Let's solve the problem step by step. --- ## **Step 1: Understanding the System** - **4 masses** (\( m \)), each connected to its neighbors by springs of spring constant \( k \). - The displacements from equilibrium are \( x_1, x_2, x_3, x_4 \). - We are interested in **longitudinal oscillations** (horizontal direction). --- ## **Step 2: Write the Equations of Motion** For each mass, Newton's second law gives: \[ m \ddot{x}_i = \text{sum of forces on } m_i \] **For mass 1:** \[ m \ddot{x}_1 = -k(x_1 - 0) + k(x_2 - x_1) = -k x_1 + k(x_2 - x_1) = -2k x_1 + k x_2 \] **For mass 2:** \[ m \ddot{x}_2 = -k(x_2 - x_1) + k(x_3 - x_2) = -k(x_2 - x_1) + k(x_3 - x_2) = k x_1 - 2k x_2 + k x_3 \] **For mass 3:** \[ m \ddot{x}_3 = -k(x_3 - x_2) + k(x_4 - x_3) = k x_2 - 2k x_3 + k x_4 \] **For mass 4:** \[ m \ddot{x}_4 = -k(x_4 - x_3) - k(x_4 - 0) = k x_3 - 2k x_4 \] --- ## **Step 3: Write in Matrix Form** Let \( \mathbf{X}(t) = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \). The equations can be written as: \[ m \ddot{\mathbf{X}} = -K \mathbf{X} \] where \( K \) is the force matrix (stiffness matrix): \[ K = k \begin{pmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{pmatrix} \] --- ## **Step 4: Find \( M^{-1}K \) Matrix** Here, \( M \) is a diagonal mass matrix: \( M = m I \), with \( I \) as the \( 4 \times 4 \) identity matrix. \[ M^{-1}K = \frac{1}{m} K \] So, \[ M^{-1} K = \frac{k}{m} \begin{pmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{pmatrix} \] --- ## **Step 5: Final Answer** ### **The \( M^{-1} K \) matrix is:** \[ \boxed{ M^{-1} K = \frac{k}{m} \begin{pmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{pmatrix} } \] This matrix describes the coupled equations for small longitudinal oscillations of the system.

Let's carefully analyze and derive the \( M^{-1} K \) matrix step by step, including detailed explanations at each stage. --- ### **Step 1: Understanding the System** - The system consists of 4 identical masses \( m \), each connected sequentially by springs with spring constant \( k \). - The displacements from equilibrium are \( x_1, x_2, x_3, x_4 \), measured to the right. - The equilibrium positions are such that the masses are equally spaced, and the springs are at their natural length when \( x_i = 0 \). - For small oscillations, the restoring forces are proportional to the displacements, leading to linear equations. --- ### **Step 2: Writing the Equations of Motion** Using Newton’s second law: \[ m \ddot{x}_i = \text{Sum of forces on mass } i \] Let's analyze each mass separately. --- ### **Mass 1 (leftmost):** - Connected to fixed end (say, wall at the left) via spring with constant \( k \), and to mass 2 via spring with constant \( k \). - Restoring force from fixed end: \( -k x_1 \). - Force from spring between mass 1 and 2: \( k(x_2 - x_1) \). Summing forces: \[ m \ddot{x}_1 = -k x_1 + k(x_2 - x_1) = -k x_1 + k x_2 - k x_1 = -2k x_1 + k x_2 \] --- ### **Mass 2:** - Connected to mass 1 and mass 3 via springs. - Forces: from mass 1: \( k(x_1 - x_2) \), from mass 3: \( k(x_3 - x_2) \). Sum: \[ m \ddot{x}_2 = k(x_1 - x_2) + k(x_3 - x_2) = k x_1 - 2k x_2 + k x_3 \] --- ### **Mass 3:** - Connected to mass 2 and mass 4. Sum: \[ m \ddot{x}_3 = k(x_2 - x_3) + k(x_4 - x_3) = k x_2 - 2k x_3 + k x_4 \] --- ### **Mass 4 (rightmost):** - Connected to mass 3 and a fixed boundary (assuming fixed boundary at the right, or mass 4 connected to a fixed point, which exerts a restoring force). - Forces: \[ k(x_3 - x_4) + (-k x_4) = k x_3 - 2k x_4 \] --- ### **Summary of equations:** \[ \begin{cases} m \ddot{x}_1 = -2k x_1 + k x_2 \\ m \ddot{x}_2 = k x_1 - 2k x_2 + k x_3 \\ m \ddot{x}_3 = k x_2 - 2k x_3 + k x_4 \\ m \ddot{x}_4 = k x_3 - 2k x_4 \end{cases} \] --- ### **Step 3: Matrix Formulation** Define the displacement vector: \[ \mathbf{X} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} \] The equations can be written as: \[ m \ddot{\mathbf{X}} = -K \mathbf{X} \] where the **stiffness matrix \( K \)** is: \[ K = k \times \begin{bmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{bmatrix} \] --- ### **Step 4: Mass matrix and \( M^{-1} K \)** Since all masses are identical: \[ M = m \times I_{4} \] and its inverse: \[ M^{-1} = \frac{1}{m} I_{4} \] The matrix \( M^{-1} K \) becomes: \[ M^{-1} K = \frac{1}{m} \times K \] which simplifies to: \[ M^{-1} K = \frac{k}{m} \begin{bmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{bmatrix} \] --- ### **Final Answer:** \[ \boxed{ \boxed{ M^{-1} K = \frac{k}{m} \begin{bmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{bmatrix} } } \] This matrix governs the coupled small longitudinal oscillations of the four masses in the system.

Great question! Let's clarify how the stiffness matrix \( K \) is constructed based on the physical setup and the forces involved. --- ### Understanding the Construction of the Stiffness Matrix \( K \): **1. Physical intuition:** - Each spring exerts a restoring force proportional to how much it is stretched or compressed from its equilibrium length. - For small oscillations, these forces are linear in the displacements. **2. How do we assign matrix elements?** - Each element \( K_{ij} \) encapsulates how the displacement of the \( j \)-th mass affects the force on the \( i \)-th mass. - The diagonal elements \( K_{ii} \) represent the total "restoring strength" acting on the \( i \)-th mass due to its neighboring springs. - The off-diagonal elements \( K_{ij} \) (for \( i \neq j \)) are negative and represent the coupling between masses due to the springs connecting them. --- ### **Step-by-step derivation of \( K \):** **Mass 1:** - Connected to the wall (fixed boundary) via spring \( k \), and to mass 2 via spring \( k \). - Restoring forces: - From fixed boundary: \( -k x_1 \). - From spring to mass 2: \( k (x_2 - x_1) \). Total: \[ F_1 = -k x_1 + k (x_2 - x_1) = -2k x_1 + k x_2 \] **Mass 2:** - Connected to mass 1 and mass 3: \[ F_2 = -k (x_2 - x_1) + k (x_3 - x_2) = k x_1 - 2k x_2 + k x_3 \] **Mass 3:** - Connected to mass 2 and mass 4: \[ F_3 = -k (x_3 - x_2) + k (x_4 - x_3) = k x_2 - 2k x_3 + k x_4 \] **Mass 4:** - Connected to mass 3 and fixed boundary at the right: \[ F_4 = -k (x_4 - x_3) - k x_4 = k x_3 - 2k x_4 \] --- ### **3. Constructing matrix \( K \):** Expressing forces as matrix-vector multiplication: \[ \mathbf{F} = -K \mathbf{X} \] with: \[ K = k \times \begin{bmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{bmatrix} \] **Why these entries?** - For \( F_1 \): \[ F_1 = - (2k x_1 - k x_2) \] Matching \( K_{11} = 2k \), \( K_{12} = -k \). - For \( F_2 \): \[ F_2 = - (-k x_1 + 2k x_2 - k x_3) \implies \text{matrix entries as above} \] And similarly for others. --- ### **Summary:** - The diagonal entries \( 2k \) are the sum of the spring constants connected to that mass. - The off-diagonal entries \( -k \) are the negative of the spring constant for springs connecting neighboring masses. --- ### **Conclusion:** The stiffness matrix \( K \) is constructed based on the linear restoring forces due to springs connecting masses and fixed points, with the entries determined by the number and nature of these connections. --- **Let me know if you'd like a more detailed explanation of any specific part!**