Peak and rise time from 40% overshoot

Overshoot fixes $\zeta$. The settling time fixes $\zeta\omega_n$ (2% criterion, $t_s = 4/\zeta\omega_n$). Then $\omega_d$ gives $t_p$ and $t_r$.

Step 1: Damping ratio from $M_p = 40\%$

$$\zeta = \frac{-\ln 0.40}{\sqrt{\pi^2 + (\ln 0.40)^2}} = \frac{0.9163}{\sqrt{9.8696 + 0.8396}} = \frac{0.9163}{3.2725} = 0.280$$

Step 2: Natural frequency from $t_s = 4$ s (2% criterion)

$$t_s = \frac{4}{\zeta\omega_n} = 4 \;\Rightarrow\; \zeta\omega_n = 1 \;\Rightarrow\; \omega_n = \frac{1}{0.280} = 3.571\ \text{rad/s}$$

Step 3: Damped frequency

$$\omega_d = \omega_n\sqrt{1-\zeta^2} = 3.571\sqrt{1 - 0.0784} = 3.571(0.960) = 3.429\ \text{rad/s}$$

Step 4: Peak time

$$t_p = \frac{\pi}{\omega_d} = \frac{3.1416}{3.429} = 0.916\ \text{s}$$

Step 5: Rise time (0 to 100%)

$$\theta = \cos^{-1}\zeta = \cos^{-1}(0.280) = 1.287\ \text{rad}$$ $$t_r = \frac{\pi - \theta}{\omega_d} = \frac{3.1416 - 1.287}{3.429} = \frac{1.8546}{3.429} = 0.541\ \text{s}$$

Answer: $t_p \approx 0.916$ s and $t_r \approx 0.541$ s ($\zeta = 0.28$, $\omega_n = 3.57$ rad/s).

(With the 5% criterion, $t_s = 3/\zeta\omega_n$, you get $\omega_n = 2.679$ rad/s, $t_p = 1.222$ s and $t_r = 0.721$ s.)