A particle executes SHM on a straight line. At two positions, its velocities are u and v while accelerations are α and ß respectively $[\mathrm{\beta }>\mathrm{\alpha }>0]$ . The distance between these two positions is

At y₁ displacement acceleration α and velocity u
$\mathrm{\alpha }={\mathrm{\omega }}^2{\mathrm{y}}_1$......(1)    $\mathrm{u}=\mathrm{\omega }\sqrt{{\mathrm{A}}^2-{\mathrm{y}}_1^2}$........ (2)
At y₂ displacement acceleration ß and velocity v
$\begin{array}{l}\mathrm{\beta }={\mathrm{\omega }}^2{\mathrm{y}}_2\cdots \dots \dots .\left(3\right)\mathrm{v}=\mathrm{\omega }\sqrt{{\mathrm{A}}^2-{\mathrm{y}}_2^2}\\ \mathrm{eq}\left(2\right)\Rightarrow {\mathrm{u}}^2={\mathrm{\omega }}^2\left({\mathrm{A}}^2-{\mathrm{y}}_1^2\right)\\ \mathrm{eq}\left(4\right)\Rightarrow {\mathrm{v}}^2={\mathrm{\omega }}^2\left({\mathrm{A}}^2-{\mathrm{y}}_2^2\right)\\ \Rightarrow {\mathrm{u}}^2-{\mathrm{v}}^2={\mathrm{\omega }}^2\left({\mathrm{y}}_2^2-{\mathrm{y}}_1^2\right)\to \left(5\right)\\ \text{ eq }\left(3\right)+\mathrm{eq}\left(1\right)\Rightarrow \mathrm{\beta }+\mathrm{\alpha }={\mathrm{\omega }}^2\left({\mathrm{y}}_2+{\mathrm{y}}_1\right)\to \left(6\right)\\ \frac{\mathrm{eq}\left(5\right)}{\mathrm{eq}\left(6\right)}\Rightarrow \frac{{\mathrm{u}}^2-{\mathrm{v}}^2}{\mathrm{\beta }+\mathrm{\alpha }}=\frac{{\mathrm{y}}_2^2-{\mathrm{y}}_1^2}{{\mathrm{y}}_2+{\mathrm{y}}_1}={\mathrm{y}}_2-{\mathrm{y}}_1\\ \therefore {\mathrm{y}}_2-{\mathrm{y}}_1=\frac{{\mathrm{u}}^2-{\mathrm{v}}^2}{\mathrm{\beta }+\mathrm{\alpha }}\end{array}$