Example 1 - Slider-crank mechanism is used to impart motion to the base of a spring-mass-damper system as shown in Figure 1. The base motion \(y(t)\) is a series of harmonic functions. Find the response of the mass for \(m = 10 \left[\mathrm{kg}\right]\), \(c = 100 \left[\frac{Ns}{m}\right]\), \(k = 1000 \left[\frac{\mathrm{N}}{\mathrm{m}}\right]\), \(r = 100 \left[\mathrm{cm}\right]\), \(l = 3 \left[\mathrm{m}\right]\), and \(\omega = 100 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\).
Solution
Torque that is transmitted to driven gear is shown in Figure XXX.Figure 1 - Slider crank mechanism.
Solution - The opposite angle will be named \(\alpha\). The first step is to mathematically describe the base motion \(y_(t)\).
\begin{eqnarray}
y(t) &=& r + l - r\cos\omega t - l\cos \alpha\nonumber\\
y(t) &=& r + l - r\cos \omega t - l \sqrt{1-\sin^2\alpha}\nonumber\\
\end{eqnarray}
Looking at the figure the right angle triangle consisting of \(r\) and \(l\) can be divided on two smaller traingles by drawing the vertical line from right angle to the horizontal dashed line. By doing so the right angle traingle is subdivided on two right angle triangles. Those two traingles share the same vertical line from which the following expression could be obtained:
\begin{eqnarray}
l\sin \alpha &=& r\sin\omega t\nonumber
\end{eqnarray}
Then the \(y(t)\) becomes:
\begin{eqnarray}
y(t) &=& r+l-r\cos \omega t - l \sqrt{1-\frac{r^2}{l^2}\sin^2 \omega t}.\nonumber
\end{eqnarray}
If the approximation is applied to the square root term as:
\begin{eqnarray}
\sqrt{1-\frac{r^2}{l^2}\sin^2 \omega t} &\approx& 1 - \frac{r^2}{2l^2} \sin^2 \omega t\nonumber
\end{eqnarray}
Applying the previous approximation to \(y(t)\) to \(y(t)\) then \(y(t)\) becomes:
\begin{eqnarray}
y(t) &=& r + l - r\cos \omega t - l \left(1-\frac{1}{2}\frac{r^2}{l^2} \sin^2 \omega t\right)\nonumber\\
y(t) &=& r - r\cos \omega t + \frac{l}{4}\left(\frac{r}{l}\right)^2 - \frac{l}{4} \left(\frac{r}{l}\right)^2 \cos 2\omega t\nonumber \\
\end{eqnarray}
Now that the function of base motion is mathematical defined the next step is to insert it into equation of motion (differential equation) and solve it. The equation of motion can be written as:
\begin{eqnarray}
m\ddot{x} + c\dot{x} + kx &=& ky + c\dot{y}\nonumber\\
&=& kr-krcos\omega t + \frac{kl}{4}\left(\frac{r}{l}\right)^2 - \frac{kl}{4} \left(\frac{r}{l}\right)^2 \cos 2\omega t \nonumber \\
&+& cr\omega \sin\omega t +\frac{cl}{4}\left(\frac{r}{l}\right)^2 (2\omega) \sin 2\omega t\nonumber
\end{eqnarray}
The solution of previous equation can be found by adding the solutions to each term on the right hand side of aforementioned equation. The solution of equation due to constant term \(F_0\) - first and third terms of right hand side of equation.
\begin{eqnarray}
x(t) &=& \frac{F_0}{k}\left[1-\frac{1}{\sqrt{1-\zeta^2}} e^{-\zeta \omega_n t} \cos(\omega_d t - \phi)\right]\nonumber\\
\theta = \mathrm{arctan}^{-1}\left(\frac{\zeta}{\sqrt{1-\zeta^2}}\right)\nonumber
\end{eqnarray}
Solution due to sinusoidal term \(F_0 \sin \Omega t\) (terms 5 and 6 on the right hand side of equation).
\begin{eqnarray}
x(t) &=& X\sin (\Omega t - \theta_0)\nonumber\\
X &=& \frac{F_0}{\left[(k-m\Omega^2)^2 + c^2 \Omega^2\right]^{\frac{1}{2}}} \nonumber\\
\theta_0 &=& \mathrm{arctan}^{-1}\left(\frac{c\Omega}{k-m\Omega^2}\right) \nonumber
\end{eqnarray}
Solution due to cosine term, \(F_0\cos\Omega t\) (terms 2 and 4 of equation).
