Example 1 Calculate the natural frequencies of the system shown in Figure 1, with \(m_1=m\), \(m_2 = 2m\), \(k_1 = k\), \(k_2=2k.\)
Figure 1 - System with two springs and two bodies with masses \(m_1\) and \(m_2\).
Determine the response of the system when \(k=1000 \left[\frac{\mathrm{N}}{\mathrm{m}}\right]\), \(m = 20\left[\mathrm{kg}\right]\), and initial values of the displacements of the masses \(m_1\) and \(m_2\) are 1 and -1, respectively.
Solution Differential equations that describe the motion of the system shown in Figure 1 can be written as: \begin{eqnarray} m_1\ddot{x}_1 + (k_1+k_2)x_1 - k_2x_2 &=& 0\nonumber\\ m_2\ddot{x}_2 + k_2 x_2 - k_2 x_1 &=& 0\nonumber \end{eqnarray} The assumed solutions for previous two equations are: \begin{eqnarray} x_1(t) &=& X_1 \cos(\omega t + \phi)\nonumber\\ x_2(t) &=& X_2 \cos(\omega t + \phi)\nonumber. \end{eqnarray} Substituting previous two solutions into differential equations the following expression is obtained: \begin{eqnarray} -m_1X_1\omega^2 \cos(\omega t +\phi) +(k_1 + k_2)X_1\cos(\omega t + \phi) - k_2 X_2 \cos(\omega t+ \phi) &=& 0\nonumber\\ -m_2X_2\omega^2 \cos(\omega t + \phi) + k_2 X_2 \cos(\omega t + \phi) - k_2 X_1\cos(\omega t + \phi) &=& 0\nonumber \end{eqnarray} Dividing previous expression with \(\cos(\omega t + \phi)\) the previous expressions are reduced to: \begin{eqnarray} -m_1X_1\omega^2 + (k_1+k_2)X_1 -k_2X_2 &=& 0\nonumber\\ -m_2X_2\omega^2 + k_2X_2 -k_2X_1 &=& 0\nonumber \end{eqnarray} The previous expression can be written in matrix form: \begin{eqnarray} \begin{bmatrix} -m_1\omega^2 + k_1 + k_2 & -k_2 \\ -k_2 & -\omega^2 m_2 +k_2 \end{bmatrix} \begin{Bmatrix} X_1 \\X_2 \end{Bmatrix} &=& 0\nonumber \end{eqnarray} Solving determinant of first matrix. \begin{eqnarray} \begin{vmatrix} -m_1\omega^2 + k_1 + k_2 & -k_2 \\ -k_2 & -\omega^2 m_2 +k_2 \end{vmatrix} &=& 0\nonumber\\ (-m_1\omega^2 +k_1 + k_2)(-\omega^2m_2 + k_2) - (-k_2)(-k_2)&=& 0\nonumber\\ m_1m_2\omega^4 -m_1\omega^2 k_2 -\omega^2 m_2 k_1 + k_1 k_2 - \omega^2 m_2k_2 + k_2^2 -k_2^2 &=&0\Big/:m_1m_2\nonumber\\ \omega^4 - \frac{\omega^2}{m_1m_2}\left(m_1k_2 + m_2k_1+ m_2 k_2\right) + \frac{k_1k_2}{m_1m_2} &=& 0\nonumber\\ \omega^4 - \omega^2\left(\frac{m_2(k_1+k_2)}{m_1m_2} + \frac{m_1k_2}{m_1m_2}\right) + \frac{k_1k_2}{m_1m_2} &=& 0\nonumber\\ \omega^4 -\omega^2\left(\frac{k_1+k_2}{m_1} + \frac{k_2}{m_2}\right) + \frac{k_1k_2}{m_1m_2} &=& 0\nonumber \end{eqnarray} The next step is to find roots of the frequency equation (previous expression): \begin{eqnarray} \omega_{1,2} &=& \frac{\left(\frac{k_1+k_2}{m_1} + \frac{k_2}{m_2}\right)\pm\sqrt{\left(\frac{k_1+k_2}{m_1} + \frac{k_2}{m_2}\right)^2 - 4\frac{k_1k_2}{m_1m_2}}}{2}\nonumber\\ \omega_{1,2} &=& \frac{\left(\frac{k_1+k_2}{m_1} + \frac{k_2}{m_2}\right)\pm\sqrt{\left(\frac{k_1+k_2}{m_1}\right)^2 + 2\frac{k_1+k_2}{m_1}\frac{k_2}{m_2} + \left(\frac{k_2}{m_2}\right)^2 - 4\frac{k_1k_2}{m_1m_2}}}{2}\nonumber\\ &=& \frac{\left(\frac{k_1+k_2}{m_1} + \frac{k_2}{m_2}\right)\pm\sqrt{\frac{k_1^2 + 2k_1k_2 +k_2^2}{m_1^2} + 2\frac{k_2(k_1+k_2)}{m_1m_2} + \frac{k_2^2}{m_2^2} - 4\frac{k_1k_2}{m_1m_2}}}{2}\nonumber\\ \omega_1,\omega_2 &=& \frac{k_1+k_2}{2m_1} + \frac{k_2}{2m}\pm \sqrt{\frac{1}{4}\left(\frac{k_1 + k_2}{m_1} + \frac{k_2}{m_2}\right)^2 - \frac{k_1k_2}{m_1m_2} }\nonumber\\ \end{eqnarray} If: \begin{eqnarray} \vec{X}^{(1)} &=& \begin{Bmatrix}X_1^{(1)}\\X_2^{(1)} = r_1 X_1^{(1)}\end{Bmatrix}\nonumber\\ \vec{X}^{(2)} &=& \begin{Bmatrix}X_1^{(2)}\\X_2^{(2)} = r_1 X_1^{(2)}\end{Bmatrix} \end{eqnarray} then: \begin{eqnarray} r_1 &=& \frac{X_2^{(1)}}{X_1^{(1)}} = \frac{-m_1\omega_1^2 +k_1 +k_2}{k_2} = \frac{k_2}{-m_2\omega_1^2 + k_2}\nonumber\\ r_2 &=& \frac{X_2^{(2)}}{X_1^{(2)}} = \frac{m_1 \omega_2^2 +k_1 + k_2}{k_2} = \frac{k_2}{-m_2\omega_2^2 +k_2}\nonumber \end{eqnarray} General solution i.e. \(x_1(t)\), and \(x_2(t)\) can be written as: \begin{eqnarray} x_1(t) &=& X_1^{(1)} \cos(\omega_1 t + \phi_1) + X_1^{(2)} \cos(\omega_2 t + \phi_2)\nonumber\\ x_2(t) &=& r_1X_1^{(1)}\cos(\omega_1 t + \phi_1) + r_2 X_1^{(2)} \cos(\omega_2 t + \phi)\nonumber \end{eqnarray} In case were \(m_1 = m\), \(m_2 = 2m\), \(k_1 = k\), and \(k_2 = 2k\) gives the solution: \begin{eqnarray} \omega_1^2 &=& (2-\sqrt{3})\frac{k}{m}\nonumber\\ \omega_2^2 &=& (2+\sqrt{3})\frac{k}{m}.\nonumber \end{eqnarray} In case where \(k=1000 \left[\frac{\mathrm{N}}{\mathrm{m}}\right]\), and \(m = 20\left[\mathrm{kg}\right]\) then: \begin{eqnarray} \omega_1 &=& \sqrt{(2-\sqrt{3})\frac{1000}{20}}\nonumber\\ \omega_1 &=& 3.6603\left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber\\ \omega_2 &=& \sqrt{(2+\sqrt{3})}\frac{1000}{20}\nonumber\\ \omega_2 &=& 13.6603 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber \end{eqnarray} The values of \(r_1\) and \(r_2\) are equal to: \begin{eqnarray} r_1 &=& \frac{k_2}{-m_2\omega_1^2 + k_2}\nonumber\\ &=& \frac{1000}{-20\cdot 3.6603^2 + 1000} \nonumber\\ &=& 1.36604\nonumber\\ r_2 &=& \frac{k_2}{-m_2\omega_2^2 + k_2}\nonumber\\ &=& \frac{1000}{-20\cdot 13.6603^2 + 1000} \nonumber\\ &=& -0.36602\nonumber \end{eqnarray} The initial conditions \(x_1(0) = 1\), \(\dot{x}_1(0) = 0\), \(x_2(0) = -1\), and \(\dot{x}_2(0) = 0\) give parameter value of solution funcitons \(x_1(t)\), and \(x_2(t)\). \begin{eqnarray} X_1^{(1)} &=& -0.36602\quad X_1^{(2)} = -1.36603,\quad \phi_1 =\phi_2 = 0\nonumber \end{eqnarray} The response of the system can be written as: \begin{eqnarray} x_1(t) &=& -0.36602\cos3.6604 t - 1.36603 \cos 13.6603 t\nonumber\\ x_2(t) &=& -0.5 \cos 3.6603 t+ 0.5\cos13.66603 t\nonumber \end{eqnarray} Example 2Set up the differential equations of motion for the double pendulum shown in Figure 2, using the coordinates \(x_1\) and \(x_2\) and assuming small amplitudes. Find the natural frequencies, the ratios of amplitudes, and the locations of nodes for the two modes of vibration when \(m_1 = m_2 = m\) and \(l_1 = l_2 = l\).
