Let's solve the problem step by step: --- # **(a) Differential Equations of Motion** Define: - \( x_1, x_2, x_3 \): Displacements of \( m_1, m_2, m_3 \) from equilibrium. - \( y = Y_ \sin \omega_f t \): Base excitation. ## **For \( m_1 \):** Forces on \( m_1 \): - Spring \( k_1 \) (relative to base): \( -k_1(x_1 - y) \) - Spring \( k_2 \) (relative to \( m_2 \)): \( -k_2 (x_1 - x_2) \) Equation: \[ m_1 \ddot{x}_1 = -k_1(x_1 - y) - k_2(x_1 - x_2) \] ## **For \( m_2 \):** Forces on \( m_2 \): - Spring \( k_2 \): \( -k_2(x_2 - x_1) \) - Damper \( c \): \( -c (\dot{x}_2 - \dot{x}_3) \) - Spring \( k_3 \): \( -k_3(x_2 - x_3) \) Equation: \[ m_2 \ddot{x}_2 = -k_2(x_2 - x_1) - c(\dot{x}_2 - \dot{x}_3) - k_3(x_2 - x_3) \] ## **For \( m_3 \):** Forces on \( m_3 \): - Damper \( c \): \( c(\dot{x}_2 - \dot{x}_3) \) - Spring \( k_3 \): \( k_3(x_2 - x_3) \) Equation: \[ m_3 \ddot{x}_3 = c(\dot{x}_2 - \dot{x}_3) + k_3(x_2 - x_3) \] --- # **(b) Natural Frequencies and Mode Shapes (Free Vibration, \( c=, y= \))** Given: \( m_1 = m_2 = m_3 = m \), \( k_1 = k_2 = k_3 = k \) ### **Write Equations in Matrix Form** \[ m\ddot{x}_1 = -k x_1 - k(x_1 - x_2) \] \[ m\ddot{x}_2 = -k(x_2 - x_1) - k(x_2 - x_3) \] \[ m\ddot{x}_3 = k(x_2 - x_3) \] Arrange: \[ m \begin{bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \\ \ddot{x}_3 \end{bmatrix} + k \begin{bmatrix} 2 & -1 & \\ -1 & 2 & -1 \\ & -1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \] Or, \( m \ddot{\mathbf{x}} + k \mathbf{K}\mathbf{x} = \) ### **Assume Solution:** \[ \mathbf{x} = \mathbf{X} e^{i\omega t} \] So, \[ (-m\omega^2 \mathbf{I} + k\mathbf{K})\mathbf{X} = \] \[ \left( \frac{k}{m}\mathbf{K} - \omega^2 \mathbf{I} \right)\mathbf{X} = \] Non-trivial solution: \( \det\left( \mathbf{K} - \lambda \mathbf{I} \right) = \), where \( \lambda = \frac{m\omega^2}{k} \). ### **Find Eigenvalues (\( \lambda \)):** \[ \mathbf{K} = \begin{bmatrix} 2 & -1 & \\ -1 & 2 & -1 \\ & -1 & 1 \end{bmatrix} \] So, \[ \left| \begin{array}{ccc} 2-\lambda & -1 & \\ -1 & 2-\lambda & -1 \\ & -1 & 1-\lambda \end{array} \right| = \] Expand determinant: First, expand along the first row: \[ (2-\lambda) \left| \begin{array}{cc} 2-\lambda & -1 \\ -1 & 1-\lambda \end{array} \right| - (-1) \left| \begin{array}{cc} -1 & -1 \\ & 1-\lambda \end{array} \right| \] Calculate: \[ (2-\lambda)[(2-\lambda)(1-\lambda) - (-1)(-1)] + 1[(-1)(1-\lambda) - ] \] \[ = (2-\lambda)[(2-\lambda)(1-\lambda) - 1] + [-(1-\lambda)] \] Now, expand \((2-\lambda)(1-\lambda)\): \[ (2-\lambda)(1-\lambda) = 2 - 2\lambda - \lambda + \lambda^2 = 2 - 3\lambda + \lambda^2 \] So, the determinant is: \[ (2-\lambda)[(2 - 3\lambda + \lambda^2) - 1] - (1-\lambda) \] \[ = (2-\lambda)(1 - 3\lambda + \lambda^2) - (1-\lambda) \] \[ = (2-\lambda)(1 - 3\lambda + \lambda^2) - (1-\lambda) \] Now expand \((2-\lambda)(1 - 3\lambda + \lambda^2)\): \[ = 2(1 - 3\lambda + \lambda^2) - \lambda(1 - 3\lambda + \lambda^2) \] \[ = 2 - 6\lambda + 2\lambda^2 - \lambda + 3\lambda^2 - \lambda^3 \] \[ = 2 - 6\lambda + 2\lambda^2 - \lambda + 3\lambda^2 - \lambda^3 \] \[ = 2 - 7\lambda + 5\lambda^2 - \lambda^3 \] So the determinant is: \[ 2 - 7\lambda + 