A particle executes SHM of amplitude A. The d | vedclass

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(C) The kinetic energy $(KE)$ of a particle in $SHM$ at a distance $x$ from the mean position is given by $KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$.Th

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$\frac{1}{\sqrt{2}} A$

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(C) The kinetic energy $(KE)$ of a particle in $SHM$ at a distance $x$ from the mean position is given by $KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$.The potential energy $(PE)$ of the particle at the same position is given by $PE = \frac{1}{2} m \omega^2 x^2$.According to the problem,$KE = PE$:$\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 x^2$.Canceling the common terms $\frac{1}{2} m \omega^2$ from both sides,we get:$A^2 - x^2 = x^2$.Rearranging the terms,we get $A^2 = 2x^2$.Solving for $x$,we find $x^2 = \frac{A^2}{2}$,which implies $x = \pm \frac{A}{\sqrt{2}}$.

$A$ particle executes $SHM$ of amplitude $A$. The distance from the mean position when its kinetic energy becomes equal to its potential energy is:

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