*10.2. Stating Research Thought*

Consider the sinusoidal responses to three classes fractional oscillators based on the equivalent oscillation equation in the form

$$\begin{cases} \begin{array}{c} m\_{eq} \frac{d^2 x\_j(t)}{dt^2} + c\_{eq} \frac{dx\_j(t)}{dt} + kx\_j(t) = A \cos \omega t \\\ x\_j(0) = x\_{j0}, \left. \frac{dx\_j(t)}{dt} \right|\_{t=0} = v\_{j0} \end{array}, j = 1, 2, 3. \end{cases} \tag{260}$$

The complete response *xj*(*t*) consists of the zero state response, denoted by *xjzs*(*t*), and zero input response denoted by *xjzi*(*t*), according to the theory of differential equations. Therefore,

$$\mathbf{x}\_{j}(t) = \mathbf{x}\_{j\mathbf{z}s}(t) + \mathbf{x}\_{j\mathbf{z}i}(t),\tag{261}$$

where *xjzi*(*t*) is solved from

$$\begin{cases} m\_{eqj} \frac{d^2 x\_{j\bar{z}i}(t)}{dt^2} + c\_{eqj} \frac{dx\_{j\bar{z}i}(t)}{dt} + kx\_{j\bar{z}i}(t) = 0\\ x\_j(0) = x\_{j0}, \frac{dx\_j(t)}{dt} \Big|\_{t=0} = v\_{j0} \end{cases}, j = 1, 2, 3. \tag{262}$$

On the other hand, *xjzs*(*t*) is the solution to 

$$\begin{cases} \left. m\_{\text{eq}\overline{\rangle}} \frac{d^2 x\_{\text{iso}}(t)}{dt^2} + c\_{\text{eq}\overline{\}} \frac{d x\_{\text{iso}}(t)}{dt} + k x\_{\text{jes}}(t) = A \cos \omega t\\ \left. x\_{\text{j}}(0) = 0, \left. \frac{d x\_{\text{j}}(t)}{dt} \right|\_{t=0} = 0 \end{cases}, j = 1, 2, 3. \end{cases} , j = 1, 2, 3. \tag{263}$$

Note that *xjzi*(*t*) is actually the free response to the fractional oscillators in Class *j*. It has been solved in Section 6. Thus, the focus of this section is on (263).

*10.3. General Form of Sinusoidal Responses to Three Classes of Fractional Oscillators*

The solution to (263) in the general form, for *t* > 0, *j* = 1, 2, 3, is given by

$$\chi\_{jzs}(t) = \frac{A}{m\_{eqj}\omega\_{eqd,j}} \left\{ \frac{\left(\omega\_{equ,j}^2 - \omega^2\right)\cos\omega t + 2\xi\_{eqj}\omega\_{equ,j}\omega\sin\omega t}{ + e^{-\xi\_{eqj}\omega\_{equ,j}t} \left[ \left(\omega\_{equ,j}^2 - \omega^2\right)\cos\omega\_{eqd,j}t - \frac{\xi\_{eqj}\left(\omega\_{equ,j}^2 + \omega^2\right)\sin\omega\_{eqd,j}t}{\sqrt{1 - \epsilon\_{eqj}^2}} \right]}{\left(\omega\_{equ,j}^2 - \omega^2\right)^2 + \left(2\xi\_{eqj}\omega\_{equ,j}\omega\right)^2} . \tag{264}$$

## *10.4. Sinusoidal Response to Fractional Oscillators in Class I*

**Theorem 22** (Sinusoidal response I)**.** *Let <sup>x</sup>*1*zs*(*t*) *be the zero state sinusoidal response to a fractional oscillator in Class I. Then, for t > 0 and 1 < α* ≤ *2, it is in the form*

