Simplifying logarithmic decrement equation

Question

I'm currently writing a paper on underdamped oscillatory systems where I'm using the logarithmic decrement equation:

$\delta = \ln\frac{x(t_n)}{x(t_n+T)}$

Where $T$ is the period of the system. I then substituted the equation for initial displacement "$x(t_0)$":

$x(t) = Ae^{-\zeta\omega_n t} \sin(\omega t + \Phi)$

(where $A$ is the amplitude, $\zeta$ is the damping coefficient, $\omega_n$ is the natural frequency, $\omega$ is the angular frequency, and $\Phi$ is the angle phase.)

into the logarithmic decrement equation, and thus formed the following equation:

$\delta = \ln\frac{Ae^{-\zeta\omega_nt_n} \sin(\omega t_n + \Phi)}{Ae^{-\zeta\omega_n(t_n+T)} \sin(\omega (t_n+T) + \Phi)}$

I know this equation can be simplified into the form: $\delta = \ln(e^{-\zeta\omega_n\frac{t_n}{t_n+T}})$ and further simplified into $\delta = -\zeta\omega_nT$. But I'm confused as to how these equations can be simplified in this way; I don't understand how the $\sin$ can cancel each other out and how the $\frac{t_n}{t_n+T}$ can be simplified into $T$.

Answer 1

Is it not simply that $\omega T=2\pi$, so that $$\sin (\omega t_n + \Phi)=\sin (\omega (t_n + T)+\Phi)\ \ \ ?$$

We can then divide the top and bottom of your fraction by $Ae^{-\zeta \omega_n t_n}$ leaving $$\delta=\ln [e^{\zeta \omega_n T}] = \zeta \omega_n T.$$

(Though I must admit that I don't understand exactly what's going on, for example the relationship between $\omega$ and $\omega_n$.)