Q4E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Two springs, two masses, and a dashpot are attached in a straight line on a horizontal frictionless surface as shown in Figure\(5.33\). The dashpot provides a damping force on mass\({{\bf{m}}_{\bf{2}}}\), given by\({\bf{F =  - by'}}\). Derive the system of differential equations for the displacements \({\bf{x}}\) and\({\bf{y}}\).

Step-by-Step Solution

Answer

The differential equations for the displacements \({\bf{x}}\) and \({\bf{y}}\) are:

 

\(\begin{aligned}{l}{\bf{mx'' =  - }}{{\bf{k}}_{\bf{1}}}{\bf{x - }}{{\bf{k}}_{\bf{2}}}{\bf{(x - y)}}\\{\bf{my'' = }}{{\bf{k}}_{\bf{2}}}{\bf{(x - y) - by'}}\end{aligned}\)

Step 1: Identifying the forces

In order to derive the system, we will assume that \({\bf{x, y > 0}}\) and\({\bf{x > y}}\). Equations will be true for any \({\bf{x}}\) and\({\bf{y}}\). The Force is proportional to the change of the length of the strings, so we have\({{\bf{F}}_{\bf{1}}}{\bf{ =  - }}{{\bf{k}}_{\bf{1}}}{\bf{x,}}\;\;{{\bf{F}}_{\bf{2}}}{\bf{ =  - }}{{\bf{k}}_{\bf{2}}}{\bf{(x - y),}}\;\;{{\bf{F}}_{\bf{3}}}{\bf{ = }}{{\bf{k}}_{\bf{2}}}{\bf{(x - y)}}.\)

 

The Forces that act on mass \({{\bf{m}}_{\bf{1}}}\) are \({{\bf{F}}_{\bf{1}}}\) and\({{\bf{F}}_{\bf{2}}}\), and the Forces that act on mass \({{\bf{m}}_{\bf{2}}}\) are \({{\bf{F}}_{\bf{3}}}\) and\({\bf{F =  - by'}}\).

Step 2: Using the Newton’s second law

By the second Newton's law we have that the system of differential equations for the displacements \({\bf{x}}\) and \({\bf{y}}\) is;

\(\begin{aligned}{c}{\bf{mx'' = }}{{\bf{F}}_{\bf{1}}}{\bf{ + }}{{\bf{F}}_{\bf{2}}}\\{\bf{my'' = }}{{\bf{F}}_{\bf{3}}}{\bf{ + F}}\;\\{\bf{mx'' =   - }}{{\bf{k}}_{\bf{1}}}{\bf{x - }}{{\bf{k}}_{\bf{2}}}{\bf{(x - y)}}\\{\bf{my'' = }}{{\bf{k}}_{\bf{2}}}{\bf{(x - y) - by'}}\end{aligned}\)

 

Therefore, differential equations for the displacements \({\bf{x}}\) and \({\bf{y}}\) are:

 

\(\begin{aligned}{c}{\bf{mx'' =  - }}{{\bf{k}}_{\bf{1}}}{\bf{x - }}{{\bf{k}}_{\bf{2}}}{\bf{(x - y)}}\\{\bf{my'' = }}{{\bf{k}}_{\bf{2}}}{\bf{(x - y) - by'}}\end{aligned}\)