\begin{eqnarray}
x(t) &=& X\cos(\Omega t - \theta_0)\nonumber\\
X &=& \frac{F_0}{\left[(k-m\Omega^2)^2 + c^2 \Omega^2\right]^{\frac{1}{2}}} \nonumber\\
\theta_0 &=& \mathrm{arctan}^{-1}\left(\frac{c\Omega}{k-m\Omega^2}\right) \nonumber
\end{eqnarray}
Example 2 - The torsional vibrations of a driven gear mounted on a shaft as shown in Figure 2 under steady conditions is descrived with differential equation which can be written as \(J_0\ddot{\theta} + k_t\theta = M_t,\) where \(k_t\) is the torsional stiffnes of the driven shaft, \(M_t\) is the torque transmitted, \(J_0\) is the mass moment of inertia, and \(\theta\) is the angular deflection of the driven gear. If one of the 16 teeth on the driving gear breaks, determine the resulting torsional vibration of the driven gear for the following data. \(J_0 = 0.1 \left[\mathrm{Nms}^2\right]\), \(v = 1000 \left[\mathrm{rpm}\right]\), \(M_t = 1000 \left[\mathrm{Nm}\right]\).Figure 2 Two gears with shafts in contact.
Torque transmitted to driven gear which is shown in the previous figure can be expressed as: \begin{eqnarray} M_t(t) &=& \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos n\omega t + b_n \sin n \omega t\right)\nonumber \\ \omega &=& 2\pi \left(\frac{1000}{60}\right) = 104.72 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber\\ \tau &=& \frac{2\pi}{\omega} = 0.06 \left[\mathrm{s}\right]\nonumber \\ \frac{15}{16}\tau &=& 0.05625 \left[\mathrm{s}\right]\nonumber\\ a_0 &=& \frac{2}{\tau}\int_0^r M_t(t) dt = \frac{2}{0.06}\int_0^{0.05625} (1000) dt = \frac{2000}{0.06} (0.05625) = 1875 \left[\mathrm{Nm}\right]\nonumber\\ a_n &=& \frac{2}{\tau} \int_0^r M_t(t)\cos n \omega t dt = \frac{2}{\tau} M_{t0}\int_0^{0.05626} \cos n \omega t dt \nonumber\\ a_n &=& \frac{2M_{t0}}{\tau} \left(\frac{\sin n\omega t}{n\omega}\right)\Bigg|_0^{0.05625} = \frac{318.3091}{n}\sin 5.8905 n \left[\mathrm{Nm}\right]\nonumber\\ b_n &=& \frac{2}{\tau} \int_0^r M_t(t) \sin n\omega t dt = \frac{2M_{t0}}{\tau} \int_0^{0.05625}\sin n\omega t dt \nonumber\\ b_n &=& \frac{2M_{t0}}{\tau}\left(-\frac{\cos n\omega t}{n\omega}\right)\Bigg|_0^{0.05625} = \frac{318.3091}{n}\left(1-\cos 5.8905 n\right)\left[\mathrm{Nm}\right]\nonumber\\ k_n &=& \frac{GJ}{l} = G\left(\frac{\pi d^4}{4}\right)\frac{1}{l} = (80\times 10^9)\cdot\left(\frac{\pi \cdot(0.05)^4}{4}\right)\frac{1}{1} = 392.7 \left[\frac{\mathrm{Nm}}{\mathrm{rad}}\right]\nonumber\\ \omega_n &=& \sqrt{\frac{k_t}{J_0}} = \sqrt{\frac{392700}{0.1}} = 1981.666 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber \end{eqnarray} The equation of motion can be written as: \begin{eqnarray} J_0\ddot{\theta} + k_t \theta = M_t(t) =\frac{a_0}{2} +\sum_{n=1}^\infty \left(a_n \cos n\omega t + b_n \sin n\omega t\right)\nonumber\\ \end{eqnarray} The response can be extracted from previous equation: \begin{eqnarray} \theta(t) &=& \frac{a_0}{2k_t} + \sum_{n=1}^\infty \left(\frac{a_n\cos n\omega t + b_n \sin n \omega t}{k_t - J_0(n\omega)^2}\right)\nonumber\\ &=& 0.0023873 + \sum_{n=1}^\infty\left(\frac{318.3091 \sin 5.8905n\cos\omega t + 318.3091 (1-\cos 5.8905 n) \sin n\omega t}{n(392700.0 - 1096.6279 n^2)}\right) \end{eqnarray}
Example 3 - Find the total response of a viscously damped single-degree-of-freedom system subjected to a harmonic base excitation for the following data: \(m = 10 \left[\mathrm{kg}\right]\), \(c = 20 \left[\frac{\mathrm{Ns}}{\mathrm{m}}\right]\), \(k = 8000 \left[\frac{\mathrm{N}}{\mathrm{m}}\right]\), \(y(t) = 0.1 \cos 5t\), \(x_0 = 0.1 \left[\mathrm{m}\right]\), \(\dot{x}_0 = 1 \left[\frac{\mathrm{m}}{\mathrm{s}}\right]\).
