5. (12th "Hope Cup" Senior High School Competition) Given the function $f(x)=A \sin (\omega x+\varphi), A, \varphi \in \mathbf{R}$. To make the smallest positive period $T \in\left(\frac{1}{100}, \frac{1}{50}\right)$ of $f(x)$, the number of elements in the set of positive integer values that $\omega$ can take is ( )
A. 311
B. 312
C. 313
D. 314
5. D) Because $f(x)=\Lambda \sin (\omega x+\varphi)$ has the smallest positive period $T$. And $T \in\left(\frac{1}{100}, \frac{1}{50}\right)$, and also $T=\frac{2 \pi}{\omega}$.

So $\frac{\omega}{2 \pi} \in(50,100)$.
So $\omega \in(100 \pi, 200 \pi)$.
So $315 \leqslant \omega \leqslant 628$ and $\omega \in \mathbf{N}$.
So the number of elements in the set of possible values of $\omega$ is $628-315+1=314$. Hence, the answer is D.