Chapter 9: Problem 20
For the coupled mass-spring system governed by system \((10),\) assume \(m_{1}=m_{2}=1 \mathrm{kg}, k_{1}=k_{2}=k_{3}=1 \mathrm{N} / \mathrm{m}\) and assume initially that \(x_{1}(0)=0 \mathrm{m}, x_{1}^{\prime}(0)=0 \mathrm{m} / \mathrm{sec}\) \(x_{2}(0)=2 \mathrm{m},\) and \(x_{2}^{\prime}(0)=0 \mathrm{m} / \mathrm{sec} .\) Using matrix alge- bra techniques, solve this initial value problem.
Short Answer
Step by step solution
Derive the Differential Equations
Convert to Matrix Form
Solve the System of Differential Equations
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Algebra Techniques
In our given exercise, we first define a vector \( X \) to represent the positions of the masses, \( x_1 \) and \( x_2 \). This is depicted as a column vector:
- \( X = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \)
- \( A = \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} \)
Eigenvalues and Eigenvectors
For our matrix \( A \), identifying eigenvalues and eigenvectors helps solve our system of equations. We compute these:
- Eigenvalues: \( \lambda_1 = 2 \), \( \lambda_2 = 0 \)
- Eigenvectors: \( v_1 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \ 1 \end{bmatrix} \), \( v_2 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \ -1 \end{bmatrix} \)
Initial Value Problem
In this scenario, we are given:
- \( x_1(0) = 0 \)
- \( x_1'(0) = 0 \)
- \( x_2(0) = 2 \)
- \( x_2'(0) = 0 \)
The goal is to apply these initial values to solve for the constants \( C_1 \) and \( C_2 \) in the general form of the solution. This process transforms the general solution into a specific one that accurately reflects the initial system setup.
Newton's Second Law
Applying this law to the coupled mass-spring system, we derive the differential equations dictating the motion of each mass. Each spring exerts a force proportional to its displacement, giving rise to equations of the form:
- \(-k_1x_1 + k_2(x_2 - x_1) = m_1x_1'' \)
- \(-k_2(x_2 - x_1) + k_3(-x_2) = m_2x_2''\)
Converting these into their simplified form, \( x_1'' + x_1 - x_2 = 0 \) and \( x_2'' + x_2 - x_1 = 0 \), allows us to engage matrix algebra for further analysis. Newton’s law thus acts as the foundation for formulating the problem in a way that enables the use of algebraic techniques for further solutions.