Solution - From the given expression of base excitaiton the amplitude and frequency can be determined: \begin{eqnarray} y(t) &=& Y\cos \omega t\nonumber \\ y(t) &=& 0.1 \cos \omega t\nonumber\\ Y &=& 0.1; \omega = 5\nonumber \end{eqnarray} Equation of motion of the system can be written as: \begin{eqnarray} m\ddot{x} + c\dot{x} + kx &=& ky + c\dot{y}\nonumber \end{eqnarray} By inserting the y function inside differential equation the equation of motion can be written as: \begin{eqnarray} m\ddot{x} + c\dot{x} + kx &=& kY\cos\omega t - c\omega Y \sin \omega t\nonumber \\ \end{eqnarray} The steady state response of the system can be written as: \begin{eqnarray} x_p(t) &=& \frac{1}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}\left(\frac{a_1}{k} \cos(\omega t - \phi_1) + \frac{b_1}{k}\sin (\omega t - \phi_1)\right) \end{eqnarray} Where \begin{eqnarray} \omega_n &=& \sqrt{\frac{k}{m}} = \sqrt{\frac{8000}{10}} = 28.28\left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber\\ r &=& \frac{\omega}{\omega_n} = \frac{5}{28.28} = 0.1768 \nonumber\\ \zeta &=& \frac{c}{c_c} = \frac{c}{2\sqrt{km}} = \frac{20}{2\sqrt{8000\cdot 10}} = 0.03535 \nonumber \\ \omega_d &=& \sqrt{1-\zeta^2}\omega_n = \sqrt{1-0.03535^2}\cdot 28.28 = 28.26232 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber\\ a_1 &=& kY = 8000\cdot 0.1 = 800 \nonumber\\ b_1 &=& -c\cdot \omega \cdot Y =-20 \cdot 5 \cdot 0.1 = -10 \nonumber\\ \phi_1 &=& \mathrm{arctan} \left(\frac{2\zeta r}{1-r^2}\right) = \mathrm{arctan} \left(\frac{2\cdot 0.03535\cdot 0.1768 }{1-0.1768^2}\right)\nonumber \\ &=& \mathrm{arctan} \left(0.0129031\right) = 0.0129024 \left[\mathrm{rad}\right]\nonumber\\ x_P (t) &=& \frac{1}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}\left(\frac{a_1}{k} \cos(\omega t - \phi_1) + \frac{b_1}{k}\sin (\omega t - \phi_1)\right)\nonumber\\ &=& \frac{1}{\sqrt{(1-0.1768^2)^2 + (2\cdot 0.03535\cdot 0.1768)^2}}\left(\frac{800}{8000}\cos(5t-0.0129024) - \frac{10}{8000}\sin(5t-0.0129024)\right)\nonumber\\ &=& \frac{1}{0.968822} \left(0.1\cdot\cos(5t-0.0129024) - 0.00125\cdot\sin(5t-0.0129024)\right)\nonumber \end{eqnarray} Solution of the homogenous part of differential equation can be written as: \begin{eqnarray} x_H (t) &=& X_0 e^{-\zeta \omega_n t} \cos (\omega_d t - \phi_0)\nonumber\\ &=& X_0 e^{-0.999698t}\cdot\cos (28.26232 t- \phi_0),\nonumber \end{eqnarray} where \(X_0\) and \(\phi_0\) are unknown constants that will be determined. The total solution it