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 $.........$
JEE MAIN 2023, Diffcult
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$x = A \sin (\omega t )$
$U _{( x )}=\frac{1}{2} kx ^2$
$\frac{ dU }{ dt }=\frac{1}{2} k 2 x \frac{ dx }{ dt }$
$= kA ^2 \omega \sin \omega t\,\cos \omega t \times \frac{2}{2}$
$\left(\frac{ dU }{ dt }\right)_{\max }=\frac{ kA ^2 \omega}{2}(\sin 2 \omega t )_{\max }$
$2 \omega t =\frac{\pi}{2} \Rightarrow t =\frac{\pi}{4} \omega=\frac{ T }{8} \Rightarrow \beta=8$
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