loading may be an interesting research topic.

11. Conclusions

For a conventional vibration system, it is well-known to obtain its sinusoidal stationary response by using phasor. However, for a fractional vibration system, it may not be so easy. As a matter of fact, an equation of motion of a fractional vibrator is a fractional differential one. In order to apply phasor method to treat the issue of sinusoidal stationary response to a fractional vibrator, fractional phasor may be necessary to be used. Based on fractional phasor (Lemma 1), we can transform a motion equation of a fractional vibrator to a fractional phasor equation of motion or fractional phasor motion equation. From that fractional phasor motion equation, we can algebraically obtain its response phasor, which is actually a fractional phasor, and then transform it to a sinusoidal stationary response in time. In this way, a sinusoidal stationary response in time to a fractional vibrator is obtained algebraically.

In this paper, we have introduced seven fractional phasor motion equations of seven classes of fractional vibrators in Theorems 2, 5, 8, 11, 14, 17, and 20, respectively. From those fractional phasor motion equations, we have presented seven analytical expressions of fractional phasor responses for seven classes of fractional vibrators in Theorems 3, 6, 9, 12, 15, 18, and 21, respectively. Then, seven expressions of sinusoidal stationary responses in time to seven classes of fractional vibration systems have been proposed in Theorems 4, 7, 10, 13, 16, 19, and 22, respectively. One thing worth noting is that those sinusoidal stationary responses in time are only expressed by elementary functions. Effects of fractional orders on stationary sinusoidal responses are considerable, as can be seen from illustrations.