*3.2. Asymptotic Solution*

The following assumptions are introduced [12,17,77]: (1) The locations of IMDs are very close to the ends, i.e., *x*1, *x*∗2 *L*; (2) the wavenumber β*n* of each mode *n* (*n* = 1, 2, ... ) has a small perturbations Δβ*n* = β*n* − β0*n* from the undamped value β0*n* = *<sup>n</sup>*π/*L*. The assumptions above lead to the following approximations:

$$\tan(\beta\_n L) \cong \beta\_n L - n\pi \sin(\beta\_n \mathbf{x}\_1) \cong \beta\_n^0 \mathbf{x}\_1 \sin(\beta\_n \mathbf{x}\_2^\*) \cong \beta\_n^0 \mathbf{x}\_2^\* \cos(\beta\_n \mathbf{x}\_1) \cong \cos(\beta\_n \mathbf{x}\_2^\*) = 1. \tag{10}$$

The asymptotic formula for the wavenumber β*n* takes the form:

$$
\beta\_n \beta\_n L \cong n\pi + \beta\_n^0 \frac{E\_1 + iF\_1}{G\_1 + iH\_1},
\tag{11}
$$

$$E\_1 = -\overline{b\_1}\mathbf{x}\_1 - \overline{b\_2}\mathbf{x}\_2^\* + (\overline{b\_1 b\_2} - \overline{c\_1 c\_2})(\mathbf{x}\_1 + \mathbf{x}\_2^\*),\tag{11a}$$

$$F\_1 = \overline{c\_1}\mathbf{x}\_1 + \overline{c\_2}\mathbf{x}\_2^\* - (\overline{b\_1}\overline{c\_2} + \overline{b\_2}\overline{c\_1})(\mathbf{x}\_1 + \mathbf{x}\_2^\*),\tag{11b}$$

$$G\_1 = 1 - \overline{b\_1} - \overline{b\_2} + (\overline{b\_1 b\_2} - \overline{c\_1 c\_2})(\mathbf{x}\_1 + \mathbf{x}\_2^\*),\tag{11c}$$

$$H\_1 = \overline{c\_1} + \overline{c\_2} - (\overline{b\_1}\overline{c\_2} + \overline{b\_2}\overline{c\_1})(\mathbf{x}\_1 + \mathbf{x}\_2^\*),\tag{11d}$$

where *c*1 = <sup>η</sup>1β<sup>0</sup>*nx*<sup>1</sup> and *c*2 = <sup>η</sup>1β<sup>0</sup>*nx*<sup>∗</sup>2 represent dimensionless damping coefficient groups, while *b*1 = <sup>χ</sup>1β<sup>0</sup>*nx*<sup>1</sup> and *b*2 = <sup>χ</sup>2β<sup>0</sup>*nx*<sup>∗</sup>2 represent dimensionless inertial mass groups.

The complex eigen-frequency corresponding to the wavenumber β*n* is denoted as ω*n*. The *nth* supplemental modal damping radio of a cable ξ*n* can be obtained by [17]:

$$\xi\_n = \frac{\operatorname{Im}[\omega\_n]}{|\omega\_n|} = \frac{\operatorname{Im}[\beta\_n]}{|\beta\_n|} \cong \frac{\operatorname{Im}[\Delta \beta\_n]}{|\beta\_n^0|}. \tag{12}$$

Substituting Equation (11) into Equation (12), the asymptotic supplemental modal damping ratio of the cable is finally derived as:

$$\chi\_n \cong \frac{\overline{c\_1}}{\left(1 - \overline{b\_1}\right)^2 + \left(\overline{c\_1}\right)^2} \frac{x\_1}{L} + \frac{\overline{c\_2}}{\left(1 - \overline{b\_2}\right)^2 + \left(\overline{c\_2}\right)^2} \frac{x\_2}{L}.\tag{13}$$

If two identical IMDs are symmetrically installed at the cable for practical implementation, some simplifications in the notation can be introduced, i.e., *x* = *x*1 = *L* − *x*2 = *<sup>x</sup>*<sup>∗</sup>2, *b*1 = *b*2 = *b*, χ1 = χ2 = χ, *b*1 = *b*2 = *b*, *c*1 = *c*2 = *c*, η1 = η2 = η, and *c*1 = *c*2 = *c*. Equation (13) can be further simplified as:

$$\xi\_n = \frac{2\tilde{c}}{\left(1 - \overline{b}\right)^2 + \left(\overline{c}\right)^2} \frac{\mathbf{x}\_1}{L}. \tag{14}$$

