I'm trying to better understand this simple situation: If I have two functions
$$ \begin{aligned} y_1 &= \sin(\omega t) \\ y_2 &= \sin(\omega t) + at \end{aligned} $$
Will they reach local maxima at the same time? On one hand, both oscillate at the same frequency, however, in the same $\Delta t$ the function $y_2$ is also changing due to the second term (so the full function is not a sine anymore). Numerically it looks like there is a time delay, however, I want to try and quantifying it analytically and not sure how to.
The derivatives are $y_1'(t) = \omega \cdot \cos( \omega t)$ and $y_2'(t) = \omega \cos( \omega t ) + a$. So, aside from the case of $\omega = 0$, this implies that
$$ y_1'(t) = 0 \Rightarrow t = { {1} \over {\omega} } \left( {\pi \over 2} + k\pi \right), \,\,\, k \in \mathbb{Z} $$
$$ y_2'(t) = 0 \Rightarrow \cos(\omega t) = { {-a} \over {\omega} } $$
$y_1$ has the local extrema at the usual spots. For $y_2'(t) = 0$, however, for there to be a solution requires that $-1 \leq -a / \omega \leq +1.$ If this inequality does not hold, then $y_2(t)$ has no local extrema. If it does hold, then you can work out where they will be starting with:
$$ \cos(\omega t) = { {-a} \over {\omega} } \Rightarrow t = {1 \over \omega} \cos^{-1}({-a \over \omega}) $$
It's gets a little messy here with $\cos^{-1}$, but you can get the information you want. You still need to evaluate the local extrema to determine if they are maxima, minima, or neither.