General second order system and mass spring damper in control theory I am studying control theory and most textbooks and web resources define a general second order system in the $s$ domain as
$$ G(s) = \frac{\omega_n^2}{s^2+2\,\zeta\,\omega_n\,s+\omega_n^2} \, .\tag1$$
However, the mass spring damper which is clearly a second order system reduces to this equation
$$ G(s) = \frac{1}{m\,s^2+b\,s+k}\tag2$$
which further reduces to
$$ G(s) = \frac{1/m}{s^2 + (b/m)s + k/m} \, .\tag3$$
The problem is I cannot come to terms with the equivalence between the equation above and the general second order problem which seems to contain the same term on the numerator as the constant coefficient in the denominator.
I've tried some books namely Ogata, Nise and I even managed to get my hands on an old book by Franklin and Powell. The three of them present the general second order system as the top one.
Can someone clarify this for me, and explain why the mass-spring-damper does not correspond to the general equation?
 A: The easiest way for you to understand (and you should do this to convince yourself) is to write each of these two systems as block diagrams using integrators, gain blocks and feedback loops. Note that in the general system there is unity gain (1-in, 1-out) which means that there is unity feedback gain in the outermost feedback loop. 
For the first system just let
$$G(s)=\frac{y(s)}{x(s)} \, ,$$
where $x$ is the input, $y$ the output.
For the specific spring mass system you do not have a 1-in, 1-out relationship. You will see from your block diagram for this system that the outer-most loop will have a feedback gain of $k$. This makes the input-output gain $\frac{1}{k}$. For the spring-mass-damper system
$$G(s)=\frac{x(s)}{F(s)} \, ,$$
where $x$ is displacement and $F$ is force. You can further see that if the input is a step force that at steady state (using the final value theorem)
$$\lim\limits_{s \to 0} s\frac{1}{s}G(s) = \frac{1}{k} \, ,$$
which is just Hooke's law, explaining why you do not have unity gain.
A: Equation 1 above is very correct - it is the generalized equation of a second order system.
In specific terms as it relates to a mass spring damper system, Equation 2 above is also very correct, when you apply newton law of motion to mass spring damper system you will arrive at equation 2.
Equation 3 is the normalized version of equation 2 and so both equation 2 and equation 3 are the same.
If you compare equation 3 and the generalized second order system (i.e. equation 1) you will see that they have the same pattern.
Comparing the coefficients, $(b/m)$ is equal to  $2\zeta\omega n$ and $(k/m)$ is equal to $\omega^2_n$.
The numerator $(1/m)$ is equal to gain times $\omega^2_n$.
For the generalized second order system in Equation 1 above the gain is assumed to be unity (1). That is why you did not see a constant in the form of a letter multiplying $\omega^2_n$.
