Given here, displacement $x=A \sin \left(\omega t\right)$

Now, potential energy of a simple harmonic oscillator is $U=\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}m{\omega }^2{A}^2{\sin }^2\left(\omega t\right)$,

So, slope of potential energy and time curve is $\frac{\mathrm{d}U}{\mathrm{d}t}=\frac{1}{2}\frac{m{\omega }^3{A}^2}{2}\sin 2\omega t$

Now, ${\left(\frac{dU}{dt}\right)}_{max}$ will be maximum when $\sin 2\omega t=1$

 $$\begin{gathered} \Rightarrow 2\omega t=\frac{\pi }{2} \\ \Rightarrow 2\left(\frac{2\pi }{T}\right)t=\frac{\pi }{2} \\ \Rightarrow t=\frac{T}{8} \\ \Rightarrow \beta =8 \end{gathered}$$