A particle performing SHM has time period $\frac{2\pi }{\sqrt{3}}$ and path length 4cm. The distance from mean position (in cm) at acceleration is equal to velocity is

Velocity $\mathrm{v}=\mathrm{\omega }\sqrt{{\mathrm{A}}^2-{\mathrm{x}}^2}$ And acceleration  $={\omega }^{2}x$

Given,

 $$\begin{gathered} \mathrm{\omega }\sqrt{{\mathrm{A}}^2-{\mathrm{x}}^2}={\mathrm{\omega }}^2\mathrm{x} \mathrm{or} \sqrt{{\mathrm{A}}^2-{\mathrm{x}}^2}=\mathrm{\omega x} \\ \mathrm{Given}, \mathrm{T}=\frac{2\mathrm{\pi }}{\sqrt{3}}\mathrm{And} \mathrm{\omega }=\frac{2\mathrm{\pi }}{\mathrm{T}}=\sqrt{3} \\ \mathrm{Substituting} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{\omega } \mathrm{in} \mathrm{Eq}. \left(\mathrm{i}\right), \mathrm{we} \mathrm{get} \\ \sqrt{{\mathrm{A}}^2-{\mathrm{x}}^2}=\sqrt{3}\mathrm{x} \end{gathered}$$

$\Rightarrow \mathrm{A}=2\mathrm{x}$

$\mathrm{As} \mathrm{amplitude} =\frac{\text{ path length }}{2}=\frac{4}{2}=2\mathrm{cm}$

$\Rightarrow \mathrm{x}=1\mathrm{cm}$