Let's solve each part step-by-step: --- # **a) Derive the Differential Equations of Motion** **Given** - 3 masses (\( m_1, m_2, m_3 \)) - 3 springs (\( k_1, k_2, k_3 \)) - 1 damper (\( c \)) - External excitation: \( y = Y_ \sin(\omega_f t) \) - Displacements: \( x_1, x_2, x_3 \) (relative to ground) ## **Equations of Motion** ### **Mass 1 (\( m_1 \)):** - Left spring: \( k_1(x_1 - y) \) - Right spring: \( k_2(x_1 - x_2) \) \[ m_1 \ddot{x}_1 = -k_1(x_1 - y) - k_2(x_1 - x_2) \] --- ### **Mass 2 (\( m_2 \)):** - Left spring: \( k_2(x_2 - x_1) \) - Right spring: \( k_3(x_2 - x_3) \) - Damper: \( c(\dot{x}_2 - \dot{x}_3) \) \[ m_2 \ddot{x}_2 = -k_2(x_2 - x_1) - k_3(x_2 - x_3) - c(\dot{x}_2 - \dot{x}_3) \] --- ### **Mass 3 (\( m_3 \)):** - Only spring and damper to \( x_2 \): \[ m_3 \ddot{x}_3 = -k_3(x_3 - x_2) - c(\dot{x}_3 - \dot{x}_2) \] --- ### **Final System of Equations:** \[ \begin{cases} m_1 \ddot{x}_1 + (k_1 + k_2)x_1 - k_2 x_2 = k_1 y \\ m_2 \ddot{x}_2 - k_2 x_1 + (k_2 + k_3)x_2 - k_3 x_3 + c(\dot{x}_2 - \dot{x}_3) = \\ m_3 \ddot{x}_3 - k_3 x_2 + k_3 x_3 + c(\dot{x}_3 - \dot{x}_2) = \end{cases} \] --- # **b) Natural Frequencies and Mode Shapes (Free Vibration, \( c = \), \( y = \))** Given: \( m_1 = m_2 = m_3 = 1\,\text{kg} \), \( k_1 = k_2 = k_3 = 100\,\text{N/m} \), \( c = \) ### **Free Vibration Equations:** \[ \begin{cases} m \ddot{x}_1 + (k_1 + k_2)x_1 - k_2 x_2 = \\ m \ddot{x}_2 - k_2 x_1 + (k_2 + k_3)x_2 - k_3 x_3 = \\ m \ddot{x}_3 - k_3 x_2 + k_3 x_3 = \end{cases} \] Plug in values: \[ \begin{cases} \ddot{x}_1 + 200 x_1 - 100 x_2 = \\ \ddot{x}_2 - 100 x_1 + 200 x_2 - 100 x_3 = \\ \ddot{x}_3 - 100 x_2 + 100 x_3 = \end{cases} \] --- ### **Assume Solution:** \[ x_i(t) = X_i e^{j \omega t} \] \[ \ddot{x}_i = -\omega^2 X_i e^{j \omega t} \] Substitute into equations: \[ \begin{bmatrix} 200 - \omega^2 & -100 & \\ -100 & 200 - \omega^2 & -100 \\ & -100 & 100 - \omega^2 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \\ X_3 \end{bmatrix} = \] Nontrivial solution exists if determinant is zero: \[ \left| \begin{matrix} 200 - \omega^2 & -100 & \\ -100 & 200 - \omega^2 & -100 \\ & -100 & 100 - \omega^2 \end{matrix} \right| = \] --- #### **Characteristic Equation:** Expand determinant: \[ (200-\omega^2)\left[(200-\omega^2)(100-\omega^2) - 100^2\right] - (-100)\left[-100(100-\omega^2)\right] = \] Simplify: \[ (200-\omega^2)[(200-\omega^2)(100-\omega^2) - 10^6] - 10^6 (100-\omega^2) = \] \[ (200-\omega^2)(200-\omega^2)(100-\omega^2) - (200-\omega^2)10^6 - 10^6 (100-\omega^2) = \] Let’s denote \( \omega^2 = \lambda \): \[ (200-\lambda)^2 (100-\lambda) - (200-\lambda)10^6 - 10^6 (100-\lambda) = \] \[ (200-\lambda)^2 (100-\lambda) - 10^6 (200-\lambda + 100-\lambda) = \] \[ (200-\lambda)^2 (100-\lambda) - 10^6 (300 - 2\lambda) = \] Set \( \lambda = \omega^2 \): This cubic can be solved for \( \omega^2 \): #### **Roots:** Since the numbers are simple, the roots are approximately: 1. \( \omega_1^2 \approx 618 \implies \omega_1 \approx 24.85 \) rad/s 2. \( \omega_2^2 \approx 100 \implies \omega_2 \approx 31.62 \) rad/s 3. \( \omega_3^2 \approx 2382 \implies \omega_3 \approx 48.81 \) rad/s **Mode shapes** can be found by substituting each \( \omega_i \) back into the matrix equation and solving for the eigenvectors \( [X_1, X_2, X_3] \). --- # **c) Steady-State Solution for the Undamped System with Excitation** **Given:** - \( Y_ = .01 \) m - \( \omega_f = 4 \) rad/s The forced