3. Solution to the EOM

The general solution to the high voltage transmission line equation of motion was derived using Euler-Bernoulli equation. The particular solution to the equation of motion was derived using a product of two functions. The two functions were first separated using the principle of variable separation as expressed in Eq. (14) [19]:

$$Y(\mathbf{x}, t) = \mathbf{X}(\mathbf{x}) T(t) \tag{14}$$

Where X xð Þ is the normalized function representing the mode shape of the equation of motion. The normalized function ensures that orthogonality condition was satisfied in the derivation of the EOM model solution. Applying the normalized function in the EOM yields Eqs. (15) and (16):

$$EI\stackrel{\textstyle \longrightarrow \textstyle \longrightarrow}{X}(\mathbf{x}) - S\stackrel{\textstyle \longrightarrow}{X}(\mathbf{x}) - \omega^2 \rho AX(\mathbf{x}) = \mathbf{0} \tag{15}$$

$$
\ddot{T}(t) + \omega^2 T(t) = 0\tag{16}
$$

==== == <sup>d</sup>4<sup>y</sup> <sup>d</sup>2<sup>y</sup> <sup>d</sup><sup>2</sup> € <sup>y</sup> Where X x <sup>x</sup> T tðÞ¼ and <sup>ω</sup><sup>2</sup> ð Þ¼ <sup>ð</sup> Þ ¼ is <sup>a</sup> constant that equates <sup>x</sup> and <sup>t</sup>. Assuming dx4, <sup>X</sup> dx2, dt<sup>2</sup> that X x<sup>ð</sup> Þ ¼ Ze<sup>Ψ</sup><sup>x</sup> , the model is expressed in Eq. (17) as:

$$\mathcal{Z}e^{\Psi \mathbf{x}} \left( EI\Psi^4 - S\Psi^2 - \rho A\omega^2 \right) = 0 \tag{17}$$

� � Considering that Ze<sup>Ψ</sup><sup>x</sup> ¼6 0, hence EIΨ<sup>4</sup> � <sup>S</sup>Ψ<sup>2</sup> <sup>þ</sup> <sup>r</sup>Aω<sup>2</sup> <sup>¼</sup> 0. The general solution of the Euler-Bernoulli equation which represents the solution to the equation of the motion of the transmission line is expressed in Eqs. (18) and (19) as [9]:

$$
\Omega^2, \Psi^2 = -\frac{(-\mathbf{S}) \pm \sqrt{\mathbf{S}^2 - 4(\mathbf{E}\mathbf{I})(-\rho A \omega^2)}}{2\mathbf{E}\mathbf{I}}\tag{18}
$$

$$
\Omega, \Psi = (\pm) \sqrt{\frac{S \pm \sqrt{S^2 + 4EI(\rho A \omega^2)}}{2EI}} \tag{19}
$$

The values of Ω and Ψ represents the general solution of the equation of motion. The practical implication of the derived solution is that it represents the transverse vibration of the high voltage transmission line. The derived solution has infinite number of solutions and the solution is indexed to accommodate all the possible solutions from the model. The indexed solution is expressed in Eqs. (20) and (21) as:

$$
\Omega\_n = \sqrt{\frac{S}{2EI} + \sqrt{\frac{S^2}{\left(2EI\right)^2} + m\_L \frac{\left(2\pi f\_n\right)^2}{EI}}} \tag{20}
$$

$$\Psi\_n = \sqrt{-\frac{S}{2EI} + \sqrt{\frac{S^2}{\left(2EI\right)^2} + m\_L \frac{\left(2\pi f\_n\right)^2}{EI}}}\tag{21}$$

Where ω<sup>n</sup> ¼ 2πf and for n ¼ 1, 2, 3, …. <sup>n</sup>

In Eqs. (22)–(24), the infinite natural frequencies of the power conductor were derived while considering that the mode shape is the same as a pinned-pinned beam eigenfunction model with no external force. Hence,

$$Y\_n(\mathbf{x}, t) = \sin \frac{n \pi \mathbf{x}}{l} \cos \omega\_n t \tag{22}$$

$$EI\left(\frac{n\pi\pi}{l}\right)^4 \sin\frac{n\pi\pi}{l} \cos\omega\_n t - S\left(-\frac{n\pi}{l}\right)^2 \sin\frac{n\pi\pi}{l} \cos\omega\_n t + \rho A(-\omega\_n) \sin\frac{n\pi x}{l} \cos\omega\_n t = 0\tag{23}$$

$$\sin\frac{n\pi x}{l}\cos\omega\_n t \left[\frac{EI}{\rho A} \left(\frac{n\pi}{l}\right)^4 + \frac{S}{\rho A} \left(\frac{n\pi}{l}\right)^2 - \omega\_n^2\right] = 0\tag{24}$$

The natural frequency of the power conductor in rad/s is expressed in Eqs. (25) and (26) as:

$$
\omega\_n^2 = \left(\frac{n\pi\tau}{l}\right)^2 \frac{S}{A\rho} + \left(\frac{n\pi\tau}{l}\right)^4 \frac{EI}{A\rho} \tag{25}
$$

$$
\omega\_n = \sqrt{\left(\frac{n\pi}{L}\right)^2 \frac{S}{m\_L} \left[1 + \left(\frac{n\pi}{L}\right)^2 \frac{EI}{S}\right]}\tag{26}
$$

The natural frequency in Hz is expressed in Eq. (27) as:

$$F\_n = \frac{1}{2\pi} \sqrt{\left(\frac{n\pi}{L}\right)^2 \frac{S}{m\iota} \left[1 + \left(\frac{n\pi}{L}\right)^2 \frac{EI}{S}\right]}\tag{27}$$
