*6.3. Free Response to Fractional Oscillators in Class II*

We state the free response to a fractional oscillator in Class II by Theorem 11.

**Theorem 11** (Free response II)**.** *Denote by <sup>x</sup>*2(*t*) *the free response to a fractional oscillator of Class II type. Then, it is, for t* ≥ *0 and 1 < β* ≤ *2, in the form*

$$\mathbf{x}\_{2}(t) = e^{-\frac{\zeta\omega\_{\text{max}}\beta - 1}{1 - \frac{\pi}{4}\omega^{\beta - 2}\cos\frac{\beta\pi}{\pi}t}} \begin{bmatrix} \begin{matrix} \begin{matrix} \frac{\omega\_{n}}{4} \sqrt{1 - \frac{\omega^{2}\omega^{2}(\beta - 1)\sin^{2}\frac{\beta\pi}{\pi}}{4}}t \\ \sqrt{1 - \frac{\omega\_{n}}{4}\omega^{\beta - 2}\cos\frac{\beta\pi}{\pi}t} \end{matrix} \\\\ \begin{matrix} \frac{\omega\_{n}}{4} \sqrt{1 - \frac{\omega\_{n}}{4}\omega^{\beta - 2}\cos\frac{\beta\pi}{\pi}}t \\\\ \end{matrix} \\\\ \begin{matrix} \begin{matrix} -\frac{\omega\_{n}\omega^{2}(\beta - 1)\sin^{2}\frac{\beta\pi}{\pi}}{4} \\ \end{matrix} \\\\ \sin\left(\frac{\omega\_{n}\sqrt{1 - \frac{\omega^{2}\omega^{2}(\beta - 1)\sin^{2}\frac{\beta\pi}{\pi}}}t}{4\left(n - \omega\beta^{-2}\cos\frac{\beta\pi}{\pi}\right)}t\end{matrix} \\\\ \end{matrix} \end{bmatrix} \tag{176}$$

*Symmetry* **2018**, *10*, 40

**Proof.** Note that, for *t* ≥ 0,

$$\mathbf{x}\_{2}(t) = e^{-\zeta\_{eq2}\omega\_{eqn2}t} \left( \mathbf{x}\_{20}\cos\omega\_{eqd,2}t + \frac{v\_{20} + \zeta\_{eq2}\omega\_{eqn,2}x\_{20}}{\omega\_{eqd,2}}\sin\omega\_{eqd,2}t \right). \tag{177}$$

In the above expression, we replace *ςeq*2, *<sup>ω</sup>eqd*,2, and *<sup>ω</sup>eqn*,<sup>2</sup> by those expressed in Section 5. Then, we have (176). Thus, Theorem 11 holds. 

**Note 6.3:** If *β* = 1, *<sup>x</sup>*2(*t*) degenerates to the ordinary free response to an oscillator with the viscous damper *c*. In fact,

