Find a and b for 25% overshoot

Reduce the minor (velocity) feedback loop first, then compare the closed-loop characteristic equation with $s^2 + 2\zeta\omega_n s + \omega_n^2$.

Step 1: Reduce the minor loop

The block $\dfrac{a}{Js}$ has feedback $b$:

$$\frac{a/(Js)}{1 + ab/(Js)} = \frac{a}{Js + ab}$$

Open-loop transfer function (in series with $1/s$):

$$G(s) = \frac{a}{s(Js + ab)}$$

Step 2: Closed-loop transfer function (unity feedback, $J = 1$)

$$\frac{C(s)}{R(s)} = \frac{a}{s^2 + ab\,s + a} \;\Rightarrow\; \omega_n^2 = a,\quad 2\zeta\omega_n = ab$$

Step 3: $\zeta$ from $M_p = 25\%$

$$\zeta = \frac{-\ln 0.25}{\sqrt{\pi^2 + (\ln 0.25)^2}} = \frac{1.3863}{\sqrt{9.8696 + 1.9218}} = \frac{1.3863}{3.4339} = 0.4037$$

Step 4: $\omega_n$ from $t_p = 2$ s

$$\omega_n = \frac{\pi}{t_p\sqrt{1-\zeta^2}} = \frac{3.1416}{2\sqrt{1 - 0.4037^2}} = \frac{3.1416}{2(0.9149)} = 1.717\ \text{rad/s}$$

Step 5: Find $a$ and $b$

$$\begin{aligned} a &= \omega_n^2 = 1.717^2 = 2.948 \\ b &= \frac{2\zeta\omega_n}{a} = \frac{2(0.4037)(1.717)}{2.948} = \frac{1.386}{2.948} = 0.470 \end{aligned}$$

Answer: $a \approx 2.95$ and $b \approx 0.47$ (with $J = 1$ kg-m²).