Q 11-13-103JEE MainJEE Main 2021 (31 Aug, Shift 1)Easy
A particle of mass $1$ kg is hanging from a spring of force constant $100\ \text{N m}^{-1}$. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period $T$. The time when the kinetic energy and potential energy of the system will become equal, is $\dfrac Tn$. The value of $n$ is ______.
Numerical value type. Enter your answer.
Answer: 8
Released from the extreme position: $x = A\cos\omega t$.
K.E. $=$ P.E. when $\dfrac12kx^2 = \dfrac14kA^2 \Rightarrow x = \dfrac{A}{\sqrt2}$.
$\cos\omega t = \dfrac{1}{\sqrt2} \Rightarrow \omega t = \dfrac\pi4 \Rightarrow t = \dfrac{T}{8}$, so $n = 8$.
Solution by Sreeraj P, M.Sc Physics