Example 1 - Find the natural frequencies and mode shapes of a spring-mass system shown in Figure 1 that is constrained to move in the vertical direction only.
Figure 1 - Two degree of freedom system
Solution
\begin{eqnarray}
m\ddot{x}_1 + 2kx_1 - kx_2 &=& 0\nonumber\\
m\ddot{x}_2 - 2kx_1 + 2kx_2 &=& 0\nonumber
\end{eqnarray}
\begin{eqnarray}
x_i(t) &=& X_i\cos(\omega t + \phi); i =1,2
\end{eqnarray}
By insertig previous equation into the two differential equaitons the following is obtained:
\begin{eqnarray}
m(-\omega^2X_1\cos(\omega t + \phi)) +2kX_1\cos(\omega t + \phi) - kX_2\cos(\omega t + \phi) &=& 0\Big/:\cos(\omega t + \phi)\nonumber\\
m(-\omega^2X_2\cos(\omega t + \phi)) -kX_1\cos(\omega t + \phi) + 2kX_2\cos(\omega t + \phi) &=& 0\Big/:\cos(\omega t + \phi)\nonumber\\
-m\omega^2X_1 + 2kX_1 - kX_2 &=& 0\nonumber\\
-m\omega^2X_2 - kX_1 + 2kX_2 &=& 0\nonumber\\
(2k-m\omega^2)X_1 - kX_2 &=& 0\nonumber\\
-k X_1 + (2k-m\omega^2)X_2 &=& 0\nonumber
\end{eqnarray}
The previous two equations will be expressed in matrix form:
\begin{eqnarray}
\begin{bmatrix}2k-m\omega^2 & -k \\ -k & 2k-m\omega^2\end{bmatrix}\begin{Bmatrix}X_1\\X_2\end{Bmatrix} &=& 0\nonumber
\end{eqnarray}
The frequency equation can be obtained by solving determinant of first matrix from previous equation as:
\begin{eqnarray}
\begin{vmatrix}
2k-m\omega^2 & -k \\ -k & 2k-m
\end{vmatrix}
&=& 0
\end{eqnarray}
\begin{eqnarray}
m^2\omega^4 - 4km\omega^2 + 3k^2 &=& 0.\nonumber
\end{eqnarray}
By introducin the substitution \(\omega^2 = x\) the previous equation is reduced to:
\begin{eqnarray}
m^2 x^2 - 4kmx + 3k^2 &=& 0.\nonumber\\
x_{1,2} &=& \frac{4km \pm \sqrt{16k^2m^2 -12m^2 k^2}}{2m^2}\nonumber\\
x_{1,2} &=& \frac{4km\pm \sqrt{4k^2m^2}}{2m^2}\nonumber\\
x_{1,2} &=& \frac{2k \pm k}{m}\nonumber\\
\omega_{1} &=& \sqrt{\frac{k}{ms}}\nonumber\\
\omega_{2} &=& \sqrt{\frac{3k}{m}}
\end{eqnarray}
The amplitude ratios can be determined from:
\begin{eqnarray}
r_1 &=& \frac{X_2^1}{X_1^1} = \frac{-m\omega_1^2 + 2k}{k} = \frac{k}{-m\omega_1^2 + 2k} = 1 \nonumber\\
r_2 &=& \frac{X_2^2}{X_1^2} = \frac{-m\omega_2^2 + 2k}{k} = \frac{k}{-m\omega_2^2 + 2k} = -1\nonumber
\end{eqnarray}
Example 2 - Find the free-vibration response of the system shown in Figure 2 with \(k_1 = 30, k_2 = 5, k_3 = 0\), \(m_1 = 10, m_2 =1\), and \(c_1 = c_2 = c_3 = 0\) for the initial conditions \(x_1(0) = 1, \dot{x}_1(0) = x_2(0) = \dot{x}_2(0) = 0\).
