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Q5.6-10E

Expert-verified
Found in: Page 288

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Suppose the coupled mass-spring system of Problem (Figure 5.31) is hung vertically from support (with mass ${{\mathbf{m}}}_{{\mathbf{2}}}$ above ${{\mathbf{m}}}_{{\mathbf{1}}}$), as in Section 4.10, page 226.(a) Argue that at equilibrium, the lower spring is stretched a distance ${{\mathbf{l}}}_{{1}}$ from its natural length ${{\mathbf{l}}}_{{2}}$, given by ${{\mathbf{l}}}_{{\mathbf{1}}}{\mathbf{=}}{{\mathbf{m}}}_{{\mathbf{1}}}{\mathbf{g}}{\mathbf{/}}{{\mathbf{k}}}_{{\mathbf{1}}}$.(b) Argue that at equilibrium, the upper spring is stretched a distance ${{\mathbf{l}}}_{{\mathbf{2}}}{\mathbf{=}}\left({\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\right){\mathbf{g}}{\mathbf{/}}{{\mathbf{k}}}_{{\mathbf{2}}}$.(c) Show that if ${{\mathbf{x}}}_{{\mathbf{1}}}$ and ${{\mathbf{x}}}_{{\mathbf{2}}}$ are redefined to be displacements from the equilibrium positions of the masses ${{\mathbf{m}}}_{{\mathbf{1}}}$ and ${{\mathbf{m}}}_{{\mathbf{2}}}$, then the equations of motion are identical with those derived in Problem 1.

The system of differential equations for the displacement x and y is

${\mathrm{m}}_{1}\mathrm{x}\text{'}\text{'}={\mathrm{k}}_{1}\left(\mathrm{y}-\mathrm{x}\right)-{\mathrm{m}}_{1}\mathrm{g}\phantom{\rule{0ex}{0ex}}{\mathrm{m}}_{2}\mathrm{y}\text{'}\text{'}=-{\mathrm{k}}_{1}\left(\mathrm{y}-\mathrm{x}\right)-{\mathrm{k}}_{2}\mathrm{y}-{\mathrm{m}}_{2}\mathrm{g}$

New equilibrium is $\left({\mathrm{x}}_{0},{\mathrm{y}}_{0}\right)=\left(-\frac{{\mathrm{m}}_{1}{\mathrm{k}}_{1}\mathrm{g}+{\mathrm{m}}_{2}{\mathrm{k}}_{1}\mathrm{g}+{\mathrm{m}}_{1}{\mathrm{k}}_{2}\mathrm{g}}{{\mathrm{k}}_{1}{\mathrm{k}}_{2}},-\frac{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)\mathrm{g}}{{\mathrm{k}}_{2}}\right)$

(a) ${\mathrm{l}}_{1}=\left|{\mathrm{x}}_{0}-{\mathrm{y}}_{0}\right|={\mathrm{m}}_{1}\mathrm{g}/{\mathrm{k}}_{1}$,

(b) ${\mathrm{l}}_{2}=\left|{\mathrm{y}}_{0}\right|=\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)\mathrm{g}/{\mathrm{k}}_{2}$,

(c) Substituting $\mathrm{\xi }=\mathrm{x}-{\mathrm{x}}_{0}$ and $\mathrm{\eta }=\mathrm{y}-{\mathrm{y}}_{0}$ into the system of differential equations for displacement we obtain a system in terms of $\mathrm{\xi }$ and $\mathrm{\eta }$ which is identical to the system we derived in Problem 1.

See the step by step solution

## Finding the differential equation

First, one needs to find the system of differential equations for the displacement x and y. Assume that $\mathrm{x},\mathrm{y}<0$ and $x. In this case, one has that ${\mathrm{F}}_{1}=-{\mathrm{m}}_{1}\mathrm{g},{\mathrm{F}}_{2}={\mathrm{k}}_{1}\left(\mathrm{y}-\mathrm{x}\right),{\mathrm{F}}_{3}=-{\mathrm{k}}_{2}\left(\mathrm{y}-\mathrm{x}\right),{\mathrm{F}}_{4}=-{\mathrm{m}}_{2}\mathrm{g},{\mathrm{F}}_{5}=-{\mathrm{k}}_{2}\mathrm{y}$

and ${\mathrm{m}}_{1}\mathrm{x}\text{'}\text{'}={\mathrm{F}}_{1}+{\mathrm{F}}_{2},{\mathrm{m}}_{2}\mathrm{y}\text{'}\text{'}={\mathrm{F}}_{3}+{\mathrm{F}}_{4}+{\mathrm{F}}_{5}$