5\lambda^2 - \lambda^3 - (1-\lambda) = 2 - 7\lambda + 5\lambda^2 - \lambda^3 - 1 + \lambda \] \[ = (2-1) + (-7\lambda+\lambda) + 5\lambda^2 - \lambda^3 \] \[ = 1 - 6\lambda + 5\lambda^2 - \lambda^3 \] Set determinant to zero: \[ 1 - 6\lambda + 5\lambda^2 - \lambda^3 = \] Or \[ -\lambda^3 + 5\lambda^2 - 6\lambda + 1 = \] Multiply both sides by -1: \[ \lambda^3 - 5\lambda^2 + 6\lambda - 1 = \] ### **Solve Cubic Equation for \( \lambda \):** Let’s use the cubic formula or numerical methods. For now, let’s denote the roots as \( \lambda_1, \lambda_2, \lambda_3 \). Recall \( \lambda = \frac{m\omega^2}{k} \implies \omega = \sqrt{ \frac{k}{m} \lambda } \). Given \( m = 1 \) kg, \( k = 100 \) N/m: So, \[ \omega_n = \sqrt{ 100 \lambda } \] ### **Approximate Roots Numerically (or using rational root theorem):** Test \( \lambda = 1 \): \[ 1^3 - 5(1)^2 + 6(1) - 1 = 1 - 5 + 6 - 1 = 1 \] Test \( \lambda = .2 \): \[ (.2)^3 - 5(.2)^2 + 6(.2) - 1 = .008 - .2 + 1.2 - 1 = .008 + 1.2 - .2 - 1 = .008 \] So, one root is close to \( \lambda \approx .2 \). Try \( \lambda = 4 \): \[ 64 - 80 + 24 - 1 = 7 \] Try \( \lambda = 3 \): \[ 27 - 45 + 18 - 1 = -1 \] Try \( \lambda = 4.79 \): \[ (4.79)^3 - 5(4.79)^2 + 6(4.79) - 1 \approx 110 - 115 - 28.7 - 1 \approx \] So, the roots are approximately: - \( \lambda_1 \approx .2 \) - \( \lambda_2 \approx 1. \) - \( \lambda_3 \approx 4.8 \) Thus, \[ \omega_1 \approx \sqrt{100 \times .2} = \sqrt{200} \approx 14.1 \ \text{rad/s} \] \[ \omega_2 \approx \sqrt{100 \times 1.} = \sqrt{100} \approx 31.6 \ \text{rad/s} \] \[ \omega_3 \approx \sqrt{100 \times 4.8} = \sqrt{480} \approx 69.3 \ \text{rad/s} \] --- # **(c) Steady-State Solution for Undamped System with Excitation** Given: - \( Y_ = .01 \) m, \( \omega_f = 4 \) rad/s - \( m_1 = m_2 = m_3 = 1 \) kg, \( k_1 = k_2 = k_3 = 100 \) N/m - \( c = \) The steady-state solution can be found by assuming all responses are harmonic at the excitation frequency: \[ x_i(t) = X_i \sin(\omega_f t) \] Let’s use the equations from (a), set all time derivatives as \( \ddot{x}_i = -\omega_f^2 X_i \sin(\omega_f t) \): Substitute into the equations: ### First equation: \[ m_1 \ddot{x}_1 = -k_1(x_1 - y) - k_2(x_1 - x_2) \] \[ -1 \cdot \omega_f^2 X_1 \sin(\omega_f t) = -100 (X_1 - Y_) \sin(\omega_f t) - 100 (X_1 - X_2) \sin(\omega_f t) \] \[ -\omega_f^2 X_1 = -100 (X_1 - Y_) - 100 (X_1 - X_2) \] \[ -\omega_f^2 X_1 = -200 X_1 + 100 Y_ + 100 X_2 \] \[ (200 - \omega_f^2) X_1 - 100 X_2 = 100 Y_ \] ### Second equation: \[ m_2 \ddot{x}_2 = -k_2(x_2 - x_1) - k_3(x_2 - x_3) \] \[ -\omega_f^2 X_2 = -100(X_2 - X_1) - 100(X_2 - X_3) \] \[ -\omega_f^2 X_2 = -200 X_2 + 100 X_1 + 100 X_3 \] \[ (200 - \omega_f^2)X_2 - 100 X_1 - 100 X_3 = \] ### Third equation: \[ m_3 \ddot{x}_3 = k_3(x_2 - x_3) \] \[ -\omega_f^2 X_3 = 100(X_2 - X_3) \] \[ -\omega_f^2 X_3 = 100 X_2 - 100 X_3 \] \[ (100 - \omega_f^2) X_3 - 100 X_2 = \] --- ## **Matrix Form** \[ \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

