11. Given the function $f(x)=\frac{\sin (\pi x)-\cos (\pi x)+2}{\sqrt{x}}\left(\frac{1}{4} \leqslant x \leqslant \frac{5}{4}\right)$, then the minimum value of $f(x)$ is
$g(x)=\sin (\pi x)-\cos (\pi x)+2=\sqrt{2} \sin \pi\left(x-\frac{1}{4}\right)+2$ is monotonically increasing on $\left[\frac{1}{4}, \frac{3}{4}\right]$, and monotonically decreasing on $\left[\frac{3}{4}, \frac{5}{4}\right]$, and for any $x_{1} \in\left[\frac{1}{4}, \frac{3}{4}\right]$, there exists $x_{2} \in\left[\frac{3}{4}, \frac{5}{4}\right]$ such that $g\left(x_{1}\right)=g\left(x_{2}\right)$. Therefore, for any $x_{1} \in\left[\frac{1}{4}, \frac{3}{4}\right]$, we have $f\left(x_{1}\right)=\frac{g\left(x_{1}\right)}{\sqrt{x_{1}}} \geqslant \frac{g\left(x_{1}\right)}{\sqrt{x_{2}}}=f\left(x_{2}\right)$. Also, $f(x)$ is monotonically decreasing on $\left[\frac{3}{4}, \frac{5}{4}\right]$, so $f(x)_{\min }=f\left(\frac{5}{4}\right)=\frac{4 \sqrt{5}}{5}$.