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 Section4.10, page226.

(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.

First, one needs to find the system of differential equations for the displacement x and y. Assume that $\mathrm{x},\mathrm{y}<0$and $x<y$. In this case, one has that ${\mathrm{F}}_1=-{\mathrm{m}}_1\mathrm{g},{\mathrm{F}}_2={\mathrm{k}}_1(\mathrm{y}-\mathrm{x}),{\mathrm{F}}_3=-{\mathrm{k}}_2(\mathrm{y}-\mathrm{x}),{\mathrm{F}}_4=-{\mathrm{m}}_2\mathrm{g},{\mathrm{F}}_5=-{\mathrm{k}}_2\mathrm{y}$

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:

$\begin{array}{rcl}{\mathrm{m}}_1\mathrm{x}\text{'}\text{'}\text{'}& =& {\mathrm{k}}_1(\mathrm{y}-\mathrm{x})-{\mathrm{m}}_1\mathrm{g}\\ {\mathrm{m}}_2\mathrm{y}\text{'}\text{'}& =& -{\mathrm{k}}_1(\mathrm{y}-\mathrm{x})-{\mathrm{k}}_2\mathrm{y}-{\mathrm{m}}_2\mathrm{g}\end{array}$

At the equilibrium $(\mathrm{x},\mathrm{y})=\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)$, 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 $y_0$:

$$\begin{gathered} 0={\mathrm{k}}_1\left({\mathrm{y}}_0-{\mathrm{x}}_0\right)-{\mathrm{m}}_1\mathrm{g} \\ 0=-{\mathrm{k}}_1\left({\mathrm{y}}_0-{\mathrm{x}}_0\right)-{\mathrm{k}}_2{\mathrm{y}}_0-{\mathrm{m}}_2\mathrm{g} \end{gathered}$$

Adding those two equations together one will get

$\begin{array}{rcl}0& =& -{\mathrm{m}}_1\mathrm{g}-{\mathrm{k}}_2{\mathrm{y}}_0-{\mathrm{m}}_2\mathrm{g}\\ {\mathrm{y}}_0& =& -\frac{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{g}}{{\mathrm{k}}_2}\end{array}$

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

$\begin{array}{rcl}0& =& {\mathrm{m}}_2{\mathrm{k}}_1\left({\mathrm{y}}_0-{\mathrm{x}}_0\right)-{\mathrm{m}}_1{\mathrm{m}}_2\mathrm{g}\partial \\ 0& =& {\mathrm{m}}_1{\mathrm{k}}_1\left({\mathrm{y}}_0-{\mathrm{x}}_0\right)+{\mathrm{m}}_1{\mathrm{k}}_2{\mathrm{y}}_0+{\mathrm{m}}_1{\mathrm{m}}_2\mathrm{g}\\ 0& =& {\mathrm{m}}_2{\mathrm{k}}_1\left({\mathrm{y}}_0-{\mathrm{x}}_0\right)+{\mathrm{m}}_1{\mathrm{k}}_1\left({\mathrm{y}}_0-{\mathrm{x}}_0\right)+{\mathrm{m}}_1{\mathrm{k}}_2{\mathrm{y}}_0\\ 0& =& {\mathrm{k}}_1\left({\mathrm{y}}_0-{\mathrm{x}}_0\right)\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)+{\mathrm{m}}_1{\mathrm{k}}_2{\mathrm{y}}_0\\ {\mathrm{k}}_1{\mathrm{x}}_0\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)& =& {\mathrm{k}}_1{\mathrm{y}}_0+{\mathrm{m}}_1{\mathrm{k}}_2{\mathrm{y}}_0\end{array}$

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 $\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)=\left(-\frac{{\mathrm{m}}_1{\mathrm{k}}_1\mathrm{g}+{\mathrm{m}}_2{\mathrm{k}}_1\mathrm{g}+{\mathrm{m}}_1{\mathrm{k}}_2\mathrm{g}}{{\mathrm{k}}_1{\mathrm{k}}_2},-\frac{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{g}}{{\mathrm{k}}_2}\right)$.

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

$\begin{array}{rcl}{\mathrm{l}}_1& =& \left|{\mathrm{x}}_0-{\mathrm{y}}_0\right|\\ & =& \left|-\frac{{\mathrm{m}}_1{\mathrm{k}}_1\mathrm{g}+{\mathrm{m}}_2{\mathrm{k}}_1\mathrm{g}+{\mathrm{m}}_1{\mathrm{k}}_2\mathrm{g}}{{\mathrm{k}}_1{\mathrm{k}}_2}+\frac{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{g}}{{\mathrm{k}}_2}\right|\\ & =& \left|-\frac{{\mathrm{m}}_1{\mathrm{k}}_1\mathrm{g}+{\mathrm{m}}_2{\mathrm{k}}_1\mathrm{g}+{\mathrm{m}}_1{\mathrm{k}}_2\mathrm{g}-{\mathrm{m}}_1{\mathrm{k}}_1\mathrm{g}-{\mathrm{m}}_2{\mathrm{k}}_1\mathrm{g}}{{\mathrm{k}}_1{\mathrm{k}}_2}\right|\\ & =& \left|-\frac{{\mathrm{m}}_1{\mathrm{k}}_2\mathrm{g}}{{\mathrm{k}}_1{\mathrm{k}}_2}\right|\\ & =& \frac{{\mathrm{m}}_1\mathrm{g}}{{\mathrm{k}}_1}\end{array}$

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 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}\\ & \Rightarrow & {\mathrm{m}}_1\mathrm{\xi }\text{'}\text{'}={\mathrm{k}}_1(\mathrm{\eta }-\mathrm{\xi })+{\mathrm{m}}_1\mathrm{g}-{\mathrm{m}}_1\mathrm{g}\\ {\mathrm{m}}_2\mathrm{\eta }\text{'}\text{'}& =& -{\mathrm{k}}_1(\mathrm{\eta }-\mathrm{\xi })-{\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}\\ & \Rightarrow & {\mathrm{m}}_1\mathrm{\xi }\text{'}\text{'}={\mathrm{k}}_1(\mathrm{\eta }-\mathrm{\xi })\\ {\mathrm{m}}_2\mathrm{\eta }\text{'}\text{'}& =& -{\mathrm{k}}_1(\mathrm{\eta }-\mathrm{\xi })-{\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.