已知函数 $f(x)=\sin ^{2} \frac{\omega x}{2}+\frac{1}{2} \sin \omega x-\frac{1}{2}(\omega>0), x \in R$. 若 $f(x)$ 在区间 $(\pi, 2 \pi)$内

没有零点, 则 $\omega$ 的取值范围是 ( )
 $f(x)=\frac{1}{2}(1-\cos \omega x)+\frac{1}{2} \sin \omega x-\frac{1}{2}=\frac{1}{2} \sin \omega x-\frac{1}{2} \cos \omega x=\frac{\sqrt{2}}{2} \sin \left(\omega x-\frac{\pi}{4}\right)$, 当 $\omega=\frac{1}{2}$ 时, $f(x)=\frac{\sqrt{2}}{2} \sin \left(\frac{1}{2} x-\frac{\pi}{4}\right), x \in(\pi, 2 \pi)$ 时, $f(x) \in\left(\frac{1}{2}, \frac{\sqrt{2}}{2}\right]$, 无零点, 排除 $\mathrm{A}, \mathrm{B}$; 当 $\omega=\frac{3}{16}$ 时, $f(x)=\frac{\sqrt{2}}{2} \sin \left(\frac{3}{16} x-\frac{\pi}{4}\right), x \in(\pi, 2 \pi)$ 时, $0 \in f(x)$, 有零点, 排除 C. 故选 D.