But this equation must be true for any x and y, so the system of differential equations for the displacement x and y is:

$\begin{array}{rcl}{\mathrm{m}}_{1}\mathrm{x}\text{'}\text{'}\text{'}& =& {\mathrm{k}}_{1}\left(\mathrm{y}-\mathrm{x}\right)-{\mathrm{m}}_{1}\mathrm{g}\\ {\mathrm{m}}_{2}\mathrm{y}\text{'}\text{'}& =& -{\mathrm{k}}_{1}\left(\mathrm{y}-\mathrm{x}\right)-{\mathrm{k}}_{2}\mathrm{y}-{\mathrm{m}}_{2}\mathrm{g}\end{array}$

## Finding the length of the springs

At the equilibrium $\left(\mathrm{x},\mathrm{y}\right)=\left({\mathrm{x}}_{0},{\mathrm{y}}_{0}\right)$, one has that ${\mathrm{m}}_{1}\mathrm{x}\text{'}\text{'}=0$ and ${\mathrm{m}}_{2}\mathrm{y}\text{'}\text{'}=0$. So, to find the change in the length of the springs, one needs to find a new equilibrium first, and to do so one will solve the following system for ${\mathrm{x}}_{0}$ and ${y}_{0}$:

$0={\mathrm{k}}_{1}\left({\mathrm{y}}_{0}-{\mathrm{x}}_{0}\right)-{\mathrm{m}}_{1}\mathrm{g}\phantom{\rule{0ex}{0ex}}0=-{\mathrm{k}}_{1}\left({\mathrm{y}}_{0}-{\mathrm{x}}_{0}\right)-{\mathrm{k}}_{2}{\mathrm{y}}_{0}-{\mathrm{m}}_{2}\mathrm{g}$

Adding those two equations together one will get

$\begin{array}{rcl}0& =& -{\mathrm{m}}_{1}\mathrm{g}-{\mathrm{k}}_{2}{\mathrm{y}}_{0}-{\mathrm{m}}_{2}\mathrm{g}\\ {\mathrm{y}}_{0}& =& -\frac{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)\mathrm{g}}{{\mathrm{k}}_{2}}\end{array}$

## Finding x0

Now we need to find ${\mathrm{x}}_{0}$. Multiplying the first equation by ${\mathrm{x}}_{0}$ and the second by and then adding them together one will get

$\begin{array}{rcl}0& =& {\mathrm{m}}_{2}{\mathrm{k}}_{1}\left({\mathrm{y}}_{0}-{\mathrm{x}}_{0}\right)-{\mathrm{m}}_{1}{\mathrm{m}}_{2}\mathrm{g}\partial \\ 0& =& {\mathrm{m}}_{1}{\mathrm{k}}_{1}\left({\mathrm{y}}_{0}-{\mathrm{x}}_{0}\right)+{\mathrm{m}}_{1}{\mathrm{k}}_{2}{\mathrm{y}}_{0}+{\mathrm{m}}_{1}{\mathrm{m}}_{2}\mathrm{g}\\ 0& =& {\mathrm{m}}_{2}{\mathrm{k}}_{1}\left({\mathrm{y}}_{0}-{\mathrm{x}}_{0}\right)+{\mathrm{m}}_{1}{\mathrm{k}}_{1}\left({\mathrm{y}}_{0}-{\mathrm{x}}_{0}\right)+{\mathrm{m}}_{1}{\mathrm{k}}_{2}{\mathrm{y}}_{0}\\ 0& =& {\mathrm{k}}_{1}\left({\mathrm{y}}_{0}-{\mathrm{x}}_{0}\right)\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)+{\mathrm{m}}_{1}{\mathrm{k}}_{2}{\mathrm{y}}_{0}\\ {\mathrm{k}}_{1}{\mathrm{x}}_{0}\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)& =& {\mathrm{k}}_{1}{\mathrm{y}}_{0}+{\mathrm{m}}_{1}{\mathrm{k}}_{2}{\mathrm{y}}_{0}\end{array}$

## Finding a new equilibrium

Substituting the value for ${\mathrm{y}}_{0}$ into the previous equation, one will get