Figure 2 - Double pendulum system.
Solution The differential equations that describe the motion of the system can be written as:
\begin{eqnarray}
m_1l_1\ddot{x}_1 + \left(W_1+W_2\left(\frac{l_1+l_2}{l_2}\right)\right)x_1 -\frac{W_2}{l_2}x_2 &=& 0\nonumber\\
m_2l_2\ddot{x}_2 - W_2x_1 + W_2x_2 &=& 0\nonumber
\end{eqnarray}
When \(m_1 = m_2 = m,\) \(l_1=l_2 = l,\) and \(W_1 = W_2 = mg\) the previous system of differential equations can be written as:
\begin{eqnarray}
ml\ddot{x_1} + 3mgx_1 -mgx_2 &=& 0\nonumber\\
ml\ddot{x_2} - mgx_1 + mgx_2 &=& 0\nonumber
\end{eqnarray}
In case of harmonic motion the solutions are:
\begin{eqnarray}
x_i (t) &=& X_i \cos\omega t\quad i = 1,2
\end{eqnarray}
If substituted into differential equations the following expressions are obtained:
\begin{eqnarray}
-\omega^2mlX_1 + 3mgX_1 - mgX_2 &=& 0\nonumber\\
-\omega^2mlX_2 -mgX_1 + mgX_"2 &=& 0\nonumber
\end{eqnarray}
From preivous two expression the frequency equation can be obtained and written as:
\begin{eqnarray}
\omega^4m^2l^2 - (4m^2 lg) \omega ^2 + 2 m^2 g^2 &=& 0\nonumber
\end{eqnarray}
The solution of the previous frequency equation can be written as:
\begin{eqnarray}
\omega_1^2 ,\omega_2^2 &=& (2\pm\sqrt{2})\frac{g}{l}\nonumber\\
\omega_1 &=& 0.7654\sqrt{\frac{g}{l}}\nonumber\\
\omega_2 &=& 1.8478\sqrt{\frac{g}{l}}\nonumber
\end{eqnarray}
Ratio of amplitudes can be written as:
\begin{eqnarray}
\frac{X_1}{X_2} &=& \frac{mg}{-\omega^2 ml+3mg} = \frac{1}{-\omega^2\frac{l}{g} + 3}\nonumber
\end{eqnarray}
In mode 1:
\begin{eqnarray}
\omega_1 = 0.7654\sqrt{\frac{g}{l}}\nonumber\\
r_1 &=& \left(\frac{X_1}{X_2}\right)^{(1)} = 0.4142\nonumber
\end{eqnarray}
In mode 2:
\begin{eqnarray}
\omega_2 &=& 1.8468\sqrt{\frac{g}{l}}
r_2 &=& \left(\frac{X_1}{X_2}\right)^{(2)} = -2.4133\nonumber
\end{eqnarray}
Determine the response of the system when \(k=1000 \left[\frac{\mathrm{N}}{\mathrm{m}}\right]\), \(m = 20\left[\mathrm{kg}\right]\), and initial values of the displacements of the masses \(m_1\) and \(m_2\) are 1 and -1, respectively.