*<sup>x</sup>*2(*t*)|*β*=<sup>1</sup> = *e* − *ςωnωβ*−<sup>1</sup> sin *βπ*2 1− *cm ωβ*−<sup>2</sup> cos *βπ*2 *t* ⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣ *x*20 cos ⎛⎜⎜⎜⎝*ωn*"##\$1− *c*2*ω*<sup>2</sup>(*β*−<sup>1</sup>) sin<sup>2</sup> *βπ*2 4 *<sup>m</sup>*−*<sup>c</sup>ωβ*−<sup>2</sup> cos *βπ*2 *k t* <sup>1</sup>− *cm ωβ*−<sup>2</sup> cos *βπ*2 ⎞⎟⎟⎟⎠ + *<sup>v</sup>*20+ *cωβ*−<sup>1</sup> sin *βπ*2 2 *<sup>m</sup>*−*<sup>c</sup>ωβ*−<sup>2</sup> cos *βπ*2 *x*20 *ωn* "##\$1− *c*2*ω*<sup>2</sup>(*β*−<sup>1</sup>) sin<sup>2</sup> *βπ*2 4 *<sup>m</sup>*−*<sup>c</sup>ωβ*−<sup>2</sup> cos *βπ*2 *k* sin ⎛⎜⎜⎜⎝*ωn*"##\$1− *c*2*ω*<sup>2</sup>(*β*−<sup>1</sup>) sin<sup>2</sup> *βπ*2 4 *<sup>m</sup>*−*<sup>c</sup>ωβ*−<sup>2</sup> cos *βπ*2 *k t* <sup>1</sup>− *cm ωβ*−<sup>2</sup> cos *βπ*2 ⎞⎟⎟⎟⎠ ⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦*β*=1 = *e* <sup>−</sup>*ςωn<sup>t</sup> <sup>x</sup>*20 cos *<sup>ω</sup>n* <sup>1</sup> − *c*2 4*mk t* + *<sup>v</sup>*20+ *c*2*m x*20 *<sup>ω</sup>n* <sup>1</sup><sup>−</sup> *c*2 4*mk* sin *<sup>ω</sup>n* <sup>1</sup> − *c*2 4*mk t* = *e* <sup>−</sup>*ςωn<sup>t</sup><sup>x</sup>*20 cos*<sup>ω</sup>n* <sup>1</sup> − *ς*2*t* + *<sup>v</sup>*20+*ςωnx*<sup>20</sup> *<sup>ω</sup>n*√<sup>1</sup>−*ς*<sup>2</sup> sin*<sup>ω</sup>n* <sup>1</sup> − *<sup>ς</sup>*2*t*.

**Note 6.4:** As far as a fractional oscillator in Class II was concerned, its free response in the closed form is rarely reported. Theorem 11 gives it by using elementary functions.

Let *m* = *c* = *k* = *x*10 = *v*10 = 1, and *ω* = 30. We use Figure 24 to illustrate *<sup>x</sup>*2(*t*).

**Figure 24.** *Cont.*

**Figure 24.** Illustrating free response *<sup>x</sup>*2(*t*) with fixed *ω* when *m* = *c* = *k* = *x*10 = *v*10 = 1. (**a**) *β* = 0.3. Solid line: *ω* = 30 (*ςeq*<sup>2</sup> = 0.02). Dot line: *ω* = 10 (*ςeq*<sup>2</sup> = 0.05). (**b**) *β* = 0.6. Solid line: *ω* = 30 (*ςeq*<sup>2</sup> = 0.10). Dot line: *ω* = 10 (*ςeq*<sup>2</sup> = 0.16). (**c**) *β* = 0.9 Solid line: *ω* = 30 (*ςeq*<sup>2</sup> = 0.35). Dot line: *ω* = 10 (*ςeq*<sup>2</sup> = 0.40). (**d**) *β* = 1. Solid line: *ω* = 30 (*ςeq*<sup>2</sup> = 0.50). Dot line: *ω* = 10 (*ςeq*<sup>2</sup> = 0.50).

Similar to *<sup>x</sup>*1(*t*), *<sup>x</sup>*2(*t*) is also with the argumen<sup>t</sup> *ω*. Its plots with variable *ω* are demonstrated in Figure 25. Figure 26 shows its plots in t-ω plane.

**Figure 25.** Plots of free response *<sup>x</sup>*2(*t*) with *ω* (= 1, 2, ..., 5), *m* = *c* = *k* =1= *x*10 = *v*10 = 1. (**a**) For *β* = 0.2 (0.04 ≤ *ςeq*<sup>2</sup> ≤ 0.70). (**b**) For *β* = 0.4 (0.12 ≤ *ςeq*<sup>2</sup> ≤ 0.67). (**c**) For *β* = 0.6 (0.22 ≤ *ςeq*<sup>2</sup> ≤ 0.63). (**d**) For *β* = 0.8 (0.35 ≤ *ςeq*<sup>2</sup> ≤ 0.57).

