Certainly! Let's solve the **Coupled Oscillations** problem step by step, as outlined: --- ## **Step (a): Find the ODEs for \( x_1 \) and \( x_2 \)** ### **System Description** - Two masses: \( m_1 = m_2 = m \) - Three springs: - Left and right: \( k_1 = k_3 = k \) - Middle: \( k_2 = 3k \) - Positions: \( x_1 \) (left mass), \( x_2 \) (right mass) ### **Force Equations** #### **Left Mass (\( m_1 \))** \[ m \ddot{x}_1 = -k x_1 - 3k (x_1 - x_2) \] \[ m \ddot{x}_1 = -k x_1 - 3k x_1 + 3k x_2 = -4k x_1 + 3k x_2 \] #### **Right Mass (\( m_2 \))** \[ m \ddot{x}_2 = -k x_2 - 3k (x_2 - x_1) \] \[ m \ddot{x}_2 = -k x_2 - 3k x_2 + 3k x_1 = 3k x_1 - 4k x_2 \] --- ## **Step (b): Convert to Algebraic Equations** Write as a system: \[ \begin{cases} m \ddot{x}_1 + 4k x_1 - 3k x_2 = \\ m \ddot{x}_2 + 4k x_2 - 3k x_1 = \end{cases} \] --- ## **Step (c): Matrix Form** \[ \begin{bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \end{bmatrix} = -\frac{k}{m} \begin{bmatrix} 4 & -3 \\ -3 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \] Let \( \vec{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \): \[ \ddot{\vec{x}} = -\frac{k}{m} A \vec{x} \] with \[ A = \begin{bmatrix} 4 & -3 \\ -3 & 4 \end{bmatrix} \] --- ## **Step (d): Find Eigenvalues (Normal Modes)** Solve \( \det(A - \lambda I) = \): \[ \begin{vmatrix} 4 - \lambda & -3 \\ -3 & 4 - \lambda \end{vmatrix} = (4-\lambda)^2 - 9 = \] \[ (4-\lambda)^2 = 9 \implies 4-\lambda = \pm 3 \implies \lambda_1 = 1, \lambda_2 = 7 \] **Eigenvalues:** \( \boxed{1, \ 7} \) --- ## **Step (e): Find Eigenvectors** For \( \lambda_1 = 1 \): \[ (A - I)\vec{v}_1 = \implies \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \begin{bmatrix} v_{11} \\ v_{12} \end{bmatrix} = \] \[ 3v_{11} - 3v_{12} = \implies v_{11} = v_{12} \] **Normalized Eigenvector:** \( \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \) For \( \lambda_2 = 7 \): \[ (A - 7I)\vec{v}_2 = \implies \begin{bmatrix} -3 & -3 \\ -3 & -3 \end{bmatrix} \begin{bmatrix} v_{21} \\ v_{22} \end{bmatrix} = \] \[ -3v_{21} - 3v_{22} = \implies v_{21} = -v_{22} \] **Normalized Eigenvector:** \( \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \) --- ## **Step (f): Normalized Modes & Matrix A** Already normalized above. Matrix \( A \) is: \[ A = \begin{bmatrix} 4 & -3 \\ -3 & 4 \end{bmatrix} \] --- ## **Step (g): General Solution** Normal mode angular frequencies: \[ \omega_1^2 = \frac{k}{m} \times 1 = \frac{k}{m} \] \[ \omega_2^2 = \frac{k}{m} \times 7 = \frac{7k}{m} \] The general solution: \[ \vec{x}(t) = a_1 \vec{v}_1 \cos(\omega_1 t) + b_1 \vec{v}_1 \sin(\omega_1 t) + a_2 \vec{v}_2 \cos(\omega_2 t) + b_2 \vec{v}_2 \sin(\omega_2 t) \] --- ## **Step (h): Initial Conditions** Given: \( x_1() = x_, \ x_2() = , \ \dot{x}_1() = , \ \dot{x}_2() = v_ \) Write \( x_ \) and \( v_ \) in terms of \( k, m \): Let’s project initial conditions onto the eigenvectors to solve for \( a_1, a_2, b_1, b_2 \). Let \[ \begin{bmatrix} x_1() \\ x_2() \end{bmatrix} = a_1 \vec{v}_1 + a_2 \vec{v}_2 \] \[ = a_1 \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + a_2 \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \] Add: \[ x_ = x_1() = \frac{1}{\sqrt{2}}(a_1 + a_2) \] \[ = x_2() = \frac{1}{\sqrt{2}}(a_1 - a_2) \] From second equation: \( a_1 = a_2 \) Then \( x_ = \sqrt{2} a_1 \implies a_1 = a_2 = \frac{x_}{\sqrt{2}} \) Now for velocities: \[ \begin{bmatrix} \dot{x}_1() \\ \dot{x}_2() \end{bmatrix} = b_1 \omega_1 \vec{v}_1 + b_2 \omega_2 \vec{v}_2 \] \[ = \frac{\omega_1}{\sqrt{2}} b_1 + \frac{\omega_2}{\sqrt{2}} b_2 \] \[ v_ = \frac{\omega_1}{\sqrt{2}} b_1 - \frac{\omega_2}{\sqrt{2}} b_2 \] Add: \[ b_1 = \frac{v_}{\sqrt{2} \omega_1} \] Plug into first equation: \[ = \frac{\omega_1}{\sqrt{2}} b_1 + \frac{\omega_2}{\sqrt{2}} b_2 \implies b_2 = -\frac{\omega_1}{\omega_2} b_1 = -\frac{\omega_1}{\omega_2} \frac{v_}{\sqrt{2} \omega_1} = -\frac{v_}{\sqrt{2} \omega_2} \] --- ## **Step (i): Explicit Equations** ### **Final Explicit Solutions** \[ x_1(t) = \frac{x_}{2} [\cos(\omega_1 t) + \cos(\omega_2 t)] + \frac{v_}{2} \left[\frac{\sin(\omega_1 t)}{\omega_1} - \frac{\sin(\omega_2 t)}{\omega_2}\right] \] \[ x_2(t) = \frac{x_}{2} [\cos(\omega_1 t) - \cos(\omega_2 t)] + \frac{v_}{2} \left[\frac{\sin(\omega_1 t)}{\omega_1} + \frac{\sin(\omega_2 t)}{\omega_2}\right] \] Where: - \( \omega_1 = \sqrt{\frac{k}{m}} \) - \( \omega_2 = \sqrt{\frac{7k}{m}} \) --- ## **Step (j): Plotting** To plot \( x_1(t) \) and \( x_2(t) \) for \( t \in [, 20] \), set \( k = m = v_ = 1 \), and use the above formulas. --- ## **Summary Table** | Step | Result | |------|--------| | ODEs | \( m\ddot{x}_1 = -4kx_1 + 3k x_2 \), \( m\ddot{x}_2 = 3k x_1 - 4k x_2 \) | | Matrix | \( \ddot{\vec{x}} = -\frac{k}{m}A\vec{x} \), \( A = \begin{bmatrix}4&-3\\-3&4\end{bmatrix} \) | | Eigenvalues | \( 1, 7 \) | | Eigenvectors | \( \frac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}, \frac{1}{\sqrt{2}}\begin{bmatrix}1\\-1\end{bmatrix} \) | | General Solution | Written above | | Initial Conditions | \( x_1() = x_, x_2() = , \dot{x}_1() = , \dot{x}_2() = v_ \) | | Final Solution | Boxed formulas above | --- **Let me know if you want the plot code in Python/Matlab!**

# Coupled Oscillations Problem Solution This document outlines the step-by-step solution to the coupled oscillations problem involving a mass-spring system with two equal masses connected by springs of different strengths. ## Step (a): Find the ODEs for \( x_1 \) and \( x_2 \) ### System Description - Two masses: \( m_1 = m_2 = m \) - Springs: - Left and right: \( k_1 = k_3 = k \) - Middle: \( k_2 = 3k \) ### Force Equations #### Left Mass (\( m_1 \)) \[ m \ddot{x}_1 = -k x_1 - 3k (x_1 - x_2) \] Simplifying: \[ m \ddot{x}_1 = -4k x_1 + 3k x_2 \] #### Right Mass (\( m_2 \)) \[ m \ddot{x}_2 = -k x_2 - 3k (x_2 - x_1) \] Simplifying: \[ m \ddot{x}_2 = 3k x_1 - 4k x_2 \] ### Final ODEs \[ \begin{cases} m \ddot{x}_1 + 4k x_1 - 3k x_2 = 0 \\ m \ddot{x}_2 + 4k x_2 - 3k x_1 = 0 \end{cases} \] --- ## Step (b): Convert to Algebraic Equations The equations can be written as: \[ \begin{cases} \ddot{x}_1 + \frac{4k}{m} x_1 - \frac{3k}{m} x_2 = 0 \\ \ddot{x}_2 + \frac{4k}{m} x_2 - \frac{3k}{m} x_1 = 0 \end{cases} \] --- ## Step (c): Matrix Form Let \( \vec{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \). Then the equations can be expressed in matrix form: \[ \ddot{\vec{x}} = -\frac{k}{m} \begin{bmatrix} 4 & -3 \\ -3 & 4 \end{bmatrix} \vec{x} \] Let \( A = \begin{bmatrix} 4 & -3 \\ -3 & 4 \end{bmatrix} \). --- ## Step (d): Find Eigenvalues (Normal Modes) To find the eigenvalues, we solve: \[ \det(A - \lambda I) = 0 \] Calculating the determinant: \[ \begin{vmatrix} 4 - \lambda & -3 \\ -3 & 4 - \lambda \end{vmatrix} = (4-\lambda)^2 - 9 = 0 \] \[ (4-\lambda)^2 = 9 \implies 4-\lambda = \pm 3 \implies \lambda_1 = 1, \lambda_2 = 7 \] ### Eigenvalues \[ \boxed{1, \ 7} \] --- ## Step (e): Find Eigenvectors ### For \( \lambda_1 = 1 \) Solving: \[ (A - I)\vec{v}_1 = 0 \implies \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \begin{bmatrix} v_{11} \\ v_{12} \end{bmatrix} = 0 \] This leads to \( v_{11} = v_{12} \). Normalized eigenvector: \[ \vec{v}_1 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \] ### For \( \lambda_2 = 7 \) Solving: \[ (A - 7I)\vec{v}_2 = 0 \implies \begin{bmatrix} -3 & -3 \\ -3 & -3 \end{bmatrix} \begin{bmatrix} v_{21} \\ v_{22} \end{bmatrix} = 0 \] This leads to \( v_{21} = -v_{22} \). Normalized eigenvector: \[ \vec{v}_2 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \] --- ## Step (f): Normalized Modes & Matrix A The normalized eigenvectors are: 1. \( \vec{v}_1 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \) 2. \( \vec{v}_2 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \) Matrix \( A \): \[ A = \begin{bmatrix} 4 & -3 \\ -3 & 4 \end{bmatrix} \] --- ## Step (g): General Solution The normal mode angular frequencies are: \[ \omega_1^2 = \frac{k}{m} \times 1 = \frac{k}{m} \] \[ \omega_2^2 = \frac{k}{m} \times 7 = \frac{7k}{m} \] The general solution is: \[ \vec{x}(t) = a_1 \vec{v}_1 \cos(\omega_1 t) + b_1 \vec{v}_1 \sin(\omega_1 t) + a_2 \vec{v}_2 \cos(\omega_2 t) + b_2 \vec{v}_2 \sin(\omega_2 t) \] --- ## Step (h): Initial Conditions Given initial conditions: \[ x_1(0) = x_0, \quad x_2(0) = 0, \quad \dot{x}_1(0) = 0, \quad \dot{x}_2(0) = v_0 \] ### Projection on Eigenvectors From initial conditions: \[ \begin{bmatrix} x_0 \\ 0 \end{bmatrix} = a_1 \vec{v}_1 + a_2 \vec{v}_2 \] This leads to: \[ x_0 = \frac{1}{\sqrt{2}}(a_1 + a_2) \] \[ 0 = \frac{1}{\sqrt{2}}(a_1 - a_2) \] From the second equation, \( a_1 = a_2 \). Substituting back gives: \[ a_1 = a_2 = \frac{x_0}{\sqrt{2}} \] ### For Velocities \[ \begin{bmatrix} 0 \\ v_0 \end{bmatrix} = b_1 \omega_1 \vec{v}_1 + b_2 \omega_2 \vec{v}_2 \] This leads to: \[ 0 = \frac{\omega_1}{\sqrt{2}} b_1 + \frac{\omega_2}{\sqrt{2}} b_2 \] \[ v_0 = \frac{\omega_1}{\sqrt{2}} b_1 - \frac{\omega_2}{\sqrt{2}} b_2 \] From the first equation, rearranging gives: \[ b_2 = -\frac{\omega_1}{\omega_2} b_1 \] Substituting this into the second equation allows us to solve for \( b_1 \) and \( b_2 \). --- ## Step (i): Explicit Equations ### Final Explicit Solutions The position functions are: \[ x_1(t) = \frac{x_0}{2} [\cos(\omega_1 t) + \cos(\omega_2 t)] + \frac{v_0}{2} \left[ \frac{\sin(\omega_1 t)}{\omega_1} - \frac{\sin(\omega_2 t)}{\omega_2} \right] \] \[ x_2(t) = \frac{x_0}{2} [\cos(\omega_1 t) - \cos(\omega_2 t)] + \frac{v_0}{2} \left[ \frac{\sin(\omega_1 t)}{\omega_1} + \frac{\sin(\omega_2 t)}{\omega_2} \right] \] Where: - \( \omega_1 = \sqrt{\frac{k}{m}} \) - \( \omega_2 = \sqrt{\frac{7k}{m}} \) --- ## Step (j): Plotting To plot \( x_1(t) \) and \( x_2(t) \) for \( t \in [0, 20] \), set \( k = m = v_0 = 1 \), and use the above equations. ### Example Python Code for Plotting: ```python import numpy as np import matplotlib.pyplot as plt # Constants k = 1 m = 1 x0 = 1 # initial displacement v0 = 1 # initial velocity t = np.linspace(0, 20, 1000) # Angular frequencies omega1 = np.sqrt(k/m) omega2 = np.sqrt(7*k/m) # Position functions x1 = (x0/2) * (np.cos(omega1*t) + np.cos(omega2*t)) + (v0/2) * (np.sin(omega1*t)/omega1 - np.sin(omega2*t)/omega2) x2 = (x0/2) * (np.cos(omega1*t) - np.cos(omega2*t)) + (v0/2) * (np.sin(omega1*t)/omega1 + np.sin(omega2*t)/omega2) # Plotting plt.figure(figsize=(12, 6)) plt.plot(t, x1, label='x1(t)', color='blue') plt.plot(t, x2, label='x2(t)', color='red') plt.title('Coupled Oscillations: Mass-Spring System') plt.xlabel('Time (t)') plt.ylabel('Displacement') plt.legend() plt.grid() plt.show() ``` --- ## Summary Table | Step | Result | |------|--------| | ODEs | \( m\ddot{x}_1 = -4kx_1 + 3k x_2 \), \( m\ddot{x}_2 = 3k x_1 - 4k x_2 \) | | Matrix | \( \ddot{\vec{x}} = -\frac{k}{m}A\vec{x} \), \( A = \begin{bmatrix}4&-3\\-3&4\end{bmatrix} \) | | Eigenvalues | \( 1, 7 \) | | Eigenvectors | \( \frac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}, \frac{1}{\sqrt{2}}\begin{bmatrix}1\\-1\end{bmatrix} \) | | General Solution | Written above | | Initial Conditions | \( x_1(0) = x_0, x_2(0) = 0, \dot{x}_1(0) = 0, \dot{x}_2(0) = v_0 \) | | Final Solution | Boxed formulas above | --- This completes the solution to the coupled oscillations problem. If you need further clarification or additional details, please let me know!