## *3.3. Numerical Solution*

The numerical solution of the wavenumber β*n* to Equation (9) is obtained by the fixed point iteration [12,14], which starts from the undamped wavenumber β<sup>0</sup>*n*. Substituting β0*n* into the right side of the Equation (9), the resulting value β1*n* is obtained. Similarly, with current estimate β*kn*, *k* = 1, 2, 3..., a new estimate β*<sup>k</sup>*+<sup>1</sup> *n* will be derived. The iterative scheme is given by:

$$
\beta\_n^{k+1} L = n\pi + \arctan \frac{\overline{A}\_1 + i\overline{B}\_1}{\overline{C}\_1 + i\overline{D}\_1} \, ^\prime \tag{15}
$$

$$\overline{A}\_{1} = -\chi\_{1}\sin^{2}\beta\_{n}^{k}\mathbf{x}\_{1} - \chi\_{2}\sin^{2}\beta\_{n}^{k}\mathbf{x}\_{2}^{\*} + (\chi\_{1}\chi\_{2} - \eta\_{1}\eta\_{2})\sin\beta\_{n}^{k}\mathbf{x}\_{1}\sin\beta\_{n}^{k}\mathbf{x}\_{2}^{\*}\sin\beta\_{n}^{k}(\mathbf{x}\_{1} + \mathbf{x}\_{2}^{\*}),\tag{15a}$$

$$\overline{B}\_{1} = \eta\_{1}\sin^{2}\beta\_{n}^{k}\mathbf{x}\_{1} + \eta\_{2}\sin^{2}\beta\_{n}^{k}\mathbf{x}\_{2}^{\*} - (\chi\_{1}\eta\_{2} + \chi\_{2}\eta\_{1})\sin\beta\_{n}^{k}\mathbf{x}\_{1}\sin\beta\_{n}^{k}\mathbf{x}\_{2}^{\*}\sin\beta\_{n}^{k}(\mathbf{x}\_{1} + \mathbf{x}\_{2}^{\*}),\tag{15b}$$

$$\begin{array}{l} \overline{\mathbf{C}}\_{1} = 1 - \chi\_{1} \sin \beta\_{n}^{k} \mathbf{x}\_{1} \cos \beta\_{n}^{k} \mathbf{x}\_{1} - \chi\_{2} \sin \beta\_{n}^{k} \mathbf{x}\_{2}^{\*} \cos \beta\_{n}^{k} \mathbf{x}\_{2}^{\*} + \\ (\chi\_{1} \chi\_{2} - \eta\_{1} \eta\_{2}) \sin \beta\_{n}^{k} \mathbf{x}\_{1} \sin \beta\_{n}^{k} \mathbf{x}\_{2}^{\*} \cos \beta\_{n}^{k} (\mathbf{x}\_{1} + \mathbf{x}\_{2}^{\*}) \end{array} \tag{15c}$$

$$\begin{array}{l} \overline{D}\_{1} = \eta\_{1}\sin\beta\_{n}^{k}\mathbf{x}\_{1}\cos\beta\_{n}^{k}\mathbf{x}\_{1} + \eta\_{2}\sin\beta\_{n}^{k}\mathbf{x}\_{2}^{\*}\cos\beta\_{n}^{k}\mathbf{x}\_{2}^{\*} -\\ \left(\chi\_{1}\eta\_{2} + \chi\_{2}\eta\_{1}\right)\sin\beta\_{n}^{k}\mathbf{x}\_{1}\sin\beta\_{n}^{k}\mathbf{x}\_{2}^{\*}\cos\beta\_{n}^{k}(\mathbf{x}\_{1} + \mathbf{x}\_{2}^{\*}) \end{array} \tag{15d}$$

Finally, the supplemental modal damping ratio of a cable with two opposite IMDs can be calculated by Equation (12) after solving the wavenumber β*<sup>n</sup>*.

#### *3.4. Comparison of Asymptotic and Numerical Solutions*

Figure 2 shows the comparison of asymptotic and numerical complex wavenumbers of a cable with two symmetric identical IMDs for various inertial masses. Two IMDs are assumed to be respectively installed at distances *x*1 of 1%*L* and *x*2 of 99%*L* from the left end of the cable, i.e., *x* = *x*1 = *x*∗2 = *L* − *x*2 = 1%*L*. When inertial masses remain constant and damping coefficients of two IMDs increase from zero to infinity, the loci, which nearly trace a semicircular contour, start from the undamped wavenumber and finally attach to the real axis. According to Equation (12), the damping properties result from the imaginary part of the wavenumber. Maximum supplemental modal damping can be obtained at the top point of the semicircle [17]. The diameter of the loci is quite small but increases with the increase of the inertial masses of IMDs, indicating that two symmetric identical IMDs have slight influences on the damped frequency of the cable and can achieve higher supplemental modal damping ratios than traditional VDs. By comparing asymptotic and numerical complex wavenumbers, it is seen that two results coincide well with each other for small or moderate inertial mass (χ ≤ 0.6/(*n*π*x*/*L*)) adopted in the IMDs. Nevertheless, the results deviate significantly from each other when the big inertial mass (χ = 0.9/(*n*π*x*/*L*)) shown in Figure 2d is adopted. Hence, solutions via numerical iteration are used to accurately predict the maximum supplemental modal damping ratio of the cable and corresponding optimal damper size of the IMD in the following discussions.

