Find M, B and K from step response

For a force input the mass-spring-damper is a standard second order system. The final value gives $K$, the overshoot gives $\zeta$, and the peak time gives $\omega_n$.

Transfer function

Force balance on $M$: $F = M\ddot{x} + B\dot{x} + Kx$, so

$$\frac{X(s)}{F(s)} = \frac{1}{Ms^2 + Bs + K} = \frac{1/M}{s^2 + \frac{B}{M}s + \frac{K}{M}}$$

with $\omega_n^2 = K/M$ and $2\zeta\omega_n = B/M$.

Step 1: $K$ from the final value

For a unit step force, $F(s) = 1/s$:

$$x(\infty) = \lim_{s\to 0} s\cdot\frac{1}{s}\cdot\frac{1}{Ms^2+Bs+K} = \frac{1}{K} = 2 \;\Rightarrow\; K = 0.5\ \text{N/m}$$

Step 2: $\zeta$ from the overshoot

$$M_p = \frac{2.36 - 2}{2} = 0.18$$ $$\zeta = \frac{-\ln 0.18}{\sqrt{\pi^2 + (\ln 0.18)^2}} = \frac{1.7148}{\sqrt{9.8696 + 2.9405}} = \frac{1.7148}{3.5791} = 0.4791$$

Step 3: $\omega_n$ from the peak time

$$t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}} = 2 \;\Rightarrow\; \omega_n = \frac{\pi}{2\sqrt{1 - 0.4791^2}} = \frac{3.1416}{2(0.8778)} = 1.790\ \text{rad/s}$$

Step 4: $M$ and $B$

$$\begin{aligned} M &= \frac{K}{\omega_n^2} = \frac{0.5}{1.790^2} = \frac{0.5}{3.2025} = 0.156\ \text{kg} \\ B &= 2\zeta\omega_n M = 2(0.4791)(1.790)(0.156) = 0.268\ \text{N-s/m} \end{aligned}$$

Answer: $M \approx 0.156$ kg, $B \approx 0.268$ N-s/m, $K = 0.5$ N/m.