Figure 2 - Two degree of freedom spring mass damper system
Solution The system presented in Figure 2 can be written in mathematical form as: \begin{eqnarray} \begin{bmatrix}-m_1\omega^2 + k_1 + k_2 & -k_2\\ -k_2 & -m_2 \omega^2 + k_2 + k_3\end{bmatrix} \begin{Bmatrix}X_1\\ X_2\end{Bmatrix} &=& 0\nonumber\\ \begin{bmatrix}-10\omega^2 + 35& -5\\ -5 & -\omega^2 + 5\end{bmatrix} \begin{Bmatrix}X_1\\ X_2\end{Bmatrix} &=& 0 \nonumber\\ \end{eqnarray} If the determinant of first matrix in previous expression is solve the frequency equaiton is obtained. \begin{eqnarray} \begin{vmatrix}-10\omega^2 + 35& -5\\ -5 & -\omega^2 + 5\end{vmatrix} &=& 0\nonumber\\ (-10\omega^2 + 35)(-\omega^2+5) - (-5)(-5) &=& 0\nonumber\\ 10\omega^4 -50\omega^2 - 35\omega^2 + 175 -25 &=& 0\nonumber\\ 10\omega^4 -85\omega^2 + 150 &=& 0.\nonumber \end{eqnarray} After the frequency equation is obtained it must be solved to calculate the natural freuqency. Since the expression is fourth order the supstitution in from \(\omega^2 = t\) is introduced. \begin{eqnarray} 10t^2 - 85t +150 &=& 0\nonumber\\ t_{1,2} &=& \frac{85\pm \sqrt{85^2 - 4\cdot 10\cdot 150}}{2\cdot 10}\nonumber\\ t_{1,2} &=& \frac{85 \pm\sqrt{7225-6000}}{20}\nonumber\\ t_{1,2} &=& \frac{85 \pm \sqrt{1225}}{20}\nonumber\\ t_{1,2} &=& \frac{85\pm 35}{20}\nonumber\\ t_1 &=& \frac{50}{20} = 2.5 \nonumber\\ t_2 &=& \frac{120}{20} = 6\nonumber\\ \omega_{1} &=& \sqrt{2.5} = 1.5811\nonumber\\ \omega_{2} &=& \sqrt{6} = 2.4495\nonumber\\ \end{eqnarray} To claculate normal modes (or eigenvectors) the obtained \(\omega\) have to be inserted into original matrix equation. In first case we will insert the \(\omega_1^2 = 2.5\) \begin{eqnarray} \begin{bmatrix}-10\cdot 2.5 + 35& -5\\ -5 & -2.5 + 5\end{bmatrix} \begin{Bmatrix}X_1\\ X_2\end{Bmatrix} &=& 0 \nonumber\\ \begin{bmatrix}10& -5\\ -5 & 2.5\end{bmatrix} \begin{Bmatrix}X_1\\ X_2\end{Bmatrix} &=& 0 \nonumber\\ 10X_1 - 5X_2&=&0\Rightarrow X_2 = 2X_1\nonumber\\ -5X_1 +2.5X_2&=& 0\Rightarrow X_2 = 2X_1\nonumber\\ \end{eqnarray} In second case we will insert the \(\omega_2^2 = 6\) into the original matrix equation. \begin{eqnarray} \begin{bmatrix}-10\cdot 6 + 35& -5\\ -5 & -6 + 5\end{bmatrix} \begin{Bmatrix}X_1\\ X_2\end{Bmatrix} &=& 0 \nonumber\\ \begin{bmatrix}-25 &-5\\-5&-1\end{bmatrix} \begin{Bmatrix}X_1\\ X_2\end{Bmatrix} &=& 0 \nonumber\\ -25X_1 -5X_2 &=& 0\Rightarrow X_2 = -5X_1\nonumber\\ -5X_1 -X_2 &=& 0\Rightarrow X_2 = -5X_1 \nonumber\\ \end{eqnarray} The normal modes or eigenvectors can be written as: \begin{eqnarray} \vec{X}^{(1)} &=& \begin{Bmatrix}X_1^{(1)}\\X_2^{(2)}\end{Bmatrix} = \begin{Bmatrix}1\\2\end{Bmatrix}\vec{X}_1^{(1)}\nonumber\\ \vec{X}^{(2)} &=& \begin{Bmatrix}X_1^{(2)}\\X_2^{(2)}\end{Bmatrix} = \begin{Bmatrix}1\\-5\end{Bmatrix}\vec{X}_2^{(2)}.\nonumber \end{eqnarray} The free vibration responses of the masses \(m_1\) nad \(m_2\) can be written as: \begin{eqnarray} x_1(t) &=& X_1^{(1)}\cos(1.5811t+\phi_1) + X_1^{(2)}\cos(2.4495t + \phi_2)\nonumber\\ x_2(t) &=& 2X_1^{(2)}\cos(1.5811t+\phi_1) - 5X_1^{(2)}\cos(2.4495t + \phi_2).