Let's break down and solve the problem step by step: --- ## **Step 5a: Derive the general equation for θ₁ and θ₂** We have **two pendulums** (masses \( M_1 \) and \( M_2 \)), each of length \( L \), coupled by a spring (spring constant \( k \)), as shown in the diagram. ### **Equations of Motion** Assume small angles so \( \sin \theta \approx \theta \). #### **Restoring torque for each pendulum:** - Due to gravity: \( -M_i g L \theta_i \) (for small angles) - Due to the spring: The extension of the spring is \( L(\theta_2 - \theta_1) \). The force on mass \( M_1 \) due to the spring is \( k L (\theta_2 - \theta_1) \), and torque is \( k L^2 (\theta_2 - \theta_1) \). #### **Equations:** For \( M_1 \): \[ M_1 L^2 \ddot{\theta}_1 = -M_1 g L \theta_1 + k L^2 (\theta_2 - \theta_1) \] For \( M_2 \): \[ M_2 L^2 \ddot{\theta}_2 = -M_2 g L \theta_2 + k L^2 (\theta_1 - \theta_2) \] Divide both sides by \( M_1 L^2 \) and \( M_2 L^2 \) respectively: \[ \ddot{\theta}_1 + \frac{g}{L} \theta_1 - \frac{k}{M_1} (\theta_2 - \theta_1) = 0 \] \[ \ddot{\theta}_2 + \frac{g}{L} \theta_2 - \frac{k}{M_2} (\theta_1 - \theta_2) = 0 \] Rewrite: \[ \ddot{\theta}_1 + \left( \frac{g}{L} + \frac{k}{M_1} \right)\theta_1 - \frac{k}{M_1}\theta_2 = 0 \] \[ \ddot{\theta}_2 + \left( \frac{g}{L} + \frac{k}{M_2} \right)\theta_2 - \frac{k}{M_2}\theta_1 = 0 \] --- ## **Step 5b: Solve for θ₁ and θ₂ (Given Initial Conditions)** Given: - \( \theta_1 = \theta_2 = 0.1 \) - \( \dot{\theta}_1 = \dot{\theta}_2 = 0 \) at \( t = 0 \) - \( M_1 = 1\,\text{kg}, M_2 = 2\,\text{kg}, k = 10\,\text{N/m}, L = 1\,\text{m} \) Let's denote \( \omega_{1}^2 = \frac{g}{L} + \frac{k}{M_1} \), \( \omega_{2}^2 = \frac{g}{L} + \frac{k}{M_2} \), and \( \alpha_1 = \frac{k}{M_1} \), \( \alpha_2 = \frac{k}{M_2} \). Thus, the equations are: \[ \ddot{\theta}_1 + \omega_1^2 \theta_1 - \alpha_1 \theta_2 = 0 \] \[ \ddot{\theta}_2 + \omega_2^2 \theta_2 - \alpha_2 \theta_1 = 0 \] #### **Decoupling: Normal Modes** Assume solutions of the form \( \theta_1 = A \cos(\Omega t), \theta_2 = B \cos(\Omega t) \): Plug into equations: \[ (-\Omega^2 + \omega_1^2)A - \alpha_1 B = 0 \] \[ (-\Omega^2 + \omega_2^2)B - \alpha_2 A = 0 \] Write as matrix: \[ \begin{pmatrix} -\Omega^2 + \omega_1^2 & -\alpha_1 \\ -\alpha_2 & -\Omega^2 + \omega_2^2 \end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix} = 0 \] Non-trivial solution if determinant is zero: \[ (-\Omega^2 + \omega_1^2)(-\Omega^2 + \omega_2^2) - \alpha_1 \alpha_2 = 0 \] \[ (\Omega^2 - \omega_1^2)(\Omega^2 - \omega_2^2) - \alpha_1 \alpha_2 = 0 \] \[ \Omega^4 - (\omega_1^2 + \omega_2^2)\Omega^2 + \omega_1^2 \omega_2^2 - \alpha_1 \alpha_2 = 0 \] This is a quadratic in \( \Omega^2 \): \[ \Omega^2 = \frac{\omega_1^2 + \omega_2^2 \pm \sqrt{(\omega_1^2 - \omega_2^2)^2 + 4\alpha_1\alpha_2}}{2} \] Plug in numbers: - \( g = 9.8\,\text{m/s}^2 \), \( L = 1 \) - \( \omega_1^2 = 9.8 + 10/1 = 19.8 \) - \( \omega_2^2 = 9.8 + 10/2 = 9.8 + 5 = 14.8 \) - \( \alpha_1 = 10 \), \( \alpha_2 = 5 \) - \( \alpha_1 \alpha_2 = 50 \) \[ \Omega^2 = \frac{19.8 + 14.8 \pm \sqrt{(19.8 - 14.8)^2 + 4 \times 50}}{2} \] \[ = \frac{34.6 \pm \sqrt{25 + 200}}{2} = \frac{34.6 \pm \sqrt{225}}{2} = \frac{34.6 \pm 15}{2} \] So, - \( \Omega_+^2 = \frac{34.6 + 15}{2} = 24.8 \Rightarrow \Omega_+ = \sqrt{24.8} \approx 4.98 \) - \( \Omega_-^2 = \frac{34.6 - 15}{2} = 9.8 \Rightarrow \Omega_- = \sqrt{9.8} \approx 3.13 \) #### **Normal Modes:** For each mode frequency, find the amplitude ratio: \[ (-\Omega^2 + \omega_1^2)A - \alpha_1 B = 0 \implies \frac{A}{B} = \frac{\alpha_1}{\Omega^2 - \omega_1^2} \] **For \( \Omega_+^2 = 24.8 \):** \[ \frac{A}{B} = \frac{10}{24.8 - 19.8} = \frac{10}{5} = 2 \] So, \( A = 2B \). **For \( \Omega_-^2 = 9.8 \):** \[ \frac{A}{B} = \frac{10}{9.8 - 19.8} = \frac{10}{-10} = -1 \] So, \( A = -B \). #### **General Solution:** \[ \theta_1(t) = C_1 \cos(\Omega_+ t) + C_2 \cos(\Omega_- t) \] \[ \theta_2(t) = \frac{1}{2}C_1 \cos(\Omega_+ t) - C_2 \cos(\Omega_- t) \] #### **Apply Initial Conditions** At \( t = 0 \): - \( \theta_1(0) = 0.1 = C_1 + C_2 \) - \( \theta_2(0) = 0.1 = \frac{1}{2}C_1 - C_2 \) - \( \dot{\theta}_1(0) = 0 = -C_1 \Omega_+ - C_2 \Omega_- \) - \( \dot{\theta}_2(0) = 0 = -\frac{1}{2}C_1\Omega_+ + C_2\Omega_- \) Solve the first two: \[ 0.1 = C_1 + C_2 \tag{1} \] \[ 0.1 = \frac{1}{2}C_1 - C_2 \tag{2} \] Add (1) and (2): \[ 0.2 = C_1 + \frac{1}{2}C_1 = \frac{3}{2}C_1 \implies C_1 = \frac{0.2}{1.5} = \frac{2}{15} \approx 0.1333 \] Plug back to (1): \[ C_2 = 0.1 - 0.1333 = -0.0333 \] **Final answer for (b):** \[ \boxed{ \begin{align*} \theta_1(t) &= 0.1333 \cos(4.98 t) - 0.0333 \cos(3.13 t) \\ \theta_2(t) &= 0.0667 \cos(4.98 t) + 0.0333 \cos(3.13 t) \end{align*} } \] --- ## **Step 5c: Solve for \( M_1 = 3\,\text{kg} \), \( M_2 = 1\,\text{kg} \), \( k = 15\,\text{N/m} \), \( L = 1\,\text{m} \), \( \theta_1 = 0.1, \theta_2' = 0.1 \), \( \theta_2 = \theta_1' = 0 \) at \( t = 0 \)** **Set up the same way:** - \( \omega_1^2 = 9.8 + \frac{15}{3} = 9.8 + 5 = 14.8 \) - \( \omega_2^2 = 9.8 + \frac{15}{1} = 24.8 \) - \( \alpha_1 = \frac{15}{3} = 5 \) - \( \alpha_2 = \frac{15}{1} = 15 \) - \( \alpha_1 \alpha_2 = 75 \) \[ \Omega^2 = \frac{14.8 + 24.8 \pm \sqrt{(24.8 - 14.8)^2 + 4 \cdot 75}}{2} = \frac{39.6 \pm \sqrt{100 + 300}}{2} = \frac{39.6 \pm \sqrt{400}}{2} = \frac{39.6 \pm 20}{2} \] So, - \( \Omega_+^2 = 59.6/2 = 29.8 \implies \Omega_+ \approx 5.46 \) - \( \Omega_-^2 = 19.6/2 = 9.8 \implies \Omega_- \approx 3.13 \) Find amplitude ratios: **For \( \Omega_+^2 = 29.8 \):** \[ \frac{A}{B} = \frac{5}{29.8 - 14.8} = \frac{5}{15} = \frac{1}{3} \] So, \( A = \frac{1}{3}B \). **For \( \Omega_-^2 = 9.8 \):** \[ \frac{A}{B} = \frac{5}{9.8 - 14.8} = \frac{5}{-5} = -1 \] So, \( A = -B \). **General solution:** \[ \theta_1(t) = C_1 \cos(5.46t) + C_2 \cos(3.13t) \] \[ \theta_2(t) = 3C_1 \cos(5.46t) - C_2 \cos(3.13t) \] **Initial conditions:** - \( \theta_1(0) = 0.1 = C_1 + C_2 \) - \( \theta_2(0) = 0 = 3C_1 - C_2 \) - \( \dot{\theta}_1(0) = 0 = -C_1 \cdot 5.46 - C_2 \cdot 3.13 \) - \( \dot{\theta}_2(0) = 0.1 = -3C_1 \cdot 5.46 + C_2 \cdot 3.13 \) From the first two: \[ 0 = 3C_1 - C