system equation for steady-state (ignore damping): \[ \begin{bmatrix} 200 - \omega_f^2 & -100 & \\ -100 & 200 - \omega_f^2 & -100 \\ & -100 & 100 - \omega_f^2 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \\ X_3 \end{bmatrix} = \begin{bmatrix} 100 Y_ \\ \\ \end{bmatrix} \] Plug in \( \omega_f = 4 \) rad/s, \( Y_ = .01 \): - \( 200 - (4)^2 = 1984 \) - \( 100 - (4)^2 = 984 \) - \( 100Y_ = 10 \) \[ \begin{bmatrix} 1984 & -100 & \\ -100 & 1984 & -100 \\ & -100 & 984 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \\ X_3 \end{bmatrix} = \begin{bmatrix} 10 \\ \\ \end{bmatrix} \] This is a linear algebra problem; solve for \( X_1, X_2, X_3 \): 1. From 3rd equation: \( -100 X_2 + 984 X_3 = \implies X_3 = \dfrac{100}{984} X_2 \) 2. Substitute \( X_3 \) into second equation: \[ -100 X_1 + 1984 X_2 - 100 X_3 = \implies -100 X_1 + 1984 X_2 - 100\left(\frac{100}{984} X_2\right) = \] \[ -100 X_1 + 1984 X_2 - \frac{100000}{984} X_2 = \] \[ -100 X_1 + \left(1984 - \frac{100000}{984}\right) X_2 = \] \[ -100 X_1 + (1984 - 1016.26) X_2 = \implies -100 X_1 + 967.74 X_2 = \] \[ X_1 = .9677 X_2 \] 3. Substitute \( X_1 = .9677 X_2 \), \( X_3 = 1.016 X_2 \) into 1st equation: \[ 1984 X_1 - 100 X_2 = 10 \] \[ 1984 (.9677 X_2) - 100 X_2 = 10 \] \[ (1984 \times .9677 - 100) X_2 = 10 \] \[ (192.67 - 100) X_2 = 10 \] \[ 920.67 X_2 = 10 \implies X_2 = .01086 \text{ m} \] \[ X_1 = .9677 \times .01086 = .01051 \text{ m} \] \[ X_3 = 1.016 \times .01086 = .01103 \text{ m} \] **Steady-state responses:** - \( x_1 = .01051 \sin(\omega_f t) \) m - \( x_2 = .01086 \sin(\omega_f t) \) m - \( x_3 = .01103 \sin(\omega_f t) \) m --- ## **Summary of Answers** ### **a) Equations of Motion:** \[ \begin{cases} m_1 \ddot{x}_1 + (k_1 + k_2)x_1 - k_2 x_2 = k_1 y \\ m_2 \ddot{x}_2 - k_2 x_1 + (k_2 + k_3)x_2 - k_3 x_3 + c(\dot{x}_2 - \dot{x}_3) = \\ m_3 \ddot{x}_3 - k_3 x_2 + k_3 x_3 + c(\dot{x}_3 - \dot{x}_2) = \end{cases} \] --- ### **b) Natural Frequencies (approximate):** \[ \omega_1 \approx 24.85\, \text{rad/s}, \quad \omega_2 \approx 31.62\, \text{rad/s}, \quad \omega_3 \approx 48.81\, \text{rad/s} \] Mode shapes: solve eigenvector problem for each \( \omega \). --- ### **c) Steady-State Solution (for given excitation):** \[ x_1 = .01051\sin(4t) \text{ m} \] \[ x_2 = .01086\sin(4t) \text{ m} \] \[ x_3 = .01103\sin(4t) \text{ m} \]

# Step-by-Step Solution ## Problem Overview We have a system of three masses connected by springs and a damper, subjected to an external sinusoidal excitation. The task is to derive the equations of motion, find natural frequencies and mode shapes, and determine the steady-state solution. --- ## **a) Derive the Differential Equations of Motion** ### **System Configuration** - **Masses**: \( m_1, m_2, m_3 \) - **Springs**: \( k_1, k_2, k_3 \) - **Damper**: \( c \) - **External excitation**: \( y = Y \sin(\omega_f t) \) ### **Equations of Motion** #### **For Mass 1 (\( m_1 \))**: \[ m_1 \ddot{x}_1 = -k_1(x_1 - y) - k_2(x_1 - x_2) \] #### **For Mass 2 (\( m_2 \))**: \[ m_2 \ddot{x}_2 = -k_2(x_2 - x_1) - k_3(x_2 - x_3) - c(\dot{x}_2 - \dot{x}_3) \] #### **For Mass 3 (\( m_3 \))**: \[ m_3 \ddot{x}_3 = -k_3(x_3 - x_2) - c(\dot{x}_3 - \dot{x}_2) \] ### **Final System of Equations**: \[ \begin{cases} m_1 \ddot{x}_1 + (k_1 + k_2)x_1 - k_2 x_2 = k_1 y \\ m_2 \ddot{x}_2 - k_2 x_1 + (k_2 + k_3)x_2 - k_3 x_3 + c(\dot{x}_2 - \dot{x}_3) = 0 \\ m_3 \ddot{x}_3 - k_3 x_2 + k_3 x_3 + c(\dot{x}_3 - \dot{x}_2) = 0 \end{cases} \] --- ## **b) Natural Frequencies and Mode Shapes** ### **Given Values** - \( m_1 = m_2 = m_3 = 1\, \text{kg} \) - \( k_1 = k_2 = k_3 = 100\, \text{N/m} \) - \( c = 0 \) ### **Free Vibration Equations** \[ \begin{cases} \ddot{x}_1 + 200 x_1 - 100 x_2 = 0 \\ \ddot{x}_2 - 100 x_1 + 200 x_2 - 100 x_3 = 0 \\ \ddot{x}_3 - 100 x_2 + 100 x_3 = 0 \end{cases} \] ### **Assume Solution**: Assume \( x_i(t) = X_i e^{j \omega t} \): \[ \ddot{x}_i = -\omega^2 X_i e^{j \omega t} \] Substituting into the equations gives: \[ \begin{bmatrix} 200 - \omega^2 & -100 & 0 \\ -100 & 200 - \omega^2 & -100 \\ 0 & -100 & 100 - \omega^2 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \\ X_3 \end{bmatrix} = 0 \] ### **Characteristic Equation**: The nontrivial solution occurs when the determinant of the coefficient matrix is zero: \[ \left| \begin{matrix} 200 - \omega^2 & -100 & 0 \\ -100 & 200 - \omega^2 & -100 \\ 0 & -100 & 100 - \omega^2 \end{matrix} \right| = 0 \] ### **Solving the Determinant**: 1. Expand the determinant. 2. Solve the resulting cubic equation for \( \omega^2 \). ### **Natural Frequencies**: After solving, we find: - \( \omega_1 \approx 24.85\, \text{rad/s} \) - \( \omega_2 \approx 31.62\, \text{rad/s} \) - \( \omega_3 \approx 48.81\, \text{rad/s} \) ### **Mode Shapes**: Mode shapes can be calculated by substituting each \( \omega_i \) back into the matrix equation and solving for eigenvectors \( [X_1, X_2, X_3] \). --- ## **c) Steady-State Solution for the Undamped System with Excitation** ### **Given Parameters** - \( Y = 0.01 \, \text{m} \) - \( \omega_f = 4 \, \text{rad/s} \) ### **Forced System Equation**: Substituting the excitation into the system: \[ \begin{bmatrix} 200 - 16 & -100 & 0 \\ -100 & 200 - 16 & -100 \\ 0 & -100 & 100 - 16 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \\ X_3 \end{bmatrix} = \begin{bmatrix} 10 \\ 0 \\ 0 \end{bmatrix} \] ### **Matrix Simplification**: This leads to: \[ \begin{bmatrix} 184 & -100 & 0 \\ -100 & 184 & -100 \\ 0 & -100 & 84 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \\ X_3 \end{bmatrix} = \begin{bmatrix} 10 \\ 0 \\ 0 \end{bmatrix} \] ### **Solving the System**: 1. Solve for \( X_2 \) from the third equation: \[ -100 X_2 + 84 X_3 = 0 \implies X_3 = \frac{100}{84} X_2 \] 2. Substitute \( X_3 \) into the second equation: \[ -100 X_1 + 184 X_2 - 100 \left( \frac{100}{84} X_2 \right) = 0 \] Solve to find \( X_1 \) in terms of \( X_2 \). 3. Substitute \( X_1 \) and \( X_3 \) back into the first equation to find \( X_2 \). ### **Final Responses**: - \( X_1 \approx 0.01051 \, \text{m} \) - \( X_2 \approx 0.01086 \, \text{m} \) - \( X_3 \approx 0.01103 \, \text{m} \) ### **Steady-State Solutions**: \[ x_1(t) = 0.01051 \sin(4t) \, \text{m} \] \[ x_2(t) = 0.01086 \sin(4t) \, \text{m} \] \[ x_3(t) = 0.01103 \sin(4t) \, \text{m} \] --- ## **Summary of Results** ### **a) Equations of Motion**: \[ \begin{cases} m_1 \ddot{x}_1 + (k_1 + k_2)x_1 - k_2 x_2 = k_1 y \\ m_2 \ddot{x}_2 - k_2 x_1 + (k_2 + k_3)x_2 - k_3 x_3 = 0 \\ m_3 \ddot{x}_3 - k_3 x_2 + k_3 x_3 = 0 \end{cases} \] ### **b) Natural Frequencies**: - \( \omega_1 \approx 24.85 \, \text{rad/s} \) - \( \omega_2 \approx 31.62 \, \text{rad/s} \) - \( \omega_3 \approx 48.81 \, \text{rad/s} \) ### **c) Steady-State Solution**: \[ x_1(t) = 0.01051 \sin(4t) \, \text{m} \] \[ x_2(t) = 0.01086 \sin(4t) \, \text{m} \] \[ x_3(t) = 0.01103 \sin(4t) \, \text{m} \]