**Figure 26.** Plots of free response *<sup>x</sup>*2(*t*) in t-ω plane with *m* = *k* = *x*10 = *v*10 = 1, *c* = 0.5, for *t* = 0, 1, ..., 30; *ω* = 1, 2, ..., 5. (**a**) *β* = 0.3 (0.04 ≤ *ςeq*<sup>2</sup> ≤ 0.15). (**b**) *β* = 0.6 (0.11 ≤ *ςeq*<sup>2</sup> ≤ 0.24). (**c**) *β* = 0.9 (0.21 ≤ *ςeq*<sup>2</sup> ≤ 0.26). (**d**) *β* =1(*ςeq*<sup>2</sup> = 0.25).

**Corollary 11** (Decrement II)**.** *Denote by <sup>x</sup>*2(*t*) *the free response to a fractional oscillator in Class II. Then, for 0 < β* ≤ *1, its logarithmic decrement* <sup>Δ</sup>*eq*2, *is in the form*

$$\Delta\_{eq2} = \frac{2\pi}{\sqrt{1 - \left(\frac{\xi\omega^{\beta - 1}\sin\frac{\beta\pi}{2}}{\sqrt{1 - \frac{\xi}{\pi}\omega^{\beta - 2}\cos\frac{\beta\pi}{2}}}\right)^2} \frac{\xi\omega^{\beta - 1}\sin\frac{\beta\pi}{2}}{\sqrt{1 - \frac{\xi}{\pi}\omega^{\beta - 2}\cos\frac{\beta\pi}{2}}}}.\tag{178}$$

**Proof.** According to (174), we have

$$
\Delta\_{eq2} = \ln \frac{x\_2(t\_i)}{x\_2(t\_{i+1})} = \frac{2\pi \varepsilon\_{eq2}}{\sqrt{1 - \varepsilon\_{eq2}^2}}.\tag{179}
$$

.

Replacing the above *ςeq*<sup>2</sup> with that in (145) produces

$$\Delta\_{cq2} = \frac{2\pi\varsigma\_{cq2}}{\sqrt{1-\varsigma\_{cq2}^2}} = \frac{2\pi}{\sqrt{1-\left(\frac{\varsigma\_{cq^{\beta-1}}\sin\frac{\beta\pi}{2}}{\sqrt{1-\frac{\varsigma\_{cq^{\beta-1}}\sin\frac{\beta\pi}{2}}{\sqrt{1-\frac{\varsigma\_{cq^{\beta-2}}\cos\frac{\beta\pi}{2}}}}}\right)^2}\frac{\varsigma\omega^{\beta^{-1}}\sin\frac{\beta\pi}{2}}{\sqrt{1-\frac{\varsigma\_{cq}}{m}\omega^{\beta-2}\cos\frac{\beta\pi}{2}}}},$$

This finishes the proof. 

> Similar to <sup>Δ</sup>*eq*1, we may write <sup>Δ</sup>*eq*<sup>2</sup> with <sup>Δ</sup>*eq*<sup>2</sup>(*<sup>ω</sup>*, *β*). Figure 27 indicates its plots.

**Figure 27.** Logarithmic decrement <sup>Δ</sup>*eq*<sup>2</sup> of fractional oscillator in Class II for *m* = *c* = 1. Solid line: *β* = 0.9. Dot line: *β* = 0.6. (**a**) For *ωn* = 1. (**b**) For *ωn* = 5. (**c**) For *ωn* = 10.

