Case 3] : General solution for under damped system ( $\xi \lt 1$) :

In this case, general solution is given by equation $x = c_1e^{s1t} + c_2e^{s2t}$

Here, $s_1 = (- \xi + j\sqrt{1-\xi ^2}) w_n$

$s_2 = ( -\xi - j \sqrt{1 - \xi ^2}) w_n$

where $j = \sqrt{-1}$ is the imaginary unit of complex roots.

Substituting these values in above equation, we get,

$x = c_1e ^{[ -\xi + j \sqrt{1-\xi ^2}] w_nt + c_2 e [ -\xi - j \sqrt{1 - \xi ^2}] w_{nt}}$

$= e^{-\xi w_nt [ c_1 e j \sqrt{1- \xi ^2} w_nt + c_2 e -j \sqrt{1 - \xi ^2} w_{nt}]}$

we know $e^{j \theta} \ = \ cos \ \theta \ + \ jsin \ \theta$ and $e^{-j \theta} = cos \ \theta \ - \ j sin \ \theta$

then above equation can be written as:

$x = e^{-\xi w_n t} [ c_1 \ cos ( \sqrt{1 - \xi ^2} w_n t ) + c_1 ^{j sin} ( \sqrt{1-\xi ^2} w_{nt} )]$

$[ c_2 cos (\sqrt{1-\xi ^2} w_{nt} )] - jc_2^{sin} (\sqrt {1-\xi ^2} w_{nt} )]$

$x = e^{-\xi w_n} [ (c_1 + c_2) cos (\sqrt{ 1 - \xi ^2} w_{nt}) + j (c_1 - c_2) \times sin ( \sqrt{1 - \xi ^2} w_{nt})]$

In above equation, the constants $(c_1 + c_2)$ and $(c_1 - c_2)$ are real quantities which make $c_1$ and $c_2$ complex conjugate quantities.

Hence, above quantities can be replaced by constants A and B respectively.

then displacement equation for under damped system is written as

$x = e^{-\xi w_n t} [ A \ cos \ ( ( \sqrt{1-\xi ^2} w_{nt} ) + B \ sin\ \sqrt{ 1 - \xi ^2} w_{nt})]$ - - - (1)

where A and B can be found out by using boundary condition.

At t = 0, $x = x_0$ - - - (1)

At t = 0, $x = x_0$ - - - - (2)

Substituting (1) boundary condition in equation (1), we get

$x_0 = A$ - - -(1)

Now differentiating equation (1), we get

$\dot{x} = e^{-\xi wnt} [ - A \ sin \ (\sqrt{1 - \xi ^2} w_{nt} ) \sqrt{1 - \xi ^2} w_n + \ B \ cos \ ( \sqrt{1 - \xi ^2} w_{nt}) \sqrt{1 - \xi ^2} w_n ]$

$+ e^{-wnt} (-\xi w_n) [ A \ cos \ ( \sqrt{1- \eta ^2} w_n + \ B \ cos \ ( \sqrt{1 - \xi ^2} w_{nt} ) \sqrt{1- \xi ^2} w_n ]$

$+ e^{-\xi w_nt} (-\xi w_n) [ A \ cos \ (\sqrt{1- \xi ^2} w_{n t} ) + B \ sin \ (\sqrt{ 1 - \xi ^2} w_{n t} ) ]$

Substituting (2) boundary condition in above equation, we get

$\dot{x_0} = [ B \sqrt{1 - \xi ^2} w_n - \xi n_n \times A $ - - - (2)

Substituting $A = x_0$, we can write

$\dot{x_0} = B \sqrt{1 - \xi ^2} w_n - \xi w_n x_0$

$\therefore$ $B = \frac{x_0 + \xi w_n x_0}{\sqrt{ 1 - \xi ^2 w_n }}$

then general solution for displacement equation is written as

$x = e^{-\xi w_n t }[ A \ cos \ ( \sqrt{1 - \xi^2} w_{nt} ) + B \ sin \ ( \sqrt{1 - \xi ^2} w_{nt} ) ]$

Above equation can be written as

$x = e^{-\xi w_nt} [ A \ cos \ (w_dt) + B \ sin \ (w_dt)]$ - - -(a)

Above equation of displacement can be written as

$x = e^{-\xi w_n t} [ X \ sin \ (w_dt + \phi)]$

Above equation can be written as

$x = e^{-\xi w_n t } [ X \ sin \ (w_dt) \ cos \ (\phi) + X \ cos \ (w_dt) \ sin \ (\phi)]$ - - -[b]

comparing (a) and (b), we get

A = X sin $\phi$

B = X cos $\phi$

$\therefore$ $tan \phi = \frac{A}{B} $ and hence $X = \sqrt{A^2 + B^2}$

Then response curve for under damped system is shown below where the displacement decreases exponentially with respect to time.

From the above graph, it can be said that maximum amplitude $x_e^{-\xi w_nt}$ decays exponentially with respect to time and also as the damping ratio $\xi$ increases for given system displacement decreases.