The mass of a spring mass da shpot system is given an initial velocity from the equilibrium position of 'Awn' where $w_n$ is the undamped natural frequency of the system, find the equation of motion for the system, for the case,

When 1] $\xi$ = 2.0

2] $\xi$ = 1.0

3] $\xi$ = 0.2

$\rightarrow$ $\xi$ = 2 [Over damped system]

$x = c_1 e^{s1t} + c_2 e^{s2t}$

where $s_1 = [ - e + \sqrt{e^2 - 1}] w_n$

$s_2 = [ - e + \sqrt{e^2 - 1}] w_n$

$\rightarrow$ t = 0, x = 0

t = 0, x = A $w_n$

Use 1st BCS,

$0 = c_1 + c_2$

$\therefore$ $c_1 = - c_2$

On differentiating w.r.t. 't'

$\frac{dx}{dt} = x = c_1 s_1 e^{s_1 t} + c_2 s_2 e^{s_2 t}$

Use 2nd BCS,

t = 0 , x =A $w_n$

$Awn = c_1 s_1 + c_2 s_2$

$= c_2 s_2 - s s_1$

$= c_2 (s_2 - s_1)$

$\therefore$ $c_2 = \frac{A$w_n$}{s_2 - s_1}$

$\therefore$ $c_2 = -c_2$

$= \frac{- A w_n}{s_2-s_1}$

$c_1 = \frac{A w_n}{s_1 - s_2}$

$\therefore$ $ s_1 = -[ 2 + \sqrt{ (2)^2 - 1} ] w_n = -3.73 w_n$

$\therefore$ $ s_2 = - [ 2 - \sqrt{2^2 - 1} ] w_n = -0.26 w_n$

$\therefore$ $x = \frac{Awn}{s_1 - s_2} e^{-3.73wnt} + \frac{Awn}{s_2 - s_1} e^{-0.26 wnt}$

$= \frac{Awn}{-3.73wn + 0.26 wn} e^{-3.73 wnt} + \frac{Awn}{-0.26wn + 3.73wn} e^{0.26wnt}$

$= -0.28A e^{-3.73wn}t + 0.28 A e^{-0.26wnt}$

$x = 0.28 A e^{-0.26wnt} - 0.28 e^{-3.73wnt}$

Case 2: eg = 1 (critically damped)

$x = (c_1 + c_2 t) e^{st}$ - - - -(2)

Here, $s = -W_n$

$\therefore$ $x = (c_1 + c_2.t) . e^{-wnt}$

$= c_1 . e^-wnt + c_2. t. e^{-wnt}$

Boundary condition

t = 0, x = 0

$\therefore$ $0 = (c_1 + 0) , e^{-0}$

t = 0, x = A. $w_n$

$x = -wn. c_1 e^{-wnt} + c_2 [ t (-wn) e^{-wnt} + e^{-wnt}]$

$Awn = 0 + c_2 [0+1]$

$C_2 = A. \ w_n$

Put $c_1$ and $c_2$ in equation (2)

$\therefore$ $x = (0 + A.wnt) .e^{-wnt}$

$x = A. wnt. e^{-wnt}$

Case 3: As $\xi$ = 0.2 [under damped system]

x = x. [ $e^{- \xi wnt } \ sin \ [wd.t+\phi ] ]$

where, $wd = \sqrt{1 = \xi^2} w_n$

$= \sqrt{1-(0.2)^2}.w_n$

Boundary condition,

As x = 0, t = 0

$0 = X \ [sin \ (wd.t) \ + \ sin \ \phi]$

$X \ sin \ \phi \ = \ 0$

As $x \neq 0$

$Sin \ \phi = 0$

$\therefore$ $\phi = 0$

$x = x [ e^{-\xi.wnt}, \ sin \ wdt]$

As $x = A \ w_n$, t = 0

$x = x [sin (wd.t) . \ (-\xi wn) e^{-\xi wnt} + e^{-\xi wnt} \ cos \ (wdt).wd]$

Awn = x [wd]

$X = \frac{A.\ wn}{wd} = \frac{A.\ wn}{\sqrt{1- \xi^2 } w_n} = \frac{A}{0.9797}$

A = X. 0.9797