**Note 6.5:** <sup>Δ</sup>*eq*<sup>2</sup> reduces to the conventional logarithmic decrement if *β* = 1, because

$$\Delta\_{eq2}|\_{\beta=1} = \left[ \frac{\frac{2\pi\xi\omega^{\beta-1}\sin\frac{\beta\pi}{2}}{\sqrt{1-\frac{\xi}{m}\omega^{\beta-2}\cos\frac{\beta\pi}{2}}}}{\sqrt{1-\left(\frac{\xi\omega^{\beta-1}\sin\frac{\beta\pi}{2}}{\sqrt{1-\frac{\xi}{m}\omega^{\beta-2}\cos\frac{\beta\pi}{2}}}\right)^2}} \right]\_{\beta=1} = \frac{2\pi\xi}{\sqrt{1-\xi^2}}.\tag{180}$$

*6.4. Free Response to Fractional Oscillators in Class III*

We now present the free response to a fractional oscillator in Class III by Theorem 12.

**Theorem 12** (Free response III)**.** *Let <sup>x</sup>*3(*t*) *be the free response to a fractional oscillator in Class III. Then, for t* ≥ *0, 1< α* ≤ *2, 0 < β* ≤ *1, it is given by*

*<sup>x</sup>*3(*t*) = *e* − *mωα*−<sup>1</sup> sin *απ*2 <sup>+</sup>*cωβ*−<sup>1</sup> sin *βπ*2 <sup>2</sup>(*mωα*−<sup>2</sup>|cos *απ*2 |−*cωβ*−<sup>2</sup> cos *βπ*2 ) *t* ⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣ *x*30 cos⎛⎜⎜⎜⎜⎜⎜⎝*ωn*"####\$1− *mωα*−1 sin *απ*2 <sup>+</sup>*cωβ*−<sup>1</sup> sin *βπ*2 2 <sup>4</sup>− *mωα*−<sup>2</sup> cos *απ*2 <sup>+</sup>*cωβ*−<sup>2</sup> cos *βπ*2 *k t* −*ωα*−<sup>2</sup> cos *απ*2 + *cm ωβ*−<sup>2</sup> cos *βπ*2 ⎞⎟⎟⎟⎟⎟⎟⎠ +⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩ *<sup>v</sup>*30+ *mωα*−<sup>1</sup> sin *απ*2 <sup>+</sup>*cωβ*−<sup>1</sup> sin *βπ*2 2 *mωα*−<sup>2</sup>|cos *απ*2 |−*cωβ*−<sup>2</sup> cos *βπ*2 *x*30 ⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩ *ωn ωα*−<sup>2</sup>|cos *απ*2 |− *cm ωβ*−<sup>2</sup> cos *βπ*2 "##\$1 − *mωα*−<sup>1</sup> sin *απ*2 +*cωβ*−<sup>1</sup> sin *βπ*2 2 <sup>4</sup>−*mωα*−<sup>2</sup> cos *απ*2 +*cωβ*−<sup>2</sup> cos *βπ*2 *k* ⎫⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎭ ⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭ sin⎛⎜⎜⎜⎜⎜⎜⎝*ωn*"####\$1− *mωα*−1 sin *απ*2 <sup>+</sup>*cωβ*−<sup>1</sup> sin *βπ*2 2 <sup>4</sup>− *mωα*−<sup>2</sup> cos *απ*2 <sup>+</sup>*cωβ*−<sup>2</sup> cos *βπ*2 *k t* −*ωα*−<sup>2</sup> cos *απ*2 + *cm ωβ*−<sup>2</sup> cos *βπ*2 ⎞⎟⎟⎟⎟⎟⎟⎠ ⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦. (181)

**Proof.** Note that, for *t* ≥ 0,

$$\mathbf{x}\_3(t) = e^{-\zeta\_{\rm eq}\omega t\_{\rm eq}\chi t} \left( \mathbf{x}\_{30} \cos \omega\_{\rm eqd,3} t + \frac{\upsilon\_{30} + \zeta\_{\rm eq}\omega \omega\_{\rm eqn,3} \mathbf{x}\_{30}}{\omega\_{\rm eqd,3}} \sin \omega\_{\rm eqd,3} t \right). \tag{182}$$

In (182), when substituting *ςeq*3, *<sup>ω</sup>eqd*,3, and *<sup>ω</sup>eqn*,<sup>3</sup> with those explained in Section 5, we have (181). The proof finishes. 