consists of a homogenous and a particular solution which can be written as: \begin{eqnarray} x(t) &=& x_H (t) + x_P(t) = X_0 e^{-0.999698t}\cdot \cos (28.26232 t - \phi_0) \nonumber\\ &+& \frac{1}{0.968822} \left(0.1\cdot\cos(5t-0.0129024) - 0.00125\cdot\sin(5t-0.0129024)\right) \nonumber\\ &=& X_0 e^{-0.999698t}\cdot (28.26232 t - \phi_0) + 0.103218\cdot\cos(5t-0.0129024) - 0.00129023\cdot\sin(5t-0.0129024)\nonumber \end{eqnarray} In previous equations the \(X_0\) and \(\phi_0\) are two unkowns that will be determined from initial conditions. Derivating previous equations with respect to variable t the following expression is obtained: \begin{eqnarray} \dot{x}(t) &=& -0.999698X_0e^{-0.999698t} \cos (28.26232 t - \phi_0) - 28.26232 X_0 e^{-0.999698t} \sin (28.26232 t - \phi_0)\nonumber\\ &+& 0.51609 \sin(5t - 0.0129024) - 0.00645115\cos(5t-0.0129024)\nonumber \end{eqnarray} Using the initial conditions we can calculate the \(X_0\) and \(\phi_0\). \begin{eqnarray} x_0 &=& x(t=0) = 0.1 = X_0 e^{-0.999698t}\cdot \cos (28.26232 t - \phi_0) + 0.103218\cdot\cos(5t-0.0129024) - 0.00129023\cdot\sin(5t-0.0129024)\nonumber\\ X_0 &=& \frac{-0.00322606}{\cos(\phi_0)}\nonumber\\ \dot{x}_0 &=& \dot{x}(t=0) = 1.0 =-0.999698X_0e^{-0.999698t} \cos (28.26232 t - \phi_0) - 28.26232 X_0 e^{-0.999698t} \sin (28.26232 t - \phi_0)\nonumber\\ &+& 0.51609 \sin(5t - 0.0129024) - 0.00645115\cos(5t-0.0129024)\nonumber\\ X_0 &=& \frac{0.0357325}{\sin(\phi_0)}\nonumber \end{eqnarray} Solution of two previous equations: \begin{eqnarray} X_0 &=& \sqrt{\left((X_0\cos\phi_0)^2 + (X_0\sin\phi_0)^2\right)} = 0.0358779 \nonumber\\ \phi_0 &=& \mathrm{arctan} \left(\frac{X_0\sin \phi_0}{X_0\cos \phi_0}\right) = -0.0900393 \end{eqnarray}
Torque that is transmitted to driven gear is shown in Figure XXX.
Example 2 - The torsional vibrations of a driven gear mounted on a shaft as shown in Figure 2 under steady conditions is descrived with differential equation which can be written as \(J_0\ddot{\theta} + k_t\theta = M_t,\) where \(k_t\) is the torsional stiffnes of the driven shaft, \(M_t\) is the torque transmitted, \(J_0\) is the mass moment of inertia, and \(\theta\) is the angular deflection of the driven gear. If one of the 16 teeth on the driving gear breaks, determine the resulting torsional vibration of the driven gear for the following data. \(J_0 = 0.1 \left[\mathrm{Nms}^2\right]\), \(v = 1000 \left[\mathrm{rpm}\right]\), \(M_t = 1000 \left[\mathrm{Nm}\right]\).