Solution Differential equations that describe the motion of the system shown in Figure 1 can be written as: \begin{eqnarray} m_1\ddot{x}_1 + (k_1+k_2)x_1 - k_2x_2 &=& 0\nonumber\\ m_2\ddot{x}_2 + k_2 x_2 - k_2 x_1 &=& 0\nonumber \end{eqnarray} The assumed solutions for previous two equations are: \begin{eqnarray} x_1(t) &=& X_1 \cos(\omega t + \phi)\nonumber\\ x_2(t) &=& X_2 \cos(\omega t + \phi)\nonumber. \end{eqnarray} Substituting previous two solutions into differential equations the following expression is obtained: \begin{eqnarray} -m_1X_1\omega^2 \cos(\omega t +\phi) +(k_1 + k_2)X_1\cos(\omega t + \phi) - k_2 X_2 \cos(\omega t+ \phi) &=& 0\nonumber\\ -m_2X_2\omega^2 \cos(\omega t + \phi) + k_2 X_2 \cos(\omega t + \phi) - k_2 X_1\cos(\omega t + \phi) &=& 0\nonumber \end{eqnarray} Dividing previous expression with \(\cos(\omega t + \phi)\) the previous expressions are reduced to: \begin{eqnarray} -m_1X_1\omega^2 + (k_1+k_2)X_1 -k_2X_2 &=& 0\nonumber\\ -m_2X_2\omega^2 + k_2X_2 -k_2X_1 &=& 0\nonumber \end{eqnarray} The previous expression can be written in matrix form: \begin{eqnarray} \begin{bmatrix} -m_1\omega^2 + k_1 + k_2 & -k_2 \\ -k_2 & -\omega^2 m_2 +k_2 \end{bmatrix} \begin{Bmatrix} X_1 \\X_2 \end{Bmatrix} &=& 0\nonumber \end{eqnarray} Solving determinant of first matrix. \begin{eqnarray} \begin{vmatrix} -m_1\omega^2 + k_1 + k_2 & -k_2 \\ -k_2 & -\omega^2 m_2 +k_2 \end{vmatrix} &=& 0\nonumber\\ (-m_1\omega^2 +k_1 + k_2)(-\omega^2m_2 + k_2) - (-k_2)(-k_2)&=& 0\nonumber\\ m_1m_2\omega^4 -m_1\omega^2 k_2 -\omega^2 m_2 k_1 + k_1 k_2 - \omega^2 m_2k_2 + k_2^2 -k_2^2 &=&0\Big/:m_1m_2\nonumber\\ \omega^4 - \frac{\omega^2}{m_1m_2}\left(m_1k_2 + m_2k_1+ m_2 k_2\right) + \frac{k_1k_2}{m_1m_2} &=& 0\nonumber\\ \omega^4 - \omega^2\left(\frac{m_2(k_1+k_2)}{m_1m_2} + \frac{m_1k_2}{m_1m_2}\right) + \frac{k_1k_2}{m_1m_2} &=& 0\nonumber\\ \omega^4 -\omega^2\left(\frac{k_1+k_2}{m_1} + \frac{k_2}{m_2}\right) + \frac{k_1k_2}{m_1m_2} &=& 0\nonumber \end{eqnarray} The next step is to find roots of the frequency equation (previous expression): \begin{eqnarray} \omega_{1,2} &=& \frac{\left(\frac{k_1+k_2}{m_1} + \frac{k_2}{m_2}\right)\pm\sqrt{\left(\frac{k_1+k_2}{m_1} + \frac{k_2}{m_2}\right)^2 - 4\frac{k_1k_2}{m_1m_2}}}{2}\nonumber\\ \omega_{1,2} &=& \frac{\left(\frac{k_1+k_2}{m_1} + \frac{k_2}{m_2}\right)\pm\sqrt{\left(\frac{k_1+k_2}{m_1}\right)^2 + 2\frac{k_1+k_2}{m_1}\frac{k_2}{m_2} + \left(\frac{k_2}{m_2}\right)^2 - 4\frac{k_1k_2}{m_1m_2}}}{2}\nonumber\\ &=& \frac{\left(\frac{k_1+k_2}{m_1} + \frac{k_2}{m_2}\right)\pm\sqrt{\frac{k_1^2 + 2k_1k_2 +k_2^2}{m_1^2} + 2\frac{k_2(k_1+k_2)}{m_1m_2} + \frac{k_2^2}{m_2^2} - 4\frac{k_1k_2}{m_1m_2}}}{2}\nonumber\\ \omega_1,\omega_2 &=& \frac{k_1+k_2}{2m_1} + \frac{k_2}{2m}\pm \sqrt{\frac{1}{4}\left(\frac{k_1 + k_2}{m_1} + \frac{k_2}{m_2}\right)^2 - \frac{k_1k_2}{m_1m_2} }\nonumber\\ \end{eqnarray} If: \begin{eqnarray} \vec{X}^{(1)} &=& \begin{Bmatrix}X_1^{(1)}\\X_2^{(1)} = r_1 X_1^{(1)}\end{Bmatrix}\nonumber\\ \vec{X}^{(2)} &=& \begin{Bmatrix}X_1^{(2)}\\X_2^{(2)} = r_1 X_1^{(2)}\end{Bmatrix} \end{eqnarray} then: \begin{eqnarray} r_1 &=& \frac{X_2^{(1)}}{X_1^{(1)}} = \frac{-m_1\omega_1^2 +k_1 +k_2}{k_2} = \frac{k_2}{-m_2\omega_1^2 + k_2}\nonumber\\ r_2 &=& \frac{X_2^{(2)}}{X_1^{(2)}} = \frac{m_1 \omega_2^2 +k_1 + k_2}{k_2} = \frac{k_2}{-m_2\omega_2^2 +k_2}\nonumber \end{eqnarray} General solution i.e. \(x_1(t)\), and \(x_2(t)\) can be written as: \begin{eqnarray} x_1(t) &=& X_1^{(1)} \cos(\omega_1 t + \phi_1) + X_1^{(2)} \cos(\omega_2 t + \phi_2)\nonumber\\ x_2(t) &=& r_1X_1^{(1)}\cos(\omega_1 t + \phi_1) + r_2 X_1^{(2)} \cos(\omega_2 t + \phi)\nonumber \end{eqnarray} In case were \(m_1 = m\), \(m_2 = 2m\), \(k_1 = k\), and \(k_2 = 2k\) gives the solution: \begin{eqnarray} \omega_1^2 &=& (2-\sqrt{3})\frac{k}{m}\nonumber\\ \omega_2^2 &=& (2+\sqrt{3})\frac{k}{m}.\nonumber \end{eqnarray} In case where \(k=1000 \left[\frac{\mathrm{N}}{\mathrm{m}}\right]\), and \(m = 20\left[\mathrm{kg}\right]\) then: \begin{eqnarray} \omega_1 &=& \sqrt{(2-\sqrt{3})\frac{1000}{20}}\nonumber\\ \omega_1 &=& 3.6603\left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber\\ \omega_2 &=& \sqrt{(2+\sqrt{3})}\frac{1000}{20}\nonumber\\ \omega_2 &=& 13.6603 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber \end{eqnarray} The values of \(r_1\) and \(r_2\) are equal to: \begin{eqnarray} r_1 &=& \frac{k_2}{-m_2\omega_1^2 + k_2}\nonumber\\ &=& \frac{1000}{-20\cdot 3.6603^2 + 1000} \nonumber\\ &=& 1.36604\nonumber\\ r_2 &=& \frac{k_2}{-m_2\omega_2^2 + k_2}\nonumber\\ &=& \frac{1000}{-20\cdot 13.6603^2 + 1000} \nonumber\\ &=& -0.36602\nonumber \end{eqnarray} The initial conditions \(x_1(0) = 1\), \(\dot{x}_1(0) = 0\), \(x_2(0) = -1\), and \(\dot{x}_2(0) = 0\) give parameter value of solution funcitons \(x_1(t)\), and \(x_2(t)\). \begin{eqnarray} X_1^{(1)} &=& -0.36602\quad X_1^{(2)} = -1.36603,\quad \phi_1 =\phi_2 = 0\nonumber \end{eqnarray} The response of the system can be written as: \begin{eqnarray} x_1(t) &=& -0.36602\cos3.6604 t - 1.36603 \cos 13.6603 t\nonumber\\ x_2(t) &=& -0.5 \cos 3.6603 t+ 0.5\cos13.66603 t\nonumber \end{eqnarray} Example 2Set up the differential equations of motion for the double pendulum shown in Figure 2, using the coordinates \(x_1\) and \(x_2\) and assuming small amplitudes. Find the natural frequencies, the ratios of amplitudes, and the locations of nodes for the two modes of vibration when \(m_1 = m_2 = m\) and \(l_1 = l_2 = l\).