Graph of $v^2 Vs x^2$ for a particle executing SHM is

$\begin{array}{l}\mathrm{v}=\omega \sqrt{{\mathrm{A}}^2-{\mathrm{x}}^2}\\ \Rightarrow {\mathrm{v}}^2={\omega }^2{\mathrm{A}}^2-{\omega }^2{\mathrm{x}}^2\\ \text{ Let }{\mathrm{v}}^2=\mathrm{Y}\text{ and }{\mathrm{x}}^2=\mathrm{X}\\ \text{ So, }Y={\omega }^2{A}^2-{\omega }^{2}X\end{array}$

This is of the form Y = mX + c  (equation of a straight line), where  $c={\omega }^2{A}^2$ and  slope, $m=-{\omega }^2$ .