periodicity of a sinusoidal function

Question

$f(t) = \sin t$ 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.

Answer 1

$$\sin(2\omega t+2\pi)=\sin(2\omega t)$$

Therefore,

$$\sin\left(2\omega\left(t+\frac{\pi}{\omega}\right)\right)=\sin(2\omega t)$$

That is,

$$h\left(t+\frac\pi\omega\right)=h(t)$$

So $\pi/\omega$ is a period of $h$.

Answer 2

A sinusoid is usually written as $\sin\omega t$ where the pulsation $\omega=\dfrac{2\pi}T$ and $T=\dfrac{2\pi}\omega$ is the period.

So $\sin2t$ denotes a sinusoid of pulsation $\omega=2$ and period $T=\dfrac{2\pi}2=\pi$.

But if you write $\sin\color{red}2\omega t$, $\omega$ no longer has the meaning of a pulsation, but is a "half-pulsation", and what you would find with $\dfrac{2\pi}\omega$ is the double period (while the true period is $\dfrac{2\pi}{\color{red}2\omega}$).

Due to the confusion this introduces, I advise you to stay away from such a notation.