Example 2 (10th "Hope Cup" Senior High School Competition Question) The function $y=\sin \omega x (\omega>0)$ has exactly 50 maximum values in the interval $[0,1]$, then the range of $\omega$ is $\qquad$ .
Let $T=\frac{2 \pi}{\omega}$ be the smallest positive period, then $\left(49+\frac{1}{4}\right) T=1$ and $\left(50+\frac{1}{4}\right) T=1$ have 50 and 51 maximum points, respectively. When $T$ increases, the number of maximum points either remains unchanged or decreases.
Therefore, $\omega \in\left[\frac{197 \pi}{2}, \frac{201 \pi}{2}\right)$.
/ Analysis: To think about the problem, start with how many maximum points $y=\sin x$ has in one period. In an interval of length one standard period $2 \pi$, there is only one maximum point $x=\frac{\pi}{2}$ in $[0,2 \pi)$, but there are two maximum points $x=\frac{\pi}{2}$ and $x=2 \pi+\frac{\pi}{2}$ in $\left[\frac{\pi}{2}, 2 \pi+\frac{\pi}{2}\right]$. To have 50 consecutive maximum points, the image must contain at least 49 periods.

For example, the function $y=\sin x$ has 50 maximum points in the interval $\left[\frac{\pi}{2}, 98 \pi+\frac{\pi}{2}\right]$ which contains 49 periods. In the interval $\left[0,98 \pi+\frac{\pi}{2}\right]$ which contains 49 and a half periods, there are exactly 50 maximum points. Generally, for the interval $[0, a], 98 \pi+\frac{\pi}{2} \leqslant a$ $0)$ to have exactly 50 maximum points in the interval $[1,2]$, the range of $\omega$ is
(2) For the function $y=\sin \omega x(\omega>0)$ to have exactly 50 maximum points in the interval (1.2], the range of $\omega$ is
(3) For the function $y=\sin \omega x(\omega>0)$ to have exactly 50 maximum points in the interval (1,2), the range of $\omega$ is
To solve, it is convenient to use the method of supplementation, first extending the interval to a complete interval $[0,2]$ for consideration. Just note whether the endpoints 1 and 2 are maximum points, and the number of maximum points in $(0,1)$ and $(1,2)$ is the same, making it not difficult to find the range of $\omega$.

The analysis uses a special case strategy, which is a commonly used strategy in problem-solving. When we face a structurally complex or difficult-to-start general problem, we should pay attention to moving from the general to the specific, first examining some simpler special problems contained in the general situation, so as to broaden the problem-solving思路, discover the solution direction or path of the original problem from the study of special problems. Specializing a general problem, by adding certain restrictions or appropriately strengthening certain conditions to the object being studied, can result in various different special problems. In actual problem-solving, one can start from special values, special situations, and special positions.