A particle performs linear S.H.M. At a partic | vedclass

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(A) Let the positions of the particle be $x_1$ and $x_2$ at the two instants.In $S.H.M.$,acceleration $a = -\omega^2 x$. Considering magnitudes,$\alph

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$\frac{u^2 - v^2}{\alpha + \beta}$

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(A) Let the positions of the particle be $x_1$ and $x_2$ at the two instants.In $S.H.M.$,acceleration $a = -\omega^2 x$. Considering magnitudes,$\alpha = \omega^2 |x_1|$ and $\beta = \omega^2 |x_2|$.Thus,$|x_1| = \frac{\alpha}{\omega^2}$ and $|x_2| = \frac{\beta}{\omega^2}$.The velocity in $S.H.M.$ is given by $v^2 = \omega^2(A^2 - x^2)$.For the first instant: $u^2 = \omega^2 A^2 - \omega^2 x_1^2$ . . . $(i)$For the second instant: $v^2 = \omega^2 A^2 - \omega^2 x_2^2$ . . . $(ii)$Subtracting $(ii)$ from $(i)$: $u^2 - v^2 = \omega^2(x_2^2 - x_1^2) = \omega^2(x_2 - x_1)(x_2 + x_1)$.Since $\alpha = \omega^2 x_1$ and $\beta = \omega^2 x_2$,we have $\alpha + \beta = \omega^2(x_1 + x_2)$.Substituting this into the equation: $u^2 - v^2 = \omega^2(x_2 - x_1) \cdot \frac{\alpha + \beta}{\omega^2}$.Therefore,the distance between the two positions is $|x_2 - x_1| = \frac{u^2 - v^2}{\alpha + \beta}$.

$A$ 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 $\beta$ $(0 < \alpha < \beta)$. The distance between the two positions is

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