User:LkNsngth/Physics problem

So we have $$a(t)=A sin(\omega t)$$

Now we know $$a(t)=\frac{d^2}{dx^2}x(t)$$ where $$x(t)$$ is the position function. Now notice taking the antiderivative of a sine or cosine wave does not change the argument: we have that

$$ \frac{d}{dx}sin(\omega t) = \omega cos(\omega t) $$ and that $$ \frac{d}{dx}cos(\omega t)=-\omega sin(\omega t) $$ Therefore we know after taking two antiderivatives, our position function $$x(t) = K sin(\omega t) + C$$ where K and C are constants; neither of the two change the angular frequency $$\omega=2\pi f$$ so we have that $$ f=\frac{\omega}{2\pi} $$ Taking its reciprocal, we have $$ T= \frac{2\pi}{\omega}$$