Case 2] : General solutions of critically damped system ($\xi = 1$) :

In this case both the roots are real and equal.

In this case, the solution is

$x = (c_1 + c_2t)e^{st} = (c_1 e^{st} + c_2te^{st})$

Differentiating

$x = c_1se^{st} + c_2te^{st} + c_2e^{st}$

$= c_1 se^{st} + c_2 (t. se^{st} + e^{st})$

Then using boundary condition,

At t = 0, x = $x_0$ and at t = 0, $x = x_0$

Here $s = - w_n$

$\therefore$ $x = -w_n c_1 e^{-wnt} - w_n tc_2 ^{-wnt} + c_2 e^{-wnt}$

$\therefore$ $x = e^{-wnt} [-w_n c_1 - c_2 (w_nt - 1)]$ - - - - (1)

Also we know displacement,

$x = (c_1 + c_{2t}) e^{-wnt}$

where $c_1$ and $c_2$ are constants to be found out by known boundary conditions.

i.e. At t = 0, x = $x_0$ and at t = 0, $x = x_0$

substituting the first boundary condition in equation (2), we get $x_0 = c_1$ and substituting the second condition in equation (1).

$\therefore$ $x = w_n c_1 + c_2 $

substituting value of $c_1$ in above equation (2), we get the general displacement equation as:

$x = e^{-wnt} [ x_0 + (x_0 + w_n x_0) t]$

From the above equation, we can say the displacement decreases exponentially with respect to time, then, the response curve for critically damped system is shown in earlier displacement time curve for $\xi = 1$

from above equation it can be said that when x = 0, $t \rightarrow \infty$

from above equation it can be said that motion of critically damped system is 'a periodic'.

then, the shortest possible time which will be taken by system under critical damping condition, so that x = 0 is given by equation.

$0 = x_0 (x_0 + w_n x_0)^t$

$\therefore$ $t = \frac{-x_0}{x_0 + w_n x_0}$

For 't' to be positive magnitude of $x_0$ > magnitude of $w_n x_0$ and direction of $x_0$ must be opposite to $x_0$