\nonumber \end{eqnarray} In previous equations four components (\(X_1^{(1)}, X_1^{(2)}, \phi_1, \phi_2 \)) must be determined. These variables are determined from initial conditions (\(x_1(0) = 1\), \(\dot{x}_1(0) = 0\), \(x_2(0) = 0\), and \(\dot{x}_2(0)=0\)). \begin{eqnarray} x_1(0) = 1 \Rightarrow X_1^{(1)} \cos\phi_1 + X_2^{(2)}\cos\phi_2 =1\nonumber\\ x_2(0) = 0 \Rightarrow 2X_1^{(2)}\cos\phi_1 - 5X_1^{(2)}\cos\phi_2 = 0 \nonumber\\ \dot{x}_1(0) = 0 \Rightarrow -1.5811X_1^{(1)}\sin\phi_1 -2.4495 X_1^{(2)}\sin\phi_2\nonumber\\ \dot{x}_2(0) = 0\Rightarrow -3.1622X_1^{(1)} \sin\phi_1 + 12.2475 X_1^{(2)}\sin\phi_2\nonumber \end{eqnarray} From first two equations the \(X_1^{(1)}\cos\phi_1\) and \(X_2^{(2)}\cos\phi_2\) components are obtained. \begin{eqnarray} 2X_1^{(1)}\cos\phi_1 - 5X_1^{(2)}\cos\phi_2 &=& 0\nonumber\\ X_1^{(1)}\cos\phi_1 &=& \frac{5}{2}X_1^{(2)}\cos\phi_2\nonumber\\ \frac{5}{2}X_1^{(2)}\cos\phi_2 + X_1^{(2)}\cos\phi_2 &=& 1\nonumber\\ \frac{7}{2}X_1^{(2)}\cos\phi_2 &=& 1\Big/\cdot \frac{2}{7}\nonumber\\ X_1^{(2)}\cos\phi_2 &=& \frac{2}{7}\nonumber \end{eqnarray} From third and fourth equations the \(X_1^{(1)}\sin\phi_1=0\), and \(X_1^{(2)}\sin\phi_2 =0\) are obtained. \begin{eqnarray} -1.5811X_1^{(1)}\sin\phi_1 &=& 2.4495X_1^{(2)}X_1^{(2)}\sin\phi_2\nonumber\\ X_1^{(1)}\sin\phi_1 &=& -1.5492X_1^{(2)}\sin\phi_2\nonumber\\ -3.1622(-1.5492X_1^{(2)}\sin\phi_2) + 12.2475X_1^{(2)}\sin\phi_2 &=& 0\nonumber\\ X_1^{(2)}\sin\phi_2 &=& 0\nonumber\\ X_1^{(1)}\sin\phi_1 &=& 0\nonumber\\ \end{eqnarray} From previous solutions \(X_1^{1}\) and \(X_1^{(2)}\) are obtained. \begin{eqnarray} X_1^{(1)} &=& \sqrt{(X_1^{(1)}\sin\phi)^2 + (X_1^{(1)}\cos\phi_1)^2}\nonumber\\ X_1^{(1)} &=& \frac{5}{7}\nonumber\\ X_1^{(2)} &=& \sqrt{(X_1^{(2)}\sin\phi)^2 + (X_1^{(2)}\cos\phi_1)^2}\nonumber\\ X_1^{(2)} &=& \frac{2}{7}\nonumber\\ \frac{X_1^{(1)}\sin\phi_1}{X_1^{(1)}\cos\phi_1} &=& \frac{0}{\frac{5}{7}} = 0\Rightarrow \phi_1 = 0\nonumber\\ \phi_2 &=& 0\nonumber \end{eqnarray} The free-vibration responses of \(m_1\) and \(m_2\) are given by: \begin{eqnarray} x_1(t) &=& \frac{5}{7}\cos 1.5811t +\frac{2}{7}\cos2.4495t\nonumber\\ x_2(t) &=& \frac{10}{7}\cos 1.5811t -\frac{10}{7}\cos2.4495t\nonumber \end{eqnarray} Example 3 - Calculate the natural frequencies and mode shapes for the torsional system that is shown in Figure 3 if \(J_1 = J_0\), \(J_2 = 2J_0\), and \(k_{t1} = k_{t2} = k_t\).Figure 3 - Torsional system
Solution - The differential equations that describes the motion of the system can be written as:
\begin{eqnarray}
J_0\ddot{\theta}_1 + 2k_t\theta_1 -k_t \theta_2 &=& 0\nonumber\\
2J_0\ddot{\theta}_2 - k_t\theta_1 + k_t \theta_2 &=& 0\nonumber
\end{eqnarray}
The solution of each equaiton can in general form be written as:
\begin{eqnarray}
\theta_1(t) &=& \Theta_i \cos(\omega t + \phi)\quad i=1,2.\nonumber\\
\end{eqnarray}
Substituting previous solution into system of differential equations the freuqency equation is obtained.
