A particle performs linear S.H.M. at a particular instant, velocity of the particle is ' \(u\) ' and acceleration is ' \(a_1\) ' while at another instant velocity is ' \(V\) ' and acceleration is \(' \mathrm{a}_2\) ' \(\left(\mathrm{o} \lt \mathrm{a}_1 \lt \mathrm{a}_2\right)\). The distance between the two position is

When velocity is \(u\) and acceleration is \(a_1\), let the position of particle be \(x_1\).
When velocity is v and acceleration is \(\mathrm{a}_2\), let the position of particle be \(x_2\).
If \(\omega\) is the angular frequency then, \(\mathrm{a}_1=\omega^2 \mathrm{x}_1\)
and \(\mathrm{a}_2=\omega^2 \mathrm{x}_2\)
\(\therefore \quad \mathrm{a}_1+\mathrm{a}_2=\omega^2\left(\mathrm{x}_1+\mathrm{x}_2\right)...(i)\)
Also, velocity of particle at particular instant can be given as,
\(u^2=\omega^2 A^2-\omega^2 x_1^2\)
and \(v^2=\omega^2 A^2-\omega^2 x_2^2\)
i.e., \(v^2-u^2=\omega^2\left(x_1^2-x_2^2\right)\)
\(v^2-u^2=\omega^2\left(x_1-x_2\right)\left(x_1+x_2\right)...(ii)\)
from equation (i) we get
\(\begin{aligned}
& \mathrm{v}^2-\mathrm{u}^2 \\
= & \left(\mathrm{x}_1-\mathrm{x}_2\right)\left(\mathrm{a}_1+\mathrm{a}_2\right) \\
\therefore \quad & \mathrm{x}_1-\mathrm{x}_2=\frac{\mathrm{v}^2-\mathrm{u}^2}{\mathrm{a}_1+\mathrm{a}_2}
\end{aligned}\)
or \(x_2-x_1=\frac{u^2-v^2}{a_1+a_2}\)