Question

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 $t=\frac{T}{\beta}$ The value of $\beta$ is

Answer 1

The correct answer is 8
x=Asin(ωt)
U(x)​=21​kx2
dtdU​=21​k2xdtdx​
=kA2ωsinωtcosωt×22​
(dtdU​)max​=2kA2ω​(sin2ωt)max​
2ωt=2π​⇒t=4π​ω=8T​⇒β=8