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

Answer: 8

$x=A\sin (\omega t)$
$U_{(x)}=\frac{1}{2}k{x}^2$
$\frac{dU}{dt}=\frac{1}{2}k2x\frac{dx}{dt}$
$=k{A}^2\omega \sin \omega t\cos \omega t\times \frac{2}{2}$
${\left(\frac{dU}{dt}\right)}_{\max }=\frac{k{A}^2\omega }{2}(\sin 2\omega t{)}_{\max }$
$2\omega t=\frac{\pi }{2}\Rightarrow t=\frac{\pi }{4}\omega =\frac{T}{8}\Rightarrow \beta =8$