Q 11-13-068JEE MainJEE Main 2023 (30 Jan, Shift 1)Medium
The general displacement of a simple harmonic oscillator is $x=A\sin\omega t$. Let $T$ be its time period. The slope of its potential energy ($U$)–time ($t$) curve will be maximum when $t=\dfrac{T}{\beta}$. The value of $\beta$ is ______.
Numerical value type. Enter your answer.
Answer: 8
$U=\tfrac12kA^2\sin^2\omega t$, so $\dfrac{dU}{dt}=\tfrac12kA^2\omega\sin2\omega t$.
This is maximum when $2\omega t=\dfrac\pi2$, i.e. $t=\dfrac{\pi}{4\omega}=\dfrac{T}{8}$. So $\beta=8$.
Solution by Sreeraj P, M.Sc Physics