Example 5 Two periodic functions $y_{1}$ and $y_{2}$ have the smallest positive periods of $a$ and $b$ respectively, and $b=n a$, where $n \geqslant 2, n \in \mathbf{N}$. If the function $y=y_{1}+y_{2}$ has the smallest positive period of $t$, then among the following 5 scenarios: (1) $t<b$;
(5) $a<t<b$, the possible scenarios are $\qquad$ (fill in the sequence number).
Analysis The period is an important property of trigonometric functions, construct a trigonometric function to answer.
Solution From the problem, we know that $b$ is the period of $y=y_{1}+y_{2}$, so $t \leqslant 6$, scenario (4) cannot occur; from $y_{2}=y-y_{1}$, if $t=a$, then $a$ is also the smallest positive period of $y_{2}$, which is a contradiction. Therefore, scenario (2) cannot occur, and the other three scenarios may occur. The following examples illustrate that the other three scenarios are possible: take $y_{2}=\sin x+\sin \frac{2 x}{3}$, then its smallest positive period is $b=6 \pi$. Let $y_{1}=-\sin \frac{2 x}{3}$, at this time $a=3 \pi, t=2 \pi, t<a$; let $y_{1}=-\sin x$, at this time $a=2 \pi, t=3 \pi, a<t<b$; let $y_{1}=\sin x$, at this time $a=2 \pi, t=6 \pi, t=b$. Therefore, the possible scenarios are (1) (3) (5)