Let's solve **question (a)** and **(b)** in **2 steps**. Explanations are given at each step. --- ## (a) Write the governing equations in matrix form ### **Step 1: Arrange the equations** Given: \[ \begin{align*} mL^2\ddot{\theta}_1 + mgL\theta_1 + (k_1 + k_2)L^2\theta_1 - k_2L^2\theta_2 &= M_1 \\ mL^2\ddot{\theta}_2 + mgL\theta_2 + (k_1 + k_2)L^2\theta_2 - k_2L^2\theta_1 &= \end{align*} \] **Explanation:** We write both equations as given, keeping all terms with \(\theta_1\) and \(\theta_2\) grouped. --- ### **Step 2: Express in matrix form** Combine coefficients and write in matrix form: \[ \begin{bmatrix} mL^2 & \\ & mL^2 \end{bmatrix} \begin{bmatrix} \ddot{\theta}_1 \\ \ddot{\theta}_2 \end{bmatrix} + \begin{bmatrix} mgL + (k_1 + k_2)L^2 & -k_2L^2 \\ -k_2L^2 & mgL + (k_1 + k_2)L^2 \end{bmatrix} \begin{bmatrix} \theta_1 \\ \theta_2 \end{bmatrix} = \begin{bmatrix} M_1 \\ \end{bmatrix} \] **Explanation:** - The mass matrix is diagonal as both pendulums have the same mass moment of inertia. - The stiffness matrix includes the gravitational and spring terms, with coupling via \(-k_2L^2\). - The right side is the forcing vector. --- ### **Final matrix equation:** \[ \boxed{ \mathbf{M}\ddot{\Theta} + \mathbf{K}\Theta = \mathbf{F} } \] where \[ \mathbf{M} = \begin{bmatrix} mL^2 & \\ & mL^2 \end{bmatrix} ,\quad \mathbf{K} = \begin{bmatrix} mgL + (k_1 + k_2)L^2 & -k_2L^2 \\ -k_2L^2 & mgL + (k_1 + k_2)L^2 \end{bmatrix} ,\quad \mathbf{F} = \begin{bmatrix} M_1 \\ \end{bmatrix} \] --- ## (b) Find the natural frequencies if \(m = L = k_1 = k_2 = 1, g = 1\) ### **Step 1: Substitute values** Substitute \(m = L = k_1 = k_2 = g = 1\): \[ \mathbf{M} = \begin{bmatrix} 1 & \\ & 1 \end{bmatrix} \] \[ \mathbf{K} = \begin{bmatrix} 1 + (1+1) & -1 \\ -1 & 1 + (1+1) \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ -1 & 3 \end{bmatrix} \] **Explanation:** All parameters are set to 1, simplifying the mass and stiffness matrices. --- ### **Step 2: Solve the eigenvalue problem** The natural frequencies are found from: \[ \det(\mathbf{K} - \omega^2 \mathbf{M}) = \] So, \[ \begin{vmatrix} 3 - \omega^2 & -1 \\ -1 & 3 - \omega^2 \end{vmatrix} = \] \[ (3 - \omega^2)^2 - 1 = \implies (3 - \omega^2)^2 = 1 \] \[ 3 - \omega^2 = \pm1 \] So, - When \(3 - \omega^2 = 1 \implies \omega^2 = 2\) - When \(3 - \omega^2 = -1 \implies \omega^2 = 4\) So, \(\omega_1 = \sqrt{2}\) \(\omega_2 = 2\) **Explanation:** Solving the determinant gives two eigenvalues, which are the squares of the natural frequencies. --- ## **Final Answers** ### **(a) Matrix form:** \[ \boxed{ \mathbf{M}\ddot{\Theta} + \mathbf{K}\Theta = \mathbf{F} } \] with \(\mathbf{M}\), \(\mathbf{K}\), and \(\mathbf{F}\) as above. ### **(b) Natural frequencies:** \[ \boxed{ \omega_1 = \sqrt{2}, \quad \omega_2 = 2 } \]