$\begin{array}{rcl}{\mathrm{k}}_{1}{\mathrm{x}}_{0}\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)& =& -\frac{{\mathrm{k}}_{1}}{{\mathrm{k}}_{2}}{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)}^{2}\mathrm{g}-{\mathrm{m}}_{1}\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)\mathrm{g}\\ {\mathrm{k}}_{1}{\mathrm{x}}_{0}& =& -\frac{{\mathrm{m}}_{1}{\mathrm{k}}_{1}\mathrm{g}+{\mathrm{m}}_{2}{\mathrm{k}}_{1}\mathrm{g}+{\mathrm{m}}_{1}{\mathrm{k}}_{2}\mathrm{g}}{{\mathrm{k}}_{2}}\\ {\mathrm{x}}_{0}& =& -\frac{{\mathrm{m}}_{1}{\mathrm{k}}_{1}\mathrm{g}+{\mathrm{m}}_{2}{\mathrm{k}}_{1}\mathrm{g}+{\mathrm{m}}_{1}{\mathrm{k}}_{2}\mathrm{g}}{{\mathrm{k}}_{1}{\mathrm{k}}_{2}}\end{array}$

The new equilibrium is $\left({\mathrm{x}}_{0},{\mathrm{y}}_{0}\right)=\left(-\frac{{\mathrm{m}}_{1}{\mathrm{k}}_{1}\mathrm{g}+{\mathrm{m}}_{2}{\mathrm{k}}_{1}\mathrm{g}+{\mathrm{m}}_{1}{\mathrm{k}}_{2}\mathrm{g}}{{\mathrm{k}}_{1}{\mathrm{k}}_{2}},-\frac{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)\mathrm{g}}{{\mathrm{k}}_{2}}\right)$.

## Finding the length I1

one can see from the Figure above that the lower spring is stretched by ${\mathrm{l}}_{1}=\left|{\mathrm{x}}_{0}-{\mathrm{y}}_{0}\right|$ which is

$\begin{array}{rcl}{\mathrm{l}}_{1}& =& \left|{\mathrm{x}}_{0}-{\mathrm{y}}_{0}\right|\\ & =& \left|-\frac{{\mathrm{m}}_{1}{\mathrm{k}}_{1}\mathrm{g}+{\mathrm{m}}_{2}{\mathrm{k}}_{1}\mathrm{g}+{\mathrm{m}}_{1}{\mathrm{k}}_{2}\mathrm{g}}{{\mathrm{k}}_{1}{\mathrm{k}}_{2}}+\frac{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)\mathrm{g}}{{\mathrm{k}}_{2}}\right|\\ & =& \left|-\frac{{\mathrm{m}}_{1}{\mathrm{k}}_{1}\mathrm{g}+{\mathrm{m}}_{2}{\mathrm{k}}_{1}\mathrm{g}+{\mathrm{m}}_{1}{\mathrm{k}}_{2}\mathrm{g}-{\mathrm{m}}_{1}{\mathrm{k}}_{1}\mathrm{g}-{\mathrm{m}}_{2}{\mathrm{k}}_{1}\mathrm{g}}{{\mathrm{k}}_{1}{\mathrm{k}}_{2}}\right|\\ & =& \left|-\frac{{\mathrm{m}}_{1}{\mathrm{k}}_{2}\mathrm{g}}{{\mathrm{k}}_{1}{\mathrm{k}}_{2}}\right|\\ & =& \frac{{\mathrm{m}}_{1}\mathrm{g}}{{\mathrm{k}}_{1}}\end{array}$

## Finding the length I2

From the Figure above we can conclude that the upper spring is stretched only by ${\mathrm{l}}_{2}=\left|{\mathrm{y}}_{0}\right|$, so

$\begin{array}{rcl}{\mathrm{l}}_{2}& =& \left|-\frac{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)\mathrm{g}}{{\mathrm{k}}_{2}}\right|\\ & =& \frac{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)\mathrm{g}}{{\mathrm{k}}_{2}}\end{array}$

Introduce new variables

$\mathrm{\xi }=\mathrm{x}-{\mathrm{x}}_{0}$ and $\mathrm{\eta }=\mathrm{y}-{\mathrm{y}}_{0}$

One has that $\mathrm{x}=\mathrm{\xi }+{\mathrm{x}}_{0},\mathrm{y}=\mathrm{\eta }+{\mathrm{y}}_{0}$ and $\mathrm{x}\text{'}\text{'}=\mathrm{\xi }\text{'}\text{'},\mathrm{y}\text{'}\text{'}=\mathrm{\eta }\text{'}\text{'}$.

