A particle executes a linear S.H.M. In two of | vedclass

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(C) For a particle in $S.H.M.$,the velocity $V$ at displacement $x$ is given by $V^2 = \omega^2(A^2 - x^2)$,and acceleration $a$ is given by $a = -\om

Vedclass · Accepted Answer

$\frac{V_1^2-V_2^2}{a_1+a_2}$

Vedclass · Answer

(C) For a particle in $S.H.M.$,the velocity $V$ at displacement $x$ is given by $V^2 = \omega^2(A^2 - x^2)$,and acceleration $a$ is given by $a = -\omega^2 x$. From the acceleration equation,we have $x = -a/\omega^2$. Substituting this into the velocity equation: $V^2 = \omega^2 A^2 - \omega^2 x^2 = \omega^2 A^2 - \omega^2 (-a/\omega^2)^2 = \omega^2 A^2 - a^2/\omega^2$. For two positions with velocities $V_1, V_2$ and accelerations $a_1, a_2$: $V_1^2 = \omega^2 A^2 - a_1^2/\omega^2$ $V_2^2 = \omega^2 A^2 - a_2^2/\omega^2$ Subtracting the two equations: $V_1^2 - V_2^2 = (a_2^2 - a_1^2)/\omega^2$. Since $a = -\omega^2 x$,the distance between two positions $x_1$ and $x_2$ is $d = |x_1 - x_2| = |(-a_1/\omega^2) - (-a_2/\omega^2)| = |a_2 - a_1|/\omega^2$. From the subtraction result,$1/\omega^2 = (V_1^2 - V_2^2) / (a_2^2 - a_1^2)$. Substituting this into the distance formula: $d = |a_2 - a_1| \cdot \frac{V_1^2 - V_2^2}{a_2^2 - a_1^2} = \frac{V_1^2 - V_2^2}{a_2 + a_1}$ (since $a_2 > a_1$). Thus,the correct option is $C$.

$A$ particle executes a linear $S.H.M.$ In two of its positions,the velocities are $V_1$ and $V_2$,and the accelerations are $a_1$ and $a_2$ respectively $(0 < a_1 < a_2)$. The distance between the positions is

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