Q 11-13-076JEE MainJEE Main 2023 (13 Apr, Shift 2)Easy
A particle executes SHM of amplitude $A$. The distance from the mean position when its kinetic energy becomes equal to its potential energy is
Answer: (A) $\dfrac1{\sqrt2}A$
$\tfrac12kx^2=\tfrac12\times\tfrac12kA^2\Rightarrow x=\dfrac{A}{\sqrt2}$.
Solution by Sreeraj P, M.Sc Physics