\begin{eqnarray}
-\omega^2\Theta_1\cos(\omega t + \phi) J_0 + 2k_t \Theta_1\cos(\omega t + \phi) - k_t \Theta_2 \cos(\omega t + \phi) &=& 0\nonumber\\
-2J_0\omega^2\Theta_2\cos(\omega t + \phi) - k_t \Theta_1\cos(\omega t +\phi) + k_t \Theta_2\cos(\omega t + \phi) &=& 0\nonumber\\
\begin{bmatrix}-\omega^2 J_0 +2k_t & -k_t \\ -k_t &-2J_0\omega^2 + k_t\end{bmatrix}\begin{Bmatrix}\Theta_1\\ \Theta_2\end{Bmatrix} &=& 0\nonumber\\
\det\begin{vmatrix}-\omega^2 J_0 +2k_t & -k_t \\ -k_t &-2J_0\omega^2 + k_t\end{vmatrix} &=& 0\nonumber\\
2\omega^4J_0^2 - \omega^2J_0k_t -4J_0k_t\omega^2 +2k_t^2 -k_t^2 &=& 0\nonumber\\
2\omega^4J_0^2 - 5J_0k_t\omega^2 +k_t^2&=& 0\nonumber
\end{eqnarray}
The frequency equation is solved but first the supstitution (\(\omega^2 = t\)) is introduced to reduce the order of polynomial.
\begin{eqnarray}
2t^2J_0^2 -5tJ_0k_t + k_t^2 &=& 0\nonumber\\
t_{1,2} &=& \frac{5J_0k_t\pm\sqrt{25J_0^2k_t^2 - 8J_0^2k_t^2}}{4J_0^2}\nonumber\\
t_{1,2} &=& \frac{5J_0k_t\pm\sqrt{17J_0^2k_t^2}}{4J_0^2}\nonumber\\
t_{1,2} &=& \frac{k_t(5\pm \sqrt{17})}{4J_0}\nonumber\\
\omega_1 &=& \sqrt{\frac{k_t}{4J_0}(5-\sqrt{17})}\nonumber\\
\omega_2 &=& \sqrt{\frac{k_t}{4J_0}(5+\sqrt{17})}\nonumber
\end{eqnarray}
The amplitude ratios are equal to:
\begin{eqnarray}
r_1 &=& \frac{\Theta_2^{(1)}}{\Theta_1^{(1)}} = 2-\frac{5-\sqrt{17}}{4}\nonumber\\
r_2 &=& \frac{\Theta_2^{(2)}}{\Theta_1^{(2)}} = 2-\frac{5+\sqrt{17}}{4}\nonumber\\
\end{eqnarray}
Solution The system presented in Figure 2 can be written in mathematical form as: \begin{eqnarray} \begin{bmatrix}-m_1\omega^2 + k_1 + k_2 & -k_2\\ -k_2 & -m_2 \omega^2 + k_2 + k_3\end{bmatrix} \begin{Bmatrix}X_1\\ X_2\end{Bmatrix} &=& 0\nonumber\\ \begin{bmatrix}-10\omega^2 + 35& -5\\ -5 & -\omega^2 + 5\end{bmatrix} \begin{Bmatrix}X_1\\ X_2\end{Bmatrix} &=& 0 \nonumber\\ \end{eqnarray} If the determinant of first matrix in previous expression is solve the frequency equaiton is obtained. \begin{eqnarray} \begin{vmatrix}-10\omega^2 + 35& -5\\ -5 & -\omega^2 + 5\end{vmatrix} &=& 0\nonumber\\ (-10\omega^2 + 35)(-\omega^2+5) - (-5)(-5) &=& 0\nonumber\\ 10\omega^4 -50\omega^2 - 35\omega^2 + 175 -25 &=& 0\nonumber\\ 10\omega^4 -85\omega^2 + 150 &=& 0.