$f(t) = \sin t$ has a period of $2\pi$, where $t$ = time. Using words: Each value of $f(t)$ is being repeated every $2\pi$. Using symbols: $\sin t = \sin(t+2\pi)$ which is the definition of a periodic function. That is understood, so far so good.

1. If I have $g(t) = \sin 2t$, **is $\omega\_0 = 2$?** Ιf yes, then $Τ = \frac{2\pi}{\omega\_0} \implies Τ= \frac{2\pi}{2} \implies Τ= \pi$. Is it correct?
2. If I have $h(t) = \sin2ωt$ now what? **Is** $\omega\_0 = 2$? Maybe $\omega\_0 = 2ω$? If yes, then $Τ = \frac{2\pi}{\omega\_0} \implies Τ= \frac{2\pi}{2\omega} \implies Τ= \frac{\pi}{\omega}$ which is not the same as $Τ = \pi$. In order to make it look like $Τ= \pi$ **how do i prove that T= π when i see sin2ωt**, without braking the math/physics rules?

**EDIT** :$\omega\_0$ is just a symbol not harmonics etc i used $\omega\_0$ to differentiate it from $\omega$, my bad i should have used an other symbol like $\omega\_b$ , $\omega\_1$ etc.