A particle performs linear S.H.M. At a particular instant, velocity of the particle is ' \(u\) ' and acceleration is ' \(\alpha\) ' while at another instant,
velocity is ' \(v\) ' and acceleration is ' \(\beta\) ' \((0 \lt \alpha \lt \beta)\). The distance between the two positions is

When velocity is \(u\) and acceleration is \(\alpha\), let the position of particle be \(\mathrm{x}_1\).
When velocity is v and acceleration is \(\beta\), let the position of particle be \(x_2\).
If \(\omega\) is the angular frequency then, \(\alpha=\omega^2 x_1\)
and \(\beta=\omega^2 x_2\)
\(\therefore \quad \alpha+\beta=\omega^2\left(\mathrm{x}_1+\mathrm{x}_2\right)\)...(i)
Also, velocity of particle at particular instant can be given as,
\(\begin{aligned}
& u^2=\omega^2 A^2-\omega^2 x_1^2 \\
& \text { and } v^2=\omega^2 A^2-\omega^2 x_2^2 \\
& \text { i.e., } v^2-u^2=\omega^2\left(x_1^2-x_2^2\right) \\
& v^2-u^2=\omega^2\left(x_1-x_2\right)\left(x_1+x_2\right)...(ii)
\end{aligned}\)
from equation (i) we get
\(\begin{aligned}
& \quad v^2-u^2=\left(x_1-x_2\right)(\alpha+\beta) \\
& \therefore \quad x_1-x_2=\frac{v^2-u^2}{\alpha+\beta} \\
& \quad \text { or } x_2-x_1=\frac{u^2-v^2}{\alpha+\beta}
\end{aligned}\)