\nonumber \end{eqnarray} After the frequency equation is obtained it must be solved to calculate the natural freuqency. Since the expression is fourth order the supstitution in from \(\omega^2 = t\) is introduced. \begin{eqnarray} 10t^2 - 85t +150 &=& 0\nonumber\\ t_{1,2} &=& \frac{85\pm \sqrt{85^2 - 4\cdot 10\cdot 150}}{2\cdot 10}\nonumber\\ t_{1,2} &=& \frac{85 \pm\sqrt{7225-6000}}{20}\nonumber\\ t_{1,2} &=& \frac{85 \pm \sqrt{1225}}{20}\nonumber\\ t_{1,2} &=& \frac{85\pm 35}{20}\nonumber\\ t_1 &=& \frac{50}{20} = 2.5 \nonumber\\ t_2 &=& \frac{120}{20} = 6\nonumber\\ \omega_{1} &=& \sqrt{2.5} = 1.5811\nonumber\\ \omega_{2} &=& \sqrt{6} = 2.4495\nonumber\\ \end{eqnarray} To claculate normal modes (or eigenvectors) the obtained \(\omega\) have to be inserted into original matrix equation. In first case we will insert the \(\omega_1^2 = 2.5\) \begin{eqnarray} \begin{bmatrix}-10\cdot 2.5 + 35& -5\\ -5 & -2.5 + 5\end{bmatrix} \begin{Bmatrix}X_1\\ X_2\end{Bmatrix} &=& 0 \nonumber\\ \begin{bmatrix}10& -5\\ -5 & 2.5\end{bmatrix} \begin{Bmatrix}X_1\\ X_2\end{Bmatrix} &=& 0 \nonumber\\ 10X_1 - 5X_2&=&0\Rightarrow X_2 = 2X_1\nonumber\\ -5X_1 +2.5X_2&=& 0\Rightarrow X_2 = 2X_1\nonumber\\ \end{eqnarray} In second case we will insert the \(\omega_2^2 = 6\) into the original matrix equation. \begin{eqnarray} \begin{bmatrix}-10\cdot 6 + 35& -5\\ -5 & -6 + 5\end{bmatrix} \begin{Bmatrix}X_1\\ X_2\end{Bmatrix} &=& 0 \nonumber\\ \begin{bmatrix}-25 &-5\\-5&-1\end{bmatrix} \begin{Bmatrix}X_1\\ X_2\end{Bmatrix} &=& 0 \nonumber\\ -25X_1 -5X_2 &=& 0\Rightarrow X_2 = -5X_1\nonumber\\ -5X_1 -X_2 &=& 0\Rightarrow X_2 = -5X_1 \nonumber\\ \end{eqnarray} The normal modes or eigenvectors can be written as: \begin{eqnarray} \vec{X}^{(1)} &=& \begin{Bmatrix}X_1^{(1)}\\X_2^{(2)}\end{Bmatrix} = \begin{Bmatrix}1\\2\end{Bmatrix}\vec{X}_1^{(1)}\nonumber\\ \vec{X}^{(2)} &=& \begin{Bmatrix}X_1^{(2)}\\X_2^{(2)}\end{Bmatrix} = \begin{Bmatrix}1\\-5\end{Bmatrix}\vec{X}_2^{(2)}.\nonumber \end{eqnarray} The free vibration responses of the masses \(m_1\) nad \(m_2\) can be written as: \begin{eqnarray} x_1(t) &=& X_1^{(1)}\cos(1.5811t+\phi_1) + X_1^{(2)}\cos(2.4495t + \phi_2)\nonumber\\ x_2(t) &=& 2X_1^{(2)}\cos(1.5811t+\phi_1) - 5X_1^{(2)}\cos(2.4495t + \phi_2).