# Step-by-Step Solution ## (a) Differential Equations of Motion ### Definitions - Let \( x_1, x_2, x_3 \) be the displacements of masses \( m_1, m_2, m_3 \) from their equilibrium positions. - The base excitation is given by \( y = Y_0 \sin(\omega_f t) \). ### Forces Acting on Each Mass #### For Mass \( m_1 \): The forces acting on \( m_1 \) are: - Spring force from the first spring \( k_1 \): \( -k_1(x_1 - y) \) - Spring force from the second spring \( k_2 \): \( -k_2(x_1 - x_2) \) The equation of motion becomes: \[ m_1 \ddot{x}_1 = -k_1(x_1 - y) - k_2(x_1 - x_2) \] #### For Mass \( m_2 \): The forces acting on \( m_2 \) are: - Spring force from \( k_2 \): \( -k_2(x_2 - x_1) \) - Damper force from \( c \): \( -c(\dot{x}_2 - \dot{x}_3) \) - Spring force from \( k_3 \): \( -k_3(x_2 - x_3) \) The equation becomes: \[ m_2 \ddot{x}_2 = -k_2(x_2 - x_1) - c(\dot{x}_2 - \dot{x}_3) - k_3(x_2 - x_3) \] #### For Mass \( m_3 \): The forces acting on \( m_3 \) are: - Damper force from \( c \): \( c(\dot{x}_2 - \dot{x}_3) \) - Spring force from \( k_3 \): \( k_3(x_2 - x_3) \) The equation becomes: \[ m_3 \ddot{x}_3 = c(\dot{x}_2 - \dot{x}_3) + k_3(x_2 - x_3) \] --- ## (b) Natural Frequencies and Mode Shapes ### Assumptions Assume \( m_1 = m_2 = m_3 = m \) and \( k_1 = k_2 = k_3 = k \). ### Matrix Form of the Equations Using the equations from part (a): 1. For \( m_1 \): \[ m \ddot{x}_1 + k(2x_1 - x_2) = k y \] 2. For \( m_2 \): \[ m \ddot{x}_2 + k(-x_1 + 2x_2 - x_3) = 0 \] 3. For \( m_3 \): \[ m \ddot{x}_3 + k(-x_2 + x_3) = 0 \] ### Eigenvalue Problem Assuming solutions of the form \( \mathbf{x} = \mathbf{X} e^{i\omega t} \): \[ \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \lambda \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \] ### Determinant Equation Calculate the determinant for the matrix: \[ \det\left(\begin{bmatrix} 2 - \lambda & -1 & 0 \\ -1 & 2 - \lambda & -1 \\ 0 & -1 & 1 - \lambda \end{bmatrix}\right) = 0 \] ### Solve for Eigenvalues This will yield a cubic polynomial in \( \lambda \). The roots of this polynomial give the eigenvalues related to the natural frequencies. ### Natural Frequencies Given \( m = 1 \) kg and \( k = 100 \) N/m, the natural frequencies are given by: \[ \omega_n = \sqrt{\frac{k}{m} \lambda} \] Calculate for each eigenvalue \( \lambda_i \). --- ## (c) Steady-State Solution for Undamped System with Excitation ### Given Parameters Assume: - \( Y_0 = 0.01 \) m, \( \omega_f = 4 \) rad/s - \( m = 1 \) kg, \( k = 100 \) N/m, \( c = 0 \) ### Formulate the Steady-State Solution Assume solutions of the form: \[ x_i(t) = X_i \sin(\omega_f t) \] ### Substitute into the Motion Equations The equations become: 1. For \( m_1 \): \[ -\omega_f^2 X_1 = -k(X_1 - Y_0) - k(X_1 - X_2) \] 2. For \( m_2 \): \[ -\omega_f^2 X_2 = -k(X_2 - X_1) - k(X_2 - X_3) \] 3. For \( m_3 \): \[ -\omega_f^2 X_3 = k(X_2 - X_3) \] ### Set Up the System of Equations This leads to a linear system of equations which can be solved for \( X_1, X_2, X_3 \) in terms of \( Y_0 \). ### Numerical Solution This system can be solved using matrix methods or numerical solvers to find the amplitudes \( X_1, X_2, X_3 \) corresponding to the steady-state response of the system. --- The final answers will require numerical computation based on the specific values of \( k \), \( m \), and \( Y_0 \).