5. (1990 National College Entrance Examination for Science) It is known that Figure 1-1-9 is the graph of the function $y=2 \sin (\omega x+\varphi)\left(|\varphi|<\frac{\pi}{2}\right)$. Then
A. $\omega=\frac{10}{11} \cdot \varphi=\frac{\pi}{6}$
B. $\omega=\frac{10}{11} \cdot \varphi=-\frac{\pi}{6}$
C. $\omega=2, \varphi=\frac{\pi}{6}$
D. $\omega=2, \varphi=-\frac{\pi}{6}$
5. C From the graph, when $x=0$, $y=1$, so
$$
\begin{array}{c}
2 \sin (\omega x+\varphi)=1, \\
\sin \varphi=\frac{1}{2}, \\
|\varphi|<\frac{\pi}{2},
\end{array}
$$

that is

Because

So
$$
\varphi=\frac{\pi}{6} \text {. }
$$

From the four given options, the value of $\omega$ is $\frac{10}{11}$ or 2. If $\omega=\frac{10}{11}$, then
$$
T=\frac{2 \pi}{10 / 11}=2.2 \pi
$$

If $\omega=2$, then
$$
T=\frac{2 \pi}{2}=\pi .
$$

From the graph, it is clear that $\omega=\frac{10}{11}$ does not meet the requirement. Choose C.