(C) The potential energy $(PE)$ of a particle in $SHM$ is given by $PE = \frac{1}{2} m \omega^2 x^2$.The kinetic energy $(KE)$ of a particle in $SHM$

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(C) The potential energy $(PE)$ of a particle in $SHM$ is given by $PE = \frac{1}{2} m \omega^2 x^2$.The kinetic energy $(KE)$ of a particle in $SHM$ is given by $KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$.Given that $PE = KE$,we have:$\frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 (A^2 - x^2)$$x^2 = A^2 - x^2$$2x^2 = A^2$$x^2 = \frac{A^2}{2}$$x = \pm \frac{A}{\sqrt{2}}$Thus,the position of the particle is $\frac{A}{\sqrt{2}}$.

Class 11 Physics Oscillations Energy of Simple Harmonic Motion

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$A$ particle is executing simple harmonic motion $(SHM)$ of amplitude $A$ along the $x$-axis about $x = 0$. When its potential energy $(PE)$ equals kinetic energy $(KE)$,the position of the particle will be

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A
$\frac{A}{2}$
B
$\frac{A}{2\sqrt{2}}$
C
$\frac{A}{\sqrt{2}}$
D
$A$

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$A$ particle is executing simple harmonic motion $(SHM)$ of amplitude $A$ along the $x$-axis about $x = 0$. When its potential energy $(PE)$ equals kinetic energy $(KE)$,the position of the particle will be

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