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(D) Let the positions of the particle be $x_1$ and $x_2$ at the two instants.The acceleration in $SHM$ is given by $a = -\omega^2 x$. Considering magn

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$\frac{u^2 - V^2}{a_1 + a_2}$

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(D) Let the positions of the particle be $x_1$ and $x_2$ at the two instants.The acceleration in $SHM$ is given by $a = -\omega^2 x$. Considering magnitudes,$a_1 = \omega^2 |x_1|$ and $a_2 = \omega^2 |x_2|$.The velocity in $SHM$ 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 $(i)$: $u^2 - V^2 = \omega^2 (x_2^2 - x_1^2) = \omega^2 (x_2 - x_1)(x_2 + x_1)$ . . . $(iii)$From the acceleration equations: $a_2 - a_1 = \omega^2 (x_2 - x_1)$ (assuming $x_2 > x_1$).However,the distance between positions is $|x_2 - x_1|$.Using $a_1 = \omega^2 x_1$ and $a_2 = \omega^2 x_2$,we have $x_1 = a_1/\omega^2$ and $x_2 = a_2/\omega^2$.Substituting into the velocity difference: $u^2 - V^2 = \omega^2 (x_2^2 - x_1^2) = \omega^2 (a_2^2/\omega^4 - a_1^2/\omega^4) = \frac{1}{\omega^2} (a_2^2 - a_1^2)$.This implies $\omega^2 = \frac{a_2^2 - a_1^2}{u^2 - V^2}$.The distance $d = |x_2 - x_1| = |\frac{a_2 - a_1}{\omega^2}| = |\frac{a_2 - a_1}{(a_2^2 - a_1^2)/(u^2 - V^2)}| = |\frac{u^2 - V^2}{a_2 + a_1}|$.Thus,the distance is $\frac{u^2 - V^2}{a_1 + a_2}$.

$A$ particle performs linear $SHM$. At a particular instant,the velocity of the particle is $u$ and the acceleration is $a_1$. At another instant,the velocity is $V$ and the acceleration is $a_2$ $(0 < a_1 < a_2)$. The distance between the two positions is:

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