A particle performs linear S.H.M. At a particular instant, velocity of the particle is ' $u$ ' and…
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
$\frac{u^2-v^2}{\alpha+\beta}$
$\frac{u^2+v^2}{\alpha+\beta}$
$\frac{u^2-v^2}{\alpha-\beta}$
$\frac{u^2+v^2}{\alpha-\beta}$
Solution
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
$\begin{aligned}
& \quad v^2-u^2=\left(x_1-x_2\right)(\alpha+\beta) \\
& \therefore \quad x_1-x_2=\frac{v^2-u^2}{\alpha+\beta} \\
& \quad \text { or } x_2-x_1=\frac{u^2-v^2}{\alpha+\beta}
\end{aligned}$