Q 11-13-071JEE MainJEE Main 2023 (31 Jan, Shift 1)Easy
The maximum potential energy of a block executing simple harmonic motion is $25\ \text{J}$. $A$ is amplitude of oscillation. At $\dfrac A2$, the kinetic energy of the block is
Answer: (C) $18.75\ \text{J}$
Total energy $E=25\ \text{J}$. At $x=\dfrac A2$, $U=\dfrac14E=6.25\ \text{J}$, so $K=25-6.25=18.75\ \text{J}$.
Solution by Sreeraj P, M.Sc Physics