Tachometer feedback for 16/[s(s+1)] to get ζ = 0.6

A second order system $\omega_n^2/(s^2 + 2\zeta\omega_n s + \omega_n^2)$ has peak overshoot $M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}}$. Tachometer (derivative output) feedback increases the $s$ coefficient, so it increases damping without changing $\omega_n$.

Without tachometer feedback

$$\frac{C(s)}{R(s)} = \frac{G}{1+G} = \frac{16}{s^2 + s + 16}$$

Comparing with $s^2 + 2\zeta\omega_n s + \omega_n^2$:

$$\begin{aligned} \omega_n^2 &= 16 \;\Rightarrow\; \omega_n = 4\ \text{rad/s} \\ 2\zeta\omega_n &= 1 \;\Rightarrow\; \zeta = \frac{1}{2\times 4} = 0.125 \end{aligned}$$ $$M_p = e^{-\pi(0.125)/\sqrt{1-0.125^2}} = e^{-0.3958} = 0.6731 = 67.31\%$$

With tachometer feedback $H(s) = 1 + ks$

$$\frac{C(s)}{R(s)} = \frac{G}{1+GH} = \frac{16}{s(s+1) + 16(1+ks)} = \frac{16}{s^2 + (1+16k)s + 16}$$

$\omega_n$ is still $4$ rad/s. For $\zeta = 0.6$:

$$\begin{aligned} 2\zeta\omega_n &= 1 + 16k \\ 2(0.6)(4) = 4.8 &= 1 + 16k \\ k &= \frac{3.8}{16} = 0.2375 \end{aligned}$$

Peak overshoot after adding feedback

$$M_p = e^{-\pi(0.6)/\sqrt{1-0.36}} = e^{-0.75\pi} = e^{-2.356} = 0.0948 = 9.48\%$$

Summary

Answer: $\omega_n = 4$ rad/s, $\zeta = 0.125$; $k = 0.2375$ s; $M_p$ falls from 67.31% to 9.48%. Tachometer feedback greatly reduces overshoot without changing $\omega_n$.