$\begin{aligned} & \mathrm{V}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \\ & \mathrm{~V}_{\max }=\mathrm{A} \omega \\ & \therefore \sqrt{\mathrm{A}^2-\mathrm{x}^2}=\frac{\mathrm{A}}{4} \\ & \therefore 16 \mathrm{~A}^2-16 \mathrm{x}^2=\mathrm{A}^2 \\ & \therefore 16 \mathrm{x}^2=15 \mathrm{~A}^2 \\ & \therefore \mathrm{x}^2=\frac{15}{16} \mathrm{~A}^2 \\ & \therefore \mathrm{x}=\frac{\mathrm{A} \sqrt{15}}{4}\end{aligned}$