In Damped Harmonic Oscillation we have:
$F = m\ddot{x} = -kx - c\dot{x}$
$\ddot{x}+\frac{c}{m}\dot{x}+\frac{k}{m}x = 0$
Where $k = m\omega^{2}$. I often see the whole equation written as:
$\ddot{x}+2\gamma\omega\dot{x}+\omega^{2}x = 0$,
where $\gamma$ is the damping factor. In literature it is often written $\gamma = c/2\sqrt{mk}$
My question is, how can we introduce $\gamma$ mathematically leading from the second equation I have written?
Addition: my question is a little vague for what I am really asking, why do we want the equation in the form of:
$\ddot{x} + 2\gamma \omega \dot{x} + \omega^2 x = 0$
why do we want $c/m=2\gamma\omega$ and how do we introduce the parameter $\gamma$ in the first place?
Let's start with
$$ \ddot{x} + \frac{c}{m} \dot{x} + \frac{k}{m} x = 0 $$
and define two new constants
$$ \omega := \sqrt{\frac{k}{m}}, \,\,\, \gamma := \frac{c}{2\sqrt{mk}}. $$
Calculating, we have
$$ \omega^2 = \frac{k}{m}, \\ 2 \gamma \omega = 2 \frac{c}{2\sqrt{mk}} \sqrt{\frac{k}{m}} = \frac{c}{\sqrt{m}\sqrt{k}} \frac{\sqrt{k}}{\sqrt{m}} = \frac{c}{m}. $$
Substituting it into the original equation, we get the equation
$$ \ddot{x} + 2\gamma \omega \dot{x} + \omega^2 x = 0. $$
The standard approach to solving second order linear ODEs is to look for a solutions of the form $x(t) = e^{\lambda t}$ for some $\lambda \in \mathbb{C}$. By plugging it into the equation, we get
$$ \lambda^2 e^{\lambda t} + 2\gamma \omega \lambda e^{\lambda t} + \omega^2 e^{\lambda t} = (\lambda^2 + 2\gamma \omega \lambda + \omega^2) e^{\lambda t} = 0.$$
The polynomial equation (in $\lambda$)
$$ \lambda^2 + 2\gamma \omega \lambda + \omega^2 = 0 $$
is called the characteristic equation of the ODE and the solutions determined the behaviour of the solutions of the ODE. Now, we can complete the square and get the equivalent equation
$$ (\lambda + \gamma \omega)^2 = (\gamma^2 - 1)\omega^2. $$
We have set up the constants in such a way that the behavior of the solutions depends on the constant $\gamma$. Assuming $\omega > 0$, we see that if $\gamma > 1$, we get exponential solutions (as the quadratic equation has two real roots), if $\gamma < 1$, we get oscillatory solutions (as the quadratic equation has two complex roots).