A particle executing SHM has velocities $u$ and $v$  and accelerations  $a$ and $b$  in two of its positions. Find the distance between these two positions.   

Let x1&x₂ be the distances of the two positions from centre
Then with usual notations
$\begin{array}{r}{u}^2={\omega }^2\left(A^2-x_1^2\right)---1\\ v^2={\omega }^2\left(A^2-x_2^2\right)---2\\ a={\omega }^2{x}_1---3\\ b={\omega }^2{x}_2---4\end{array}$

Subtracting equation 2 from equation 1 we get

$u^2-v^2={\omega }^2\left(x_2^2-x_1^2\right)---5$

Adding equations 3 and 4 we get $a+b={\omega }^2\left(x_1+x_2\right)--6$

Dividing equation 5 by equation 6

$\frac{u^2-v^2}{a+b}=x_2-x_1$