## Step-by-Step Solution ### **1. Derivation of Differential Equations** Let \( x_1(t) \) and \( x_2(t) \) be the displacements of \( m_1 \) and \( m_2 \) downward from equilibrium. #### **For \( m_1 \):** - Forces: - Downward: gravity (cancels at equilibrium), spring \( k_2 \) if \( x_1 > x_2 \) - Upward: spring \( k_1 \), spring \( k_2 \) if \( x_1 < x_2 \) The net force on \( m_1 \): \[ \begin{align*} F_{m_1} &= -k_1 x_1 - k_2(x_1 - x_2) \\ m_1 \ddot{x}_1 &= -k_1 x_1 - k_2(x_1 - x_2) \end{align*} \] #### **For \( m_2 \):** - Forces: - Downward: spring \( k_3 \) - Upward: spring \( k_2 \) The net force on \( m_2 \): \[ \begin{align*} F_{m_2} &= -k_3 x_2 + k_2(x_1 - x_2) \\ m_2 \ddot{x}_2 &= -k_3 x_2 + k_2(x_1 - x_2) \end{align*} \] Given \( k_1 = k_2 = k_3 = 1 \), \( m_1 = m_2 = 1 \): \[ \begin{align*} \ddot{x}_1 &= -x_1 - (x_1 - x_2) = -2x_1 + x_2 \\ \ddot{x}_2 &= -x_2 + (x_1 - x_2) = x_1 - 2x_2 \end{align*} \] ### **2. Initial Conditions** \[ x_1() = , \quad \dot{x}_1() = -1, \quad x_2() = , \quad \dot{x}_2() = 1 \] --- ## (a) **Laplace Transform Method** Let \( X_1(s) \) and \( X_2(s) \) be Laplace transforms of \( x_1(t) \), \( x_2(t) \). Take Laplace transforms of both equations: \[ \mathcal{L}\{\ddot{x}_1\} = s^2 X_1(s) - s x_1() - \dot{x}_1() \] \[ s^2 X_1(s) + X_1(s) - X_2(s) = -\dot{x}_1() \] Plug in initial conditions: \[ s^2 X_1(s) = -2X_1(s) + X_2(s) -1 \] \[ s^2 X_2(s) = X_1(s) - 2X_2(s) + 1 \] Rewriting: \[ (s^2 + 2) X_1(s) - X_2(s) = -1 \] \[ - X_1(s) + (s^2 + 2) X_2(s) = 1 \] Now solve this linear system. Add the first equation to the second (after multiplying the first by 1 to align coefficients): \[ (s^2 + 2) X_1(s) - X_2(s) = -1 \qquad (1) \] \[ - X_1(s) + (s^2 + 2) X_2(s) = 1 \qquad (2) \] Multiply (1) by \((s^2 + 2)\) and add to (2): From (1): \[ X_2(s) = (s^2 + 2) X_1(s) + 1 \] Plug into (2): \[ - X_1(s) + (s^2 + 2)\left[ (s^2 + 2) X_1(s) + 1 \right] = 1 \] \[ - X_1(s) + (s^2 + 2)^2 X_1(s) + (s^2 + 2) = 1 \] \[ \left[(s^2 + 2)^2 - 1\right] X_1(s) = 1 - (s^2 + 2) \] \[ (s^4 + 4 s^2 + 3) X_1(s) = -s^2 - 1 \] \[ X_1(s) = \frac{-(s^2 + 1)}{s^4 + 4s^2 + 3} \] Now \( X_2(s) \): \[ X_2(s) = (s^2 + 2) X_1(s) + 1 \] \[ X_2(s) = (s^2 + 2) \left[ \frac{-(s^2 + 1)}{s^4 + 4s^2 + 3} \right] + 1 \] \[ = \frac{-(s^2 + 2)(s^2 + 1) + (s^4 + 4s^2 + 3)}{s^4 + 4s^2 + 3} \] Calculate numerator: \[ -(s^2 + 2)(s^2 + 1) + s^4 + 4s^2 + 3 \] \[ = -(s^4 + 3s^2 + 2) + s^4 + 4s^2 + 3 \] \[ = -s^4 - 3s^2 - 2 + s^4 + 4s^2 + 3 \] \[ = () + (4s^2 - 3s^2) + (3 - 2) \] \[ = s^2 + 1 \] So, \[ X_2(s) = \frac{s^2 + 1}{s^4 + 4s^2 + 3} \] Now, factor denominator: \[ s^4 + 4s^2 + 3 = (s^2 + 1)(s^2 + 3) \] So, \[ X_1(s) = - \frac{s^2 + 1}{(s^2 + 1)(s^2 + 3)} = -\frac{1}{s^2 + 3} \] \[ X_2(s) = \frac{s^2 + 1}{(s^2 + 1)(s^2 + 3)} \] Partial fractions on \( X_2(s) \): \[ \frac{s^2 + 1}{(s^2 + 1)(s^2 + 3)} = \frac{A}{s^2 + 1} + \frac{B}{s^2 + 3} \] \[ s^2 + 1 = A(s^2 + 3) + B(s^2 + 1) \] Let \( s^2 = -1 \): \[ = A(2) \implies A = \] Let \( s^2 = -3 \): \[ -2 = B(-2) \implies B = 1 \] So, \[ X_2(s) = \frac{1}{s^2 + 3} \] Now invert Laplace transforms: \[ \mathcal{L}^{-1}\left\{ \frac{1}{s^2 + a^2} \right\} = \frac{1}{a} \sin(at) \] So, \[ X_1(s) = -\frac{1}{s^2 + 3} \implies x_1(t) = -\frac{1}{\sqrt{3}}\sin(\sqrt{3} t) \] \[ X_2(s) = \frac{1}{s^2 + 3} \implies x_2(t) = \frac{1}{\sqrt{3}}\sin(\sqrt{3} t) \] --- ## (b) **Alternative Method: Matrix Method** The system can be written as: \[ \mathbf{x}'' = \begin{bmatrix} -2 & 1 \\ 1 & -2 \end{bmatrix} \mathbf{x} \] where \( \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \). The general solution for such a system: \[ \mathbf{x}(t) = c_1 \mathbf{v}_1 \sin(\sqrt{\lambda_1} t) + c_2 \mathbf{v}_2 \sin(\sqrt{\lambda_2} t) \] where \( \lambda_i \) and \( \mathbf{v}_i \) are eigenvalues/vectors. Find eigenvalues: \[ \text{det} \begin{bmatrix} -2-\lambda & 1 \\ 1 & -2-\lambda \end{bmatrix} = \implies (-2-\lambda)^2 - 1 = \] \[ (-2-\lambda)^2 = 1 \implies -2-\lambda = \pm 1 \implies \lambda_1 = -3, \lambda_2 = -1 \] So, \( \sqrt{-\lambda_1} = \sqrt{3} \), \( \sqrt{-\lambda_2} = 1 \). Eigenvectors for \( \lambda_1 = -3 \): \[ \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \implies v_1 = -v_2 \] So eigenvector is \( \begin{bmatrix} 1 \\ -1 \end{bmatrix} \). For \( \lambda_2 = -1 \): \[ \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \implies v_1 = v_2 \] So eigenvector is \( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \). General solution: \[ \mathbf{x}(t) = A \begin{bmatrix} 1 \\ -1 \end{bmatrix} \sin(\sqrt{3} t) + B \begin{bmatrix} 1 \\ 1 \end{bmatrix} \sin(t) \] Apply initial conditions \( x_1() = x_2() = \), so both sine terms start at , no cosine terms. Now derivatives: \[ \dot{\mathbf{x}}(t) = A \sqrt{3} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \cos(\sqrt{3} t) + B \begin{bmatrix} 1 \\ 1 \end{bmatrix} \cos(t) \] At \( t = \): \[ \dot{\mathbf{x}}() = A\sqrt{3} \begin{bmatrix} 1 \\ -1 \end{bmatrix} + B \begin{bmatrix} 1 \\ 1 \end{bmatrix} \] Set equal to \( \begin{bmatrix} -1 \\ 1 \end{bmatrix} \): So, \[ A\sqrt{3} + B = -1 \\ -A\sqrt{3} + B = 1 \] Add: \[ 2B = \implies B = \] Then, \[ A\sqrt{3} = -1 \implies A = -\frac{1}{\sqrt{3}} \] So: \[ x_1(t) = -\frac{1}{\sqrt{3}} \sin(\sqrt{3} t) \\ x_2(t) = \frac{1}{\sqrt{3}} \sin(\sqrt{3} t) \] --- ## (c) **Comparison of Methods** Both the Laplace Transform and Matrix Method yield the same solution. However, the Matrix Method is more computationally efficient and conceptually straightforward for homogeneous, constant-coefficient systems like this, avoiding Laplace tables and partial fractions. **Final Answer:** \[ \boxed{ x_1(t) = -\frac{1}{\sqrt{3}}\sin(\sqrt{3} t),\qquad x_2(t) = \frac{1}{\sqrt{3}}\sin(\sqrt{3} t) } \] The Matrix Method is the most efficient and simplest for this problem.