Answer :

10.90

Given

$G_p(s)=\frac{14.4}{s(1+0.1s)}$

G_c(s) = 1

H(s) = 1

G(s) = G_c(s)G_P(s)

$G(s)=\frac{14.4}{s(1+0.1s)}$

The transfer function of the system will be,

$T\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\frac{G\left(s\right)}{1+G\left(s\right)H\left(s\right)}$

$T\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\frac{\frac{14.4}{s\left(1+0.1s\right)}}{1+\frac{14.4}{s\left(1+0.1s\right)}}$

$T\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\frac{14.4}{0.1s^2+s+14.4}$

$T\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\frac{14.4}{0.1[s^2+10s+144]}$

$T\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\frac{144}{[s^2+10s+144]}$.... (1)

We know that,

$T\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\frac{{\omega }_n^2}{[s^2+2\zeta {\omega }_{n}s+{\omega }_n^2]}$.... (2)

From equation (1) and (2),

${\omega }_n^2=144$

${\omega }_n=12$

$2\zeta {\omega }_n=10$

$\zeta =\frac{10}{24}=\frac{5}{12}$

${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}=12\sqrt{1-{\left(\frac{10}{24}\right)}^2}=10.91rad/sec$