*6.2. Free Response to Fractional Oscillators in Class I*

We state the free response to a fractional oscillator in Class I by Theorem 10.

**Theorem 10** (Free response I)**.** *Let <sup>x</sup>*1(*t*) *be the free response to a fractional oscillator in Class I. Then, for 1 < α* ≤ *2, <sup>x</sup>*1(*t*) *is given by*

$$\mathbf{x}\_{1}(t) = e^{-\frac{\omega \sin \frac{\mathbf{a} \pi}{2}}{2|\cos \frac{\mathbf{a} \pi}{2}|}t} \begin{bmatrix} x\_{10} \cos \left(\frac{\omega \imath\_{n}}{\sqrt{\omega^{a-2} \left|\cos \frac{\mathbf{a} \pi}{2}\right|}} \sqrt{1 - \frac{\omega^{a} \sin^{2} \frac{\mathbf{a} \pi}{2}}{4\omega^{2} \left|\cos \frac{\mathbf{a} \pi}{2}\right|}} t \right) \\\\ x\_{10} + \frac{\omega \not\equiv \sin \frac{\omega \imath\_{n}}{\sqrt{\omega^{a} \left|\cos \frac{\mathbf{a} \pi}{2}\right|}}} x\_{10} \\\\ + \frac{\omega \imath\_{n} \sin^{2} \frac{\omega \imath\_{n}}{2}}{\sqrt{1 - \frac{\omega^{a} \sin^{2} \frac{\mathbf{a} \pi}{2}}{4\omega^{2} \left|\cos \frac{\mathbf{a} \pi}{2}\right|}}} \sin \left(\frac{\omega\_{n}}{\sqrt{\omega^{a-2} \left|\cos \frac{\mathbf{a} \pi}{2}\right|}} \sqrt{1 - \frac{\omega^{a} \sin^{2} \frac{\mathbf{a} \pi}{2}}{4\omega^{2}\_{n} \left|\cos \frac{\mathbf{a} \pi}{2}\right|}} t \right) \end{bmatrix}. \tag{172}$$

**Proof.** For *t* ≥ 0, consider

$$\mathbf{x}\_{1}(t) = e^{-\zeta\_{eq1}\omega\_{eq1,1}t} \left( \mathbf{x}\_{10}\cos\omega\_{eqd,1}t + \frac{\upsilon\_{10} + \zeta\_{eq1}\omega\_{eqn,1}x\_{10}}{\omega\_{eqd,1}}\sin\omega\_{eqd,1}t \right). \tag{173}$$

In the above, replacing *<sup>ω</sup>eqn*,<sup>1</sup> by the one in (132), *ςeq*<sup>1</sup> with that in (141), and *<sup>ω</sup>eqd*,<sup>1</sup> by the one in (154) yields

*<sup>x</sup>*1(*t*) = *e* − *ω α* 2 sin *απ*2 <sup>2</sup>*ωn*√|cos *απ*2 | √ *ωn ωα*−<sup>2</sup>|cos *απ*2 | *t* ⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣ *x*10 cos ⎛⎜⎜⎝*ωn*!1− *ω<sup>α</sup>* sin<sup>2</sup> *απ*2 <sup>4</sup>*ω*2*n*|cos *απ*2 | *ωα*−<sup>2</sup>|cos *απ*2 | *t*⎞⎟⎟⎠ + *<sup>v</sup>*10+ *ω α* 2 sin *απ*2 <sup>2</sup>*ωn*|cos *απ*2 | *ωn ωα*−<sup>2</sup>|cos *απ*2 | *<sup>x</sup>*10 *ωn 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 | sin ⎛⎜⎜⎝*ωn*!1− *ω<sup>α</sup>* sin<sup>2</sup> *απ*2 <sup>4</sup>*ω*2*n*|cos *απ*2 | *ωα*−<sup>2</sup>|cos *απ*2 | *t*⎞⎟⎟⎠ ⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ = *e* − *ω* sin *απ*2 <sup>2</sup>|cos *απ*2 | *t*⎡⎢⎢⎢⎣*x*10 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 | ⎞⎟⎟⎠ + *<sup>v</sup>*10+ *ω α*2 sin *απ*2 <sup>2</sup>*ωn*|cos *απ*2 | *<sup>x</sup>*10 !1− *ω<sup>α</sup>* sin<sup>2</sup> *απ*2 <sup>4</sup>*ω*2*n*|cos *απ*2 | sin⎛⎜⎜⎝*<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 | *ωα*−<sup>2</sup>|cos *απ*2 | *t*⎞⎟⎟⎠⎤⎥⎥⎥⎦.

This completes the proof. 

> Figure 20 indicates *<sup>x</sup>*1(*t*) with fixed *ω*.

**Figure 20.** Indicating free response *<sup>x</sup>*1(*t*) for *x*10 = *v*10 = *ωn* = 1. (**a**) *α* = 1.9. Solid line: *ω* = 1 (*ςeq*<sup>1</sup> = 0.08). Dot line: *ω* = 0.7 (*ςeq*<sup>1</sup> = 0.04). (**b**) α = 1.6. Solid line: ω =1(*ςeq*<sup>1</sup> = 0.33). Dot line: *ω* = 0.7 (*ςeq*<sup>1</sup> = 0.16). (**c**) *α* = 1.3. Solid line: *ω* =1(*ςeq*<sup>1</sup> = 0.66). Dot line: *ω* = 0.7 (*ςeq*<sup>1</sup> = 0.42). (d) *α* = 2. Solid line: *ω* =1(*ςeq*<sup>1</sup> = 0). Dot line: *ω* = 0.7 (*ςeq*<sup>1</sup> = 0).