*<sup>x</sup>*1*zs*(*t*) = ⎛ 1 ⎜⎜⎜⎜⎝ *<sup>m</sup>ωn* <sup>−</sup>*ωα*−<sup>2</sup> cos *απ* 2 1 − *ω<sup>α</sup>* sin<sup>2</sup> *απ*2 <sup>4</sup>*ω*2*n*|cos *απ*2 | ⎞⎟⎟⎟⎟⎠ *A ω*4 *ω*2*n* −*ω<sup>α</sup>* cos *απ*2 −<sup>1</sup><sup>2</sup><sup>+</sup> *ω*4 sin<sup>2</sup> *απ*2 |cos *απ*2 |2 ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩*ω*4 *ω*2*n* −*ω<sup>α</sup>* cos *απ*2 − 12 cos *ωt* + *ω*<sup>2</sup> sin *απ*2 |cos *απ*2 | sin *ωt* <sup>+</sup>*e*<sup>−</sup> *ω* sin *απ*2 <sup>2</sup>|cos *απ*2 | *t*⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣ *ω*4 *ω*2*n* −*ω<sup>α</sup>* cos *απ*2 − 12 cos *<sup>ω</sup>n*!<sup>1</sup><sup>−</sup> *ω<sup>α</sup>* sin<sup>2</sup> *απ*2 <sup>4</sup>*ω*2*n*|cos *απ*2 | *t* √−*ωα*−<sup>2</sup> cos *απ*2 − *ω α*2 sin *απ*2 <sup>2</sup>*ωn*√− cos *απ*2 *ω*4 *ω*2*n* −*ω<sup>α</sup>* cos *απ*2 +12 sin *<sup>ω</sup>n*"##\$<sup>1</sup>− *ω<sup>α</sup>* sin<sup>2</sup> *απ*2 <sup>4</sup>*ω*2*n*|cos *απ*2 | *t* √−*ωα*−<sup>2</sup> cos *απ* " 2 ##\$1−) *ω α*2 sin *απ*2 <sup>2</sup>*ωn*√− cos *απ*2 \*2 ⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ ⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭. (265)

**Proof.** Consider the expression below

$$\chi\_{12i}(t) = \frac{1}{\frac{1}{m\_{\rm eq1}\omega\_{\rm eq1}t}} \left\{ \frac{\left(\omega\_{\rm eq1,1}^2 - \omega^2\right)\cos\omega t + 2\zeta\_{\rm eq1}\omega\_{\rm eq1,1}\omega\sin\omega t}{\left( + e^{-\zeta\_{\rm eq1}\omega\_{\rm eq1,1}t} \left[ \left(\omega\_{\rm eq1,1}^2 - \omega^2\right)\cos\omega\_{\rm eq1,1}t - \frac{\zeta\_{\rm eq1}}{\sqrt{1-\zeta\_{\rm eq1}^2}} \left(\omega\_{\rm eq1,1}^2 + \omega^2\right)\sin\omega t\_{\rm eq1,1}t \right] \right)}{\left(\omega\_{\rm eq1,1}^2 - \omega^2\right) + \left(2\zeta\_{\rm eq1}\omega\_{\rm eq1,1}\omega\right)^2} \right\}. \tag{266}$$

In the above, replacing *meq*1 by the one in Section 4, *ςeq*1, *<sup>ω</sup>eqd*,<sup>1</sup> and *<sup>ω</sup>eqn*,<sup>1</sup> by those in Section 5, respectively, produces (265). This finishes the proof. 

Denote by *<sup>x</sup>*1*zs*,*<sup>s</sup>*(*t*) and *<sup>x</sup>*1*zs*,*tr*(*t*) the steady component and the instantaneous one, respectively. Then, we have

$$\begin{split} \mathbf{x}\_{12:s,s}(t) &= \left[ \frac{1}{m\omega\_{n}\sqrt{-\omega^{a-2}\cos\frac{\alpha\pi}{2}}\sqrt{1-\frac{\omega^{a}\sin^{2}\frac{\alpha\pi}{2}}{4\omega\_{n}^{2}\left|\cos\frac{\alpha\pi}{2}\right|^{2}}}} \frac{A}{\omega^{4}\left(-\frac{\omega\_{n}^{2}}{-\omega^{a}\cos\frac{\alpha\pi}{2}}-1\right)^{2}+\frac{\omega^{4}\sin^{2}\frac{4\pi}{\omega^{2}}}{\left|\cos\frac{\alpha\pi}{2}\right|^{2}}} \right] \\ &\left[ \omega^{4}\left(\frac{\omega\_{n}^{2}}{-\omega^{a}\cos\frac{\alpha\pi}{2}}-1\right)^{2}\cos\omega t + \frac{\omega^{2}\sin\frac{\alpha\pi}{2}}{\left|\cos\frac{\alpha\pi}{2}\right|^{2}}\sin\omega t \right]. \end{split} \tag{267}$$

