Overshoot, settling and peak time of rotational system

Write the torque equations at each angle, eliminate $\theta_1$ to get $\Theta_2(s)/T(s)$, then compare with the standard second order form.

Equations of motion

At $\theta_1$ (inertia $J = 1$, damper $D = 1$ between $\theta_1$ and $\theta_2$):

$$J\ddot{\theta}_1 + D(\dot{\theta}_1 - \dot{\theta}_2) = T \;\Rightarrow\; (s^2 + s)\Theta_1 - s\Theta_2 = T(s)$$

At $\theta_2$ (no inertia; damper and spring $K = 1$ to the wall):

$$D(\dot{\theta}_2 - \dot{\theta}_1) + K\theta_2 = 0 \;\Rightarrow\; -s\Theta_1 + (s + 1)\Theta_2 = 0$$

Transfer function

From the second equation, $\Theta_1 = \dfrac{s+1}{s}\Theta_2$. Substituting in the first:

$$\begin{aligned} (s^2 + s)\frac{s+1}{s}\Theta_2 - s\Theta_2 &= T \\ \left[(s+1)^2 - s\right]\Theta_2 &= T \\ \frac{\Theta_2(s)}{T(s)} &= \frac{1}{s^2 + s + 1} \end{aligned}$$

Parameters

$$\omega_n^2 = 1 \Rightarrow \omega_n = 1\ \text{rad/s}, \qquad 2\zeta\omega_n = 1 \Rightarrow \zeta = 0.5$$

Percent overshoot

$$\%OS = 100\,e^{-\pi\zeta/\sqrt{1-\zeta^2}} = 100\,e^{-\pi(0.5)/0.866} = 100\,e^{-1.814} = 16.3\%$$

Settling time (2% criterion)

$$t_s = \frac{4}{\zeta\omega_n} = \frac{4}{0.5 \times 1} = 8\ \text{s}$$

Peak time

$$t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}} = \frac{3.1416}{0.866} = 3.63\ \text{s}$$

Answer: $\%OS = 16.3\%$, $t_s = 8$ s, $t_p = 3.63$ s.