Problem $2$ Solution:

FBD:

i$),M_1{x}_1^{}+K{x}_1+B_1{x}_1^{}-M_{1}g-K(x_3-x_1)=0$

ii$),M_2{x}_2^{}+K{x}_2-M_{2}g-K(x_3-x_2)-B_2(x_3^{}-(x_2^{}))=0$

iii$),M_3{x}_3^{}+K(x_3-x_2)+B_2(x_3^{}-x_2^{})+K(x_3-x_1)-M_{3}g=f_a(t)$

B$)$ When the masses are stationary and the system is in equilibrium, the first derivative

$($velocities$)$ and second derivates $($acceleration$)$ are zero, and let x$1\_$o$,$ x$2\_$o$,$ x$3\_$o be the

constant elongation at equilibrium.

In Problem $3$ of HWO$3($that referred to textbook P$2\mathrm{.}17)$ you derived the differential equations as a set of

three second$-$order equations. For this problem:

$($a$)$ Write the system equations in state$-$space form. This will be a set of six coupled first$-$order linear

equations where the six state variables $q_{1}dots{q}_6$ are the positions $x$ and velocities $v$ of the three

masses and the input is the forcing function $f_a(t).$

$($b$)$ Display the equation in matrix form: $q^{}=Aq+Bu.$

$($c$)$ Assume the output of the system is the force in the spring at the top left that connects M$1$ to

ground. Write the output equation in matrix form: $y=Cq.$

Tip: For an example of the matrix form of a $2-$mass $($fourth$-$order$)$ system, see textbook Example $3\mathrm{.}15.$

For an example of an output equation, see textbook Example $3\mathrm{.}14.$