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
JEE MAIN 2019, Medium
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$\mathrm{PE}=\mathrm{KE}$
$\Rightarrow \quad \frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)=\frac{1}{2} m \omega^{2} x^{2}$
$\Rightarrow \quad x=\frac{A}{\sqrt{2}}$
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