## Substituting the values into the differential equation

Substituting this into the system of differential equations one will get

$\begin{array}{rcl}{\mathrm{m}}_{1}\mathrm{\xi }\text{'}\text{'}& =& {\mathrm{k}}_{1}\left(\mathrm{\eta }+{\mathrm{y}}_{0}-\mathrm{\xi }-{\mathrm{x}}_{0}\right)-{\mathrm{m}}_{1}\mathrm{g}\\ {\mathrm{m}}_{2}\mathrm{\eta }\text{'}\text{'}& =& -{\mathrm{k}}_{1}\left(\mathrm{\eta }+{\mathrm{y}}_{0}-\mathrm{\xi }-{\mathrm{x}}_{0}\right)-{\mathrm{k}}_{2}\left(\mathrm{\eta }+{\mathrm{y}}_{0}\right)-{\mathrm{m}}_{2}\mathrm{g}\\ {\mathrm{m}}_{1}\mathrm{\xi }\text{'}\text{'}& =& {\mathrm{k}}_{1}\left(\mathrm{\eta }-\mathrm{\xi }-\frac{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)\mathrm{g}}{{\mathrm{k}}_{2}}+\frac{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right){\mathrm{k}}_{1}\mathrm{g}+{\mathrm{m}}_{1}{\mathrm{k}}_{2}\mathrm{g}}{{\mathrm{k}}_{1}{\mathrm{k}}_{2}}\right)-{\mathrm{m}}_{1}\mathrm{g}\\ {\mathrm{m}}_{2}\mathrm{\eta }\text{'}\text{'}& =& -{\mathrm{k}}_{1}\left(\mathrm{\eta }-\mathrm{\xi }-\frac{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)\mathrm{g}}{{\mathrm{k}}_{2}}+\frac{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right){\mathrm{k}}_{1}\mathrm{g}+{\mathrm{m}}_{1}{\mathrm{k}}_{2}\mathrm{g}}{{\mathrm{k}}_{1}{\mathrm{k}}_{2}}\right)-{\mathrm{k}}_{2}\left(\mathrm{\eta }-\frac{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)\mathrm{g}}{{\mathrm{k}}_{2}}\right)-{\mathrm{m}}_{2}\mathrm{g}\\ {\mathrm{m}}_{1}\mathrm{\xi }\text{'}\text{'}& =& {\mathrm{k}}_{1}\left(\mathrm{\eta }-\mathrm{\xi }+\frac{{\mathrm{m}}_{1}{\mathrm{k}}_{2}\mathrm{g}}{{\mathrm{k}}_{1}{\mathrm{k}}_{2}}\right)-{\mathrm{m}}_{1}\mathrm{g}\\ {\mathrm{m}}_{2}\mathrm{\eta }\text{'}\text{'}& =& -{\mathrm{k}}_{1}\left(\mathrm{\eta }-\mathrm{\xi }+\frac{{\mathrm{m}}_{1}{\mathrm{k}}_{2}\mathrm{g}}{{\mathrm{k}}_{1}{\mathrm{k}}_{2}}\right)-{\mathrm{k}}_{2}\mathrm{\eta }+\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)\mathrm{g}-{\mathrm{m}}_{2}\mathrm{g}\\ & ⇒& {\mathrm{m}}_{1}\mathrm{\xi }\text{'}\text{'}={\mathrm{k}}_{1}\left(\mathrm{\eta }-\mathrm{\xi }\right)+{\mathrm{m}}_{1}\mathrm{g}-{\mathrm{m}}_{1}\mathrm{g}\\ {\mathrm{m}}_{2}\mathrm{\eta }\text{'}\text{'}& =& -{\mathrm{k}}_{1}\left(\mathrm{\eta }-\mathrm{\xi }\right)-{\mathrm{m}}_{1}\mathrm{g}-{\mathrm{k}}_{2}\mathrm{\eta }+\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)\mathrm{g}={\mathrm{m}}_{2}\mathrm{g}\\ & ⇒& {\mathrm{m}}_{1}\mathrm{\xi }\text{'}\text{'}={\mathrm{k}}_{1}\left(\mathrm{\eta }-\mathrm{\xi }\right)\\ {\mathrm{m}}_{2}\mathrm{\eta }\text{'}\text{'}& =& -{\mathrm{k}}_{1}\left(\mathrm{\eta }-\mathrm{\xi }\right)-{\mathrm{k}}_{2}\mathrm{\eta }\end{array}$

One sees that the differential equations of motion in terms of $\mathrm{\xi }=\mathrm{x}-{\mathrm{x}}_{0}$ and $\mathrm{\eta }=\mathrm{y}-{\mathrm{y}}_{0}$ are identical to the equations in terms of and derived in Problem 1 when one had masses attached to the horizontal springs.