Q 11-13-129JEE MainJEE Main 2019 (9 Jan, Shift 2)Easy
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:
Answer: (D) $\dfrac{A}{\sqrt2}$
Total energy $E = \tfrac12kA^2$ and $PE = \tfrac12kx^2$. When $PE = KE$, each is $E/2$:
$$\tfrac12kx^2 = \tfrac14kA^2 \;\Rightarrow\; x = \pm\frac{A}{\sqrt2}$$
Solution by Sreeraj P, M.Sc Physics