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

$ x = A \sin (\omega t ) $
$ U _{( x )}=\frac{1}{2} kx ^2 $
$ \frac{ dU }{ dt }=\frac{1}{2} k 2 x \frac{ dx }{ dt } $
$ = kA ^2 \omega \sin \omega t \cos \omega t \times \frac{2}{2}$
$\left(\frac{ dU }{ dt }\right)_{\max }=\frac{ kA ^2 \omega}{2}(\sin 2 \omega t)_{\max } $
$ 2 \omega t=\frac{\pi}{2} \Rightarrow t=\frac{\pi}{4} \omega=\frac{T}{8} \Rightarrow \beta=8$