QJEEM23Phb49f: JEE Main 2023 Physics - Class 11 JEEM23Phb49f Solution: The general displacement of a simple harmonic oscillator is .…

$U = {1 \over 2}m{\omega ^2}{A^2}{\sin ^2}\omega t$

So, ${{dU} \over {dt}} = {{m{\omega ^3}{A^2}} \over 2}\sin 2\omega t$

This value will be maximum when $\sin 2\omega t = 1$

or $2\omega t = {\pi \over 2}$

$2 \times {{2\pi } \over T}t = {\pi \over 2}$

$\Rightarrow t = {T \over 8}$

So $\beta = 8$