Help me understand what happens from here also maybe make some more detailed computation and also if you can plot a graph using python code

Then equations $(5\mathrm{.}5\mathrm{.}1)$ and $(5\mathrm{.}5\mathrm{.}2)$ become

$m_1\frac{d^2{x}_1}{d{t}^2}=-k_1(x_1-L_1)+k_2(x_2-x_1-L_2)+b(\frac{p_2}{m_2}-\frac{p_1}{m_1})$

$m_2\frac{d^2{x}_2}{d{t}^2}=-k_2(x_2-x_1-L_2)-b(\frac{p_2}{m_2}-\frac{p_1}{m_1}),$

and our new first$-$order system is

$\frac{d{x}_1}{dt}=\frac{p_1}{m_1}$

$\frac{d{x}_2}{dt}=\frac{p_2}{m_2}$

$\frac{d{p}_1}{dt}=-k_1(x_1-L_1)+k_2(x_2-x_1-L_2)+b(\frac{p_2}{m_2}-\frac{p_1}{m_1})$

$\frac{d{p}_2}{dt}=-k_2(x_2-x_1-L_2)-b(\frac{p_2}{m_2}-\frac{p_1}{m_1}).$

Indeed,

$\frac{dH}{dt}=-b(\frac{p_2}{m_2}-\frac{p_1}{m_1}{)}^2=-b(\frac{d{x}_2}{dt}-\frac{d{x}_1}{dt}{)}^2.$

Equation $(5\mathrm{.}5\mathrm{.}3)$ tells us that $d\frac{H}{d}t0.$ Moreover, $d\frac{H}{d}t<mn\; id="EditorElement\_602045">0</mn>$ whenever the distance

between the two masses is changing. Thus, energy decreases whenever the second mass is

moving relative to the first. Thus, if the wind or an earthquake starts our building $(m_1)$

swaying back and forth, then the tuned mass$-$damper $(m_2)$ located on one of the top floors

of the building will start to move relative to the building and energy will be removed from

the system by the dampers.

Of course, $m_1$ and $k_1$ involve the building and are set by the architects and engineers.

We$,$ however, are free to choose $m_2,k_2$ and $b.$ We want to choose $b$ fairly large so that there

is a rapid loss of energy; i$.$e$.,$ the magnitude of $d\frac{H}{d}t$ is large. We should choose $m_2$ large

enough so that this mass oscillates with respect to $m_1.$ If we choose $m_2$ two small, then

the strong damper will almost serve as a rigid connection between $m_1$ and $m_2.$ Therefore,

we wish to choose $m_2$ to be large so that we guaranteed that this mass will oscillate with

respect to $m_1.$ We should, however, remember that $m_2$ is sitting on top of a very tall

building. Finally, we should choose $k_2$ to maximize the rate at which the second mass

oscillates with respect to the first and to maximize the oscillations of $m_2.$ If we choose $k_2$ so

that the second spring is in resonance with the first, the oscillations of the first mass force

the second at its resonance frequency. Thus, we will have relatively large oscillations of the

second mass and a large loss of energy.

Let us consider example. Suppose that $m_1=1,k_1=1,$ We will also choose $m_2$ to be

$0\mathrm{.}05$ and $b=0\mathrm{.}1.$ If the initial conditions for our system are

$x_1(0)=10$

$x_2(0)=0$

$p_1(0)=0$

$p_2(0)=0,$