Torque transmitted to driven gear which is shown in the previous figure can be expressed as: \begin{eqnarray} M_t(t) &=& \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos n\omega t + b_n \sin n \omega t\right)\nonumber \\ \omega &=& 2\pi \left(\frac{1000}{60}\right) = 104.72 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber\\ \tau &=& \frac{2\pi}{\omega} = 0.06 \left[\mathrm{s}\right]\nonumber \\ \frac{15}{16}\tau &=& 0.05625 \left[\mathrm{s}\right]\nonumber\\ a_0 &=& \frac{2}{\tau}\int_0^r M_t(t) dt = \frac{2}{0.06}\int_0^{0.05625} (1000) dt = \frac{2000}{0.06} (0.05625) = 1875 \left[\mathrm{Nm}\right]\nonumber\\ a_n &=& \frac{2}{\tau} \int_0^r M_t(t)\cos n \omega t dt = \frac{2}{\tau} M_{t0}\int_0^{0.05626} \cos n \omega t dt \nonumber\\ a_n &=& \frac{2M_{t0}}{\tau} \left(\frac{\sin n\omega t}{n\omega}\right)\Bigg|_0^{0.05625} = \frac{318.3091}{n}\sin 5.8905 n \left[\mathrm{Nm}\right]\nonumber\\ b_n &=& \frac{2}{\tau} \int_0^r M_t(t) \sin n\omega t dt = \frac{2M_{t0}}{\tau} \int_0^{0.05625}\sin n\omega t dt \nonumber\\ b_n &=& \frac{2M_{t0}}{\tau}\left(-\frac{\cos n\omega t}{n\omega}\right)\Bigg|_0^{0.05625} = \frac{318.3091}{n}\left(1-\cos 5.8905 n\right)\left[\mathrm{Nm}\right]\nonumber\\ k_n &=& \frac{GJ}{l} = G\left(\frac{\pi d^4}{4}\right)\frac{1}{l} = (80\times 10^9)\cdot\left(\frac{\pi \cdot(0.05)^4}{4}\right)\frac{1}{1} = 392.7 \left[\frac{\mathrm{Nm}}{\mathrm{rad}}\right]\nonumber\\ \omega_n &=& \sqrt{\frac{k_t}{J_0}} = \sqrt{\frac{392700}{0.1}} = 1981.666 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber \end{eqnarray} The equation of motion can be written as: \begin{eqnarray} J_0\ddot{\theta} + k_t \theta = M_t(t) =\frac{a_0}{2} +\sum_{n=1}^\infty \left(a_n \cos n\omega t + b_n \sin n\omega t\right)\nonumber\\ \end{eqnarray} The response can be extracted from previous equation: \begin{eqnarray} \theta(t) &=& \frac{a_0}{2k_t} + \sum_{n=1}^\infty \left(\frac{a_n\cos n\omega t + b_n \sin n \omega t}{k_t - J_0(n\omega)^2}\right)\nonumber\\ &=& 0.0023873 + \sum_{n=1}^\infty\left(\frac{318.3091 \sin 5.8905n\cos\omega t + 318.3091 (1-\cos 5.8905 n) \sin n\omega t}{n(392700.0 - 1096.6279 n^2)}\right) \end{eqnarray}
Example 3 - Find the total response of a viscously damped single-degree-of-freedom system subjected to a harmonic base excitation for the following data: \(m = 10 \left[\mathrm{kg}\right]\), \(c = 20 \left[\frac{\mathrm{Ns}}{\mathrm{m}}\right]\), \(k = 8000 \left[\frac{\mathrm{N}}{\mathrm{m}}\right]\), \(y(t) = 0.1 \cos 5t\), \(x_0 = 0.1 \left[\mathrm{m}\right]\), \(\dot{x}_0 = 1 \left[\frac{\mathrm{m}}{\mathrm{s}}\right]\).
Solution - From the given expression of base excitaiton the amplitude and frequency can be determined: \begin{eqnarray} y(t) &=& Y\cos \omega t\nonumber \\ y(t) &=& 0.1 \cos \omega t\nonumber\\ Y &=& 0.1; \omega = 5\nonumber \end{eqnarray} Equation of motion of the system can be written as: \begin{eqnarray} m\ddot{x} + c\dot{x} + kx &=& ky + c\dot{y}\nonumber \end{eqnarray} By inserting the y function inside differential equation the equation of motion can be written as: \begin{eqnarray} m\ddot{x} + c\dot{x} + kx &=& kY\cos\omega t - c\omega Y \sin \omega t\nonumber \\ \end{eqnarray} The steady state response of the system can be written as: \begin{eqnarray} x_p(t) &=& \frac{1}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}\left(\frac{a_1}{k} \cos(\omega t - \phi_1) + \frac{b_1}{k}\sin (\omega t - \phi_1)\right) \end{eqnarray} Where \begin{eqnarray} \omega_n &=& \sqrt{\frac{k}{m}} = \sqrt{\frac{8000}{10}} = 28.28\left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber\\ r &=& \frac{\omega}{\omega_n} = \frac{5}{28.28} = 0.1768 \nonumber\\ \zeta &=& \frac{c}{c_c} = \frac{c}{2\sqrt{km}} = \frac{20}{2\sqrt{8000\cdot 10}} = 0.03535 \nonumber \\ \omega_d &=& \sqrt{1-\zeta^2}\omega_n = \sqrt{1-0.03535^2}\cdot 28.28 = 28.26232 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber\\ a_1 &=& kY = 8000\cdot 0.1 = 800 \nonumber\\ b_1 &=& -c\cdot \omega \cdot Y =-20 \cdot 5 \cdot 0.1 = -10 \nonumber\\ \phi_1 &=& \mathrm{arctan} \left(\frac{2\zeta r}{1-r^2}\right) = \mathrm{arctan} \left(\frac{2\cdot 0.03535\cdot 0.1768 }{1-0.1768^2}\right)\nonumber \\ &=& \mathrm{arctan} \left(0.0129031\right) = 0.0129024 \left[\mathrm{rad}\right]\nonumber\\ x_P (t) &=& \frac{1}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}\left(\frac{a_1}{k} \cos(\omega t - \phi_1) + \frac{b_1}{k}\sin (\omega t - \phi_1)\right)\nonumber\\ &=& \frac{1}{\sqrt{(1-0.1768^2)^2 + (2\cdot 0.03535\cdot 0.1768)^2}}\left(\frac{800}{8000}\cos(5t-0.0129024) - \frac{10}{8000}\sin(5t-0.0129024)\right)\nonumber\\ &=& \frac{1}{0.968822} \left(0.1\cdot\cos(5t-0.0129024) - 0.00125\cdot\sin(5t-0.0129024)\right)\nonumber \end{eqnarray} Solution of the homogenous part of differential equation can be written as: \begin{eqnarray} x_H (t) &=& X_0 e^{-\zeta \omega_n t} \cos (\omega_d t - \phi_0)\nonumber\\ &=& X_0 e^{-0.999698t}\cdot\cos (28.26232 t- \phi_0),\nonumber \end{eqnarray} where \(X_0\) and \(\phi_0\) are unknown constants that will be determined. The total solution it consists of a homogenous and a particular solution which can be written as: \begin{eqnarray} x(t) &=& x_H (t) + x_P(t) = X_0 e^{-0.999698t}\cdot \cos (28.26232 t - \phi_0) \nonumber\\ &+& \frac{1}{0.968822} \left(0.1\cdot\cos(5t-0.0129024) - 0.00125\cdot\sin(5t-0.0129024)\right) \nonumber\\ &=& X_0 e^{-0.999698t}\cdot (28.26232 t - \phi_0) + 0.103218\cdot\cos(5t-0.0129024) - 0.00129023\cdot\sin(5t-0.0129024)\nonumber \end{eqnarray} In previous equations the \(X_0\) and \(\phi_0\) are two unkowns that will be determined from initial conditions. Derivating previous equations with respect to variable t the following expression is obtained: \begin{eqnarray} \dot{x}(t) &=& -0.999698X_0e^{-0.999698t} \cos (28.26232 t - \phi_0) - 28.26232 X_0 e^{-0.999698t} \sin (28.26232 t - \phi_0)\nonumber\\ &+& 0.51609 \sin(5t - 0.0129024) - 0.00645115\cos(5t-0.0129024)\nonumber \end{eqnarray} Using the initial conditions we can calculate the \(X_0\) and \(\phi_0\). \begin{eqnarray} x_0 &=& x(t=0) = 0.1 = X_0 e^{-0.999698t}\cdot \cos (28.26232 t - \phi_0) + 0.103218\cdot\cos(5t-0.0129024) - 0.00129023\cdot\sin(5t-0.0129024)\nonumber\\ X_0 &=& \frac{-0.00322606}{\cos(\phi_0)}\nonumber\\ \dot{x}_0 &=& \dot{x}(t=0) = 1.0 =-0.999698X_0e^{-0.999698t} \cos (28.26232 t - \phi_0) - 28.26232 X_0 e^{-0.999698t} \sin (28.26232 t - \phi_0)\nonumber\\ &+& 0.51609 \sin(5t - 0.0129024) - 0.00645115\cos(5t-0.0129024)\nonumber\\ X_0 &=& \frac{0.0357325}{\sin(\phi_0)}\nonumber \end{eqnarray} Solution of two previous equations: \begin{eqnarray} X_0 &=& \sqrt{\left((X_0\cos\phi_0)^2 + (X_0\sin\phi_0)^2\right)} = 0.0358779 \nonumber\\ \phi_0 &=& \mathrm{arctan} \left(\frac{X_0\sin \phi_0}{X_0\cos \phi_0}\right) = -0.0900393 \end{eqnarray}