**Note 6.6:** If (*<sup>α</sup>*, *β*) = (2, 1), *<sup>x</sup>*3(*t*) returns to be the free response to an ordinary oscillator with the viscous damping. As a matter of fact,

$$\left.x\_3(t)\right|\_{a=2, \emptyset=1} = e^{-\xi\omega\_n t} \left(x\_{30}\cos\omega\_d t + \frac{v\_{30} + \xi\omega\_n x\_{30}}{\omega\_d}\sin\omega\_d t\right),\tag{183}$$

where *ωd* = *<sup>ω</sup>n* <sup>1</sup> − *ς*2.

> Figure 28 indicates *<sup>x</sup>*3(*t*) for *m* = *c* = *k* = *x*10 = *v*10 = 1.

**Figure 28.** Plots of free response *<sup>x</sup>*3(*t*) for *m* = *c* = *k* = *x*10 = *v*10 = 1. Solid line: *ω* = 1.1. Dot line: *ω* = 1.5. (**a**) *α* = 1.8, *β* = 0.8 (*ςeq*<sup>3</sup> = 0.78). (**b**) *α* = 1.5, *β* = 0.8 (*ςeq*<sup>3</sup> = 1.33). (**c**) *α* = 1.8, *β* = 0.3 (*ςeq*<sup>3</sup> = 0.91). (**d**) *α* = 2, *β* =1(*ςeq*<sup>3</sup> = 0.50).

Note that the plots regarding with *<sup>x</sup>*3(*t*) in Figure 28 are with fixed *ω*. However, actual *<sup>x</sup>*3(*t*) is frequency varying. Figure 29 shows its frequency varying pictures in time domain and Figure 30 in t-ω plane.

**Figure 29.** Free response *<sup>x</sup>*3(*t*) with variable *ω* (=1, 2, ..., 10) for *m* = *c* = *k* = 1. (**a**) *α* = 1.9, *β* = 0.8 (*ςeq*<sup>3</sup> = 0.66). (**b**) *α* = 1.5, *β* = 0.8 (*ςeq*<sup>3</sup> = 1.33). (**c**) *α* = 1.8, *β* = 0.3 (*ςeq*<sup>3</sup> = 0.91). (**d**) *α* = 2, *β* =1(*ςeq*<sup>3</sup> = 0.50).

**Figure 30.** Free response *<sup>x</sup>*3(*t*) in t-ω plane with *m* = *k* = *x*10 = *v*10 = 1 and *c* = 0.1 for *t* = 0, 1, ..., 30; *ω* = 1, 2, ..., 5. (**a**) *α* = 1.8, *β* = 0.8 (0.21 ≤ *ςeq*<sup>3</sup> ≤ 0.72). (**b**) *α* = 1.8, *β* = 0.5 (0.20 ≤ *ςeq*<sup>3</sup> ≤ 0.70). (**c**) *α* = 1.5, *β* = 0.8 (0.49 ≤ *ςeq*<sup>3</sup> ≤ 1.49). (**d**) *α* = 2, *β* =1(*ςeq*<sup>3</sup> = 0.05).

**Corollary 12** (Decrement III)**.** *Denote by <sup>x</sup>*3(*t*) *the free response to a fractional oscillator in Class III. Then, for 1 < α* ≤ *2 and 0 < β* ≤ *1, its logarithmic decrement, denoted by* <sup>Δ</sup>*eq*3, *is given by*

