# Damping factor in critical damping

1. May 30, 2014

### PsychonautQQ

The position equation for a oscillator undergoing critical damping is given by

x(t) = Ate^(-γt) + Be^(-γt)
where γ = c/2m
and c is from the original force equation
ma + cv + kx = 0

γ is called the damping factor
my book then goes on to say without explanation that
γ = c/2m = (k/m)^(1/2) = initial angular velocity

I understand the units of c/2m work out to be s^-1 so that's good, but other than that I'm not understanding the relation c/2m = w_initial as well as I want to.

LaTex coming soon to a post near you

2. May 30, 2014

### vanhees71

Let's write the equation of motion in a somewhat more convenient form
$$\ddot{x}+2 \gamma \dot{x}+\omega^2 x=0.$$
Then
$$\gamma=\frac{c}{2m}, \quad \omega^2=\frac{k}{m}.$$
now consider the special case of "critical damping", where $\gamma=\omega$.

To solve the equation, we make the standard ansatz for this linear differential equation with constant coefficients:
$$x(t)=A \exp(\lambda t).$$
Plugging this into the equation leads after some simple algebra to
$$\lambda^2 + 2 \omega \lambda + \omega^2=(\omega+\lambda)^2=0.$$
This means there is only one solution for $\lambda$:
$$\lambda_1=-\omega,$$
and one solution to the ODE is
$$x(t)=A \exp(-\omega t).$$
For a complete solution, however we need one other linearly independent solution.

Also here, a standard ansatz helps, namely the "ansatz of the variation of the constant", i.e.,
$$x(t)=\exp(-\omega t) y(t).$$
Plugging this into our oscillator equation of motion yields
$$\exp(-\omega t) [\ddot{y}-2 \omega \dot{y}+\omega^2 y-2 \omega^2 y + 2 \omega \dot{y} + \omega^2 y)]=\exp(-\omega t) \ddot{y}=0,$$
which immediately implies
$$\ddot{y}=0 \; \Rightarrow \; y(t)=(A+B t).$$
The general solution of our equation is thus
$$x(t)=(A+B t) \exp(-\omega t),$$
and thus the two linear independent set solutions is given by
$$x_1(t)=\exp(-\omega t), \quad x_2(t)=t \exp(-\omega t).$$