**Note 6.1:** As indicated in Figure 20, both oscillation frequency *ω* and the fractional order *α* have affects on the damping *<sup>ς</sup>eq*<sup>1</sup>(*<sup>ω</sup>*, *<sup>α</sup>*), also see Figure 10. When *α* = 2, *<sup>x</sup>*1(*t*) reduces to the free response to the ordinary harmonic oscillation with damping free in the form (also see Figure 20d)

$$\mathbf{x}\_1(t) = \left(\mathbf{x}\_{10}\cos\omega\_n t + \frac{\upsilon\_{10}}{\omega\_n}\sin\omega\_n t\right), t \ge 0.$$

The free response to a fractional oscillator in Class I is presented in (172). It uses elementary functions instead of special functions.

Since there exists infinity of natural frequencies for a fractional oscillator, as we explained in Section 5, *<sup>x</sup>*1(*t*) is actually a function of both *t* and *ω* as can be seen from (172). In Figure 20, plots are only specifically with fixed *ω*. Its plots with varying *ω* are viewed by Figure 21.

**Figure 21.** Illustrating free response *<sup>x</sup>*1(*t*) with variable *ω* ( = 0, 0.2, 0.4, ..., 10) for *x*10 = *v*10 = 1. (**a**) For *ωn* = 1 and *α* = 1.9 (0 ≤ *ςeq*<sup>1</sup> ≤ 0.64). (**b**) For *ωn* = 3 and *α* = 1.6 (0 ≤ *ςeq*<sup>1</sup> ≤ 0.63).

When emphasizing the point of time-frequency behavior, we view it in t-ω plane as Figure 22 shows.

**Figure 22.** Indicating free response *<sup>x</sup>*1(*t*) in t-ω plane for *t* = 0, 1, ..., 30 and *ω* = 1, 2, ..., 5, with *x*10 = *v*10 = 1, *ωn* = 6. (**a**) *α* = 1.9 (0.01 ≤ *ςeq*<sup>1</sup> ≤ 0.28). (**b**) *α* = 1.6 (0.05 ≤ *ςeq*<sup>1</sup> ≤ 0.72). (**c**) *α* = 1.3 (0.11 ≤ *ςeq*<sup>1</sup> ≤ 0.89). (**d**) *α* =2(*ςeq*<sup>1</sup> = 0).

Let *ti* and *ti* + 1 be two time points where *xj*(*ti*) reaches its successive peak values of *xj*(*ti*) and *xj*(*ti*+<sup>1</sup>), respectively. Let <sup>Δ</sup>*eqj* be the logarithmic decrement of *xj*(*ti*). Then, from (178), we immediately obtain

$$\Delta\_{eqj} = \ln \frac{\mathbf{x}\_j(t\_i)}{\mathbf{x}\_j(t\_{i+1})} = \frac{2\pi \mathbf{g}\_{eqj}}{\sqrt{1 - \boldsymbol{\xi}\_{eqj}^2}}.\tag{174}$$

**Corollary 10** (Decrement I)**.** *Let <sup>x</sup>*1(*t*) *be the free response of a fractional oscillator in Class I. Then, its logarithmic decrement is given in the form*

$$\Delta\_{cq1} = \frac{\pi}{\sqrt{1 - \left(\frac{\omega^{\frac{\pi}{2}} \sin \frac{a\pi}{2}}{2\omega\_n \sqrt{-\cos \frac{a\pi}{2}}}\right)^2}} \frac{\omega^{\frac{a}{2}} \sin \frac{a\pi}{2}}{\omega\_n \sqrt{-\cos \frac{a\pi}{2}}}, 1 < a \le 2. \tag{175}$$

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

$$\Delta\_{eq1} = \ln \frac{\mathbf{x}\_1(t\_i)}{\mathbf{x}\_1(t\_{i+1})} = \frac{2\pi \zeta\_{eq1}}{\sqrt{1 - \zeta\_{eq1}^2}} = \frac{\pi}{\sqrt{1 - \left(\frac{\omega^{\frac{\mathfrak{d}}{2}} \sin \frac{a\pi}{2}}{2\omega\_n \sqrt{-\cos \frac{a\pi}{2}}}\right)^2}} \frac{\omega^{\frac{\mathfrak{d}}{2}} \sin \frac{a\pi}{2}}{\omega\_n \sqrt{-\cos \frac{a\pi}{2}}}, 1 < a \le 2.$$

The proof finishes. 

> Since <sup>Δ</sup>*eq*<sup>1</sup> is a function of *ω* and *α*, we may write it with <sup>Δ</sup>*eq*<sup>1</sup>(*<sup>ω</sup>*, *<sup>α</sup>*). Figure 23 indicates <sup>Δ</sup>*eq*1.

**Figure 23.** Plots of <sup>Δ</sup>*eq*1. Solid line: α = 1.9. Dot line: α = 1.6. (**a**) For *ωn* = 1. (**b**) For *ωn* = 5. (**c**) For *ωn* = 10.

**Note 6.2:** <sup>Δ</sup>*eq*1= 0 for *α* = 2. As a matter of fact, a fractional oscillator in Class I reduces to a harmonic one if *α* = 2. Accordingly, <sup>Δ</sup>*eq*1= 0 in that case.