\nonumber \end{eqnarray} In previous equations four components (\(X_1^{(1)}, X_1^{(2)}, \phi_1, \phi_2 \)) must be determined. These variables are determined from initial conditions (\(x_1(0) = 1\), \(\dot{x}_1(0) = 0\), \(x_2(0) = 0\), and \(\dot{x}_2(0)=0\)). \begin{eqnarray} x_1(0) = 1 \Rightarrow X_1^{(1)} \cos\phi_1 + X_2^{(2)}\cos\phi_2 =1\nonumber\\ x_2(0) = 0 \Rightarrow 2X_1^{(2)}\cos\phi_1 - 5X_1^{(2)}\cos\phi_2 = 0 \nonumber\\ \dot{x}_1(0) = 0 \Rightarrow -1.5811X_1^{(1)}\sin\phi_1 -2.4495 X_1^{(2)}\sin\phi_2\nonumber\\ \dot{x}_2(0) = 0\Rightarrow -3.1622X_1^{(1)} \sin\phi_1 + 12.2475 X_1^{(2)}\sin\phi_2\nonumber \end{eqnarray} From first two equations the \(X_1^{(1)}\cos\phi_1\) and \(X_2^{(2)}\cos\phi_2\) components are obtained. \begin{eqnarray} 2X_1^{(1)}\cos\phi_1 - 5X_1^{(2)}\cos\phi_2 &=& 0\nonumber\\ X_1^{(1)}\cos\phi_1 &=& \frac{5}{2}X_1^{(2)}\cos\phi_2\nonumber\\ \frac{5}{2}X_1^{(2)}\cos\phi_2 + X_1^{(2)}\cos\phi_2 &=& 1\nonumber\\ \frac{7}{2}X_1^{(2)}\cos\phi_2 &=& 1\Big/\cdot \frac{2}{7}\nonumber\\ X_1^{(2)}\cos\phi_2 &=& \frac{2}{7}\nonumber \end{eqnarray} From third and fourth equations the \(X_1^{(1)}\sin\phi_1=0\), and \(X_1^{(2)}\sin\phi_2 =0\) are obtained. \begin{eqnarray} -1.5811X_1^{(1)}\sin\phi_1 &=& 2.4495X_1^{(2)}X_1^{(2)}\sin\phi_2\nonumber\\ X_1^{(1)}\sin\phi_1 &=& -1.5492X_1^{(2)}\sin\phi_2\nonumber\\ -3.1622(-1.5492X_1^{(2)}\sin\phi_2) + 12.2475X_1^{(2)}\sin\phi_2 &=& 0\nonumber\\ X_1^{(2)}\sin\phi_2 &=& 0\nonumber\\ X_1^{(1)}\sin\phi_1 &=& 0\nonumber\\ \end{eqnarray} From previous solutions \(X_1^{1}\) and \(X_1^{(2)}\) are obtained. \begin{eqnarray} X_1^{(1)} &=& \sqrt{(X_1^{(1)}\sin\phi)^2 + (X_1^{(1)}\cos\phi_1)^2}\nonumber\\ X_1^{(1)} &=& \frac{5}{7}\nonumber\\ X_1^{(2)} &=& \sqrt{(X_1^{(2)}\sin\phi)^2 + (X_1^{(2)}\cos\phi_1)^2}\nonumber\\ X_1^{(2)} &=& \frac{2}{7}\nonumber\\ \frac{X_1^{(1)}\sin\phi_1}{X_1^{(1)}\cos\phi_1} &=& \frac{0}{\frac{5}{7}} = 0\Rightarrow \phi_1 = 0\nonumber\\ \phi_2 &=& 0\nonumber \end{eqnarray} The free-vibration responses of \(m_1\) and \(m_2\) are given by: \begin{eqnarray} x_1(t) &=& \frac{5}{7}\cos 1.5811t +\frac{2}{7}\cos2.4495t\nonumber\\ x_2(t) &=& \frac{10}{7}\cos 1.5811t -\frac{10}{7}\cos2.4495t\nonumber \end{eqnarray} Example 3 - Calculate the natural frequencies and mode shapes for the torsional system that is shown in Figure 3 if \(J_1 = J_0\), \(J_2 = 2J_0\), and \(k_{t1} = k_{t2} = k_t\).