A particle executes SHM of amplitude A. The distance from the mean position when its's kinetic energy becomes equal to its potential energy is on
JEE MAIN 2023, Medium
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$KE = PE$
$\frac{1}{2} M \omega^2\left( A ^2- x ^2\right)=\frac{1}{2} M \omega^2 x ^2$
$A ^2- x ^2= x ^2 \Rightarrow A ^2=2 \times 2$
$\Rightarrow x = \pm \frac{ A }{\sqrt{2}}$
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