and

$$\begin{split} \mathbf{x}\_{12,tr}(t) &= \left[ \frac{1}{m\omega\_{n}\sqrt{-\omega^{a-2}\cos\frac{\omega\pi}{4}}\sqrt{1-\frac{\omega^{a}\sin^{2}\frac{\omega\pi}{4}}{4\omega\_{n}^{2}\left|\cos\frac{\omega\pi}{2}\right|}}}} \frac{A}{\omega^{4}\left(\frac{\omega^{2}}{-\omega^{a}\cos\frac{\omega\pi}{4}}-1\right)^{2}+\frac{\omega^{a}\sin^{2}\frac{\omega\pi}{4}}{\left|\cos\frac{\omega\pi}{4}\right|^{2}}} \right] \\ &\left\{ \begin{bmatrix} \omega^{a}\left(\frac{\omega^{2}}{-\omega^{a}\cos\frac{\omega\pi}{4}}-1\right)^{2}\cos\frac{\omega\pi}{4}\sqrt{1-\frac{\omega^{a}\sin^{2}\frac{\omega\pi}{4}}{4\omega\_{n}^{2}\left|\cos\frac{\omega\pi}{4}\right|}}t \\ -\frac{\omega^{\frac{a}{2}}\sin\frac{\omega\pi}{4}}{2\omega\sqrt{-\omega^{a}\cos\frac{\omega\pi}{4}}}\omega^{a}\left(\frac{\omega^{a}}{-\omega^{a}\cos\frac{\omega\pi}{4}}+1\right)^{2}\sin\frac{\omega\pi}{4}\sqrt{1-\frac{\omega^{a}\sin^{2}\frac{\omega\pi}{4}}{4\omega\_{n}^{2}\left|\cos\frac{\omega\pi}{4}\right|^{2}}t}{\sqrt{1-\frac{\omega^{a}\sin\frac{\omega\pi}{4}}{4\omega\_{n}^{2}\left|\cos\frac{\omega\pi}{4}\right|^{2}}} + 1} \end{bmatrix} \right\}. \tag{268}$$

Figure 58 shows the plots of *<sup>x</sup>*1*zs*(*t*).

**Figure 58.** Illustrating *<sup>x</sup>*1*zs*(*t*) for *m* = 1, *k* =9(*<sup>ω</sup>n* = 3), *ω* = 1. Solid line: *α* = 1.9 (*ςeq*<sup>1</sup> = 0.03). Dot line: *α* = 1.6 (*ςeq*<sup>1</sup> = 0.11). (**a**) *t* = 0, 1, ..., 20. (**b**) *t* = 0, 1, ..., 100.

**Note 10.1:** *<sup>x</sup>*1*zs*(*t*) is not a pure harmonic function as can be seen from Figure 58.

**Remark 29.** *We found that the sinusoidal response to fractional oscillators in Class I for any value of α* ∈ *(1, 2) does have steady-state component <sup>x</sup>*1*zs*,*<sup>s</sup>*(*t*) *expressed by (267), also see Figure 59*.

**Figure 59.** Steady-state component, *<sup>x</sup>*1*zs*,*<sup>s</sup>*(*t*), of sinusoidal response to a fractional oscillator in Class I for *m* = 1, *k* =9(*<sup>ω</sup>n* = 3), *ω* = 1. Solid line: *α* = 1.9 (*ςeq*<sup>1</sup> = 0.03). Dot line: *α* = 1.6 (*ςeq*<sup>1</sup> = 0.11). Dash dot line: *α* = 1.3 (*ςeq*<sup>1</sup> = 0.22).

The illustration of *<sup>x</sup>*1*zs*,*tr*(*t*) is indicated in Figure 60.

**Figure 60.** Instantaneous component *<sup>x</sup>*1*zs*,*tr*(*t*) for *m* = 1, *k* =9(*<sup>ω</sup>n* = 3), *ω* = 1. Solid line: *α* = 1.9 (*ςeq*<sup>1</sup> = 0.03). Dot line: *α* = 1.6 (*ςeq*<sup>1</sup> = 0.11). Dash dot line: *α* = 1.3 (*ςeq*<sup>1</sup> = 0.22).

**Note 10.2:** For *α* = 2, we have

$$\begin{split} \mathbf{x}\_{1zs}(t) &= \frac{1}{m\omega\_n} \frac{A}{\omega^4 \left(\frac{\omega\_n^2}{\omega^2} - 1\right)^2} \left[ \omega^4 \left(\frac{\omega\_n^2}{\omega^2} - 1\right)^2 \cos\omega^4 t + \omega^4 \left(\frac{\omega\_n^2}{\omega^2} - 1\right)^2 \cos\omega\_n t \right] \\ &= \frac{A}{m\omega\_n} (\cos\omega^\* t + \cos\omega\_n t). \end{split} \tag{269}$$
