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 ' $a_1$ ' 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
  1. $\frac{\mathrm{V}^2-\mathrm{u}^2}{\mathrm{a}_1-\mathrm{a}_2}$
  2. $\frac{\mathrm{V}^2+\mathrm{u}^2}{\mathrm{a}_1+\mathrm{a}_2}$
  3. $\frac{u^2+V^2}{a_1-a_2}$
  4. $\frac{\mathrm{u}^2-\mathrm{V}^2}{\mathrm{a}_1+\mathrm{a}_2}$

Solution

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 $x_2-x_1=\frac{u^2-v^2}{a_1+a_2}$

Asked in: MHT CET 2024 (11 May Shift 1)

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