$$\Delta\_{eq3} = \frac{\frac{\pi \left(\omega^{\alpha-1} \sin \frac{\alpha \pi}{2} + 2\xi \omega \mu\_{\alpha} \omega^{\beta-1} \sin \frac{\beta \pi}{2}\right)}{\omega\_{1\eta} \sqrt{-\left(\omega^{\alpha-2} \cos \frac{\alpha \pi}{2} + 2\xi \omega \mu\_{\alpha} \omega^{\beta-2} \cos \frac{\beta \pi}{2}\right)}}}{\sqrt{1 - \left(\frac{\omega^{\alpha-1} \sin \frac{\alpha \pi}{2} + 2\xi \omega \mu\_{\alpha} \omega^{\beta-1} \sin \frac{\beta \pi}{2}}{2\omega\_{\alpha} \sqrt{-\left(\omega^{\alpha-2} \cos \frac{\alpha \pi}{2} + 2\xi \omega \mu\_{\alpha} \omega^{\beta-2} \cos \frac{\beta \pi}{2}\right)}}\right)^{2}}}\tag{184}$$

**Proof.** Note that

$$
\Delta\_{eq3} = \ln \frac{x\_3(t\_i)}{x\_3(t\_{i+1})} = \frac{2\pi \varepsilon\_{eq3}}{\sqrt{1 - \varepsilon\_{eq3}^2}}.\tag{185}
$$

Replacing the above *ςeq*<sup>3</sup> with that in (147) yields (184). This completes the proof.  *Symmetry* **2018**, *10*, 40

When *α* = 2 and *β* = 1, <sup>Δ</sup>*eq*<sup>3</sup> = <sup>Δ</sup>*eq*<sup>3</sup>(*<sup>ω</sup>*, *α*, *β*) becomes the conventional logarithmic decrement. In fact,

$$\Delta\_{eq3}\big|\_{a=1,\beta=1} = \frac{\frac{\pi\left(\omega^{a-1}\sin\frac{\theta\pi}{2} + 2\varphi\omega\_{n}\omega^{\beta-1}\sin\frac{\beta\pi}{2}\right)}{\omega\sqrt{-\left(\omega^{a-2}\cos\frac{\theta\pi}{2} + 2\varphi\omega\_{n}\omega^{\beta-2}\cos\frac{\beta\pi}{2}\right)}}}{\sqrt{1-\left(\frac{\omega^{a-1}\sin\frac{\theta\pi}{2} + 2\varphi\omega\_{n}\omega^{\beta-1}\sin\frac{\beta\pi}{2}}{2\omega\_{n}\sqrt{-\left(\omega^{a-2}\cos\frac{\theta\pi}{2} + 2\varphi\omega\_{n}\omega^{\beta-2}\cos\frac{\beta\pi}{2}\right)}}\right)^{2}}\bigg|\_{a=1,\beta=1} = \frac{2\pi\mathfrak{c}}{\sqrt{1-\mathfrak{c}^{2}}}.\tag{186}$$

As <sup>Δ</sup>*eq*<sup>3</sup> is a function of *ω* and (*<sup>α</sup>*, *β*), we write it by <sup>Δ</sup>*eq*<sup>3</sup>(*<sup>ω</sup>*, *α*, *β*). Figures 31 and 32 show its plots.

**Figure 31.** <sup>Δ</sup>*eq*<sup>3</sup>(*<sup>ω</sup>*, *α*, *β*) : Logarithmic decrement of fractional oscillator in Class III for *m* = *c* = *k* = 1. (**a**) Solid line: *α* = 1.7, *β* = 0.8. Dot line: *α* = 1.7, *β* = 0.5. (**b**) Solid line: *α* = 1.8, *β* = 0.8. Dot line: *α* = 1.8, *β* = 0.5.

**Figure 32.** <sup>Δ</sup>*eq*<sup>3</sup>(*<sup>ω</sup>*, *α*, *β*) for *m* = *c* = 1, *k* = 25. (**a**) Solid line: *α* = 1.7, *β* = 0.8. Dot line: *α* = 1.7, *β* = 0.5. (**b**) Solid line: *α* = 1.8, *β* = 0.8. Dot line: *α* = 1.8, *β* = 0.5.