**Figure 2.** Comparison of asymptotic and numerical complex wavenumbers of a cable with two symmetric identical inertial mass dampers (IMDs) (*x* = *x*1 = *x*∗2= 1%*L*).

## *3.5. Parametric Studies*

Figure 3 presents the supplemental modal damping ratio of a cable with two symmetric identical VDs versus damping coefficients. For the convenience of comparisons, the results of a cable with a single VD are also shown. It is observed that symmetrically attaching two VDs on the opposite end of a cable is favorable to increasing the maximum supplemental damping ratio of the cable, and its maximum supplemental modal damping ratio is asymptotically the sum of contributions from each VD separately. The findings above are quite consistent with those reported in previous studies [76,77].

**Figure 3.** The supplemental modal damping ratio curve of a cable with a single viscous damper (VD) or two symmetric identical VDs (*x* = *x*1 = *x*∗2 = 1%*L*).

Figure 4 presents the supplemental modal damping ratio of a cable with two symmetric identical IMDs or a single IMD versus damping coefficients for various inertial masses. It is clear that two opposite IMDs can provide superior control performance to the cable over a single IMD. Figure 5 directly compares the maximum supplemental modal damping ratio of a cable equipped with a single IMD or two symmetric identical IMDs. For two symmetric identical IMDs with small or medium inertial mass, similarly to the case of two symmetric identical VDs, the maximum achievable supplemental damping ratio is approximately doubled with that provided by a single IMD. It indicates that two opposite IMDs are almost independent from each other. This finding may explain why the optimal damping coefficients of the IMDs for both the single configuration and two-symmetric configuration are similar to each other in magnitude, as shown in Figure 4. Moreover, the maximum achievable supplemental modal damping ratio of a cable provided by two symmetric identical IMDs is larger than that provided by a single IMD or two symmetric identical VDs.

**Figure 4.** The supplemental modal damping ratio curve of a cable with a single IMD or two symmetric identical IMDs (*x* = *x*1 = *x*∗2= 1%*L*).

**Figure 5.** The maximum achievable supplemental modal damping ratio of a cable equipped with a single IMD or two symmetric identical IMDs (*x* = *x*1 = *x*∗2= 1%*L*).

#### **4. Two IMDs at the Same End**

#### *4.1. The Wavenumber Equation*

When two IMDs are installed at the same end of the cable, substituting Equation (8) into Equation (7) and rearranging terms gives the following expression relating to *x*1 and *x*2:

$$
\tan \beta L = \frac{A\_2 + iB\_2}{C\_2 + iD\_2} \,\,\,\tag{16}
$$

$$A\_2 = -\chi\_1 \sin^2 \beta \mathbf{x}\_1 - \chi\_2 \sin^2 \beta \mathbf{x}\_2 + (\chi\_1 \chi\_2 - \eta\_1 \eta\_2) \sin \beta \mathbf{x}\_1 \sin \beta \mathbf{x}\_2 \sin \beta (\mathbf{x}\_2 - \mathbf{x}\_1),\tag{16a}$$

$$B\_2 = \eta\_1 \sin^2 \beta \mathbf{x}\_1 + \eta\_2 \sin^2 \beta \mathbf{x}\_2 - (\chi\_1 \eta\_2 + \chi\_2 \eta\_1) \sin \beta \mathbf{x}\_1 \sin \beta \mathbf{x}\_2 \sin \beta (\mathbf{x}\_2 - \mathbf{x}\_1),\tag{16b}$$

$$\begin{aligned} \mathbf{C}\_2 &= 1 - \chi\_1 \sin \beta \mathbf{x}\_1 \cos \beta \mathbf{x}\_1 - \chi\_2 \sin \beta \mathbf{x}\_2 \cos \beta \mathbf{x}\_2 + \\\ (\chi\_1 \chi\_2 - \eta\_1 \eta\_2) \sin \beta \mathbf{x}\_1 \cos \beta \mathbf{x}\_2 \sin \beta (\mathbf{x}\_2 - \mathbf{x}\_1) \end{aligned} \tag{16c}$$

$$\begin{aligned} D\_2 &= \eta\_1 \sin \beta \mathbf{x}\_1 \cos \beta \mathbf{x}\_1 + \eta\_2 \sin \beta \mathbf{x}\_2 \cos \beta \mathbf{x}\_2 - \\ &\quad \left(\chi\_1 \eta\_2 + \chi\_2 \eta\_1\right) \sin \beta \mathbf{x}\_1 \cos \beta \mathbf{x}\_2 \sin \beta (\mathbf{x}\_2 - \mathbf{x}\_1) \end{aligned} \tag{16d}$$
