Given here, displacement x=Asin(ωt)
Now, potential energy of a simple harmonic oscillator is U=21mω2x2=21mω2A2sin2(ωt),
So, slope of potential energy and time curve is dtdU=212mω3A2sin2ωt
Now, (dtdU)max will be maximum when sin2ωt=1
⇒2ωt=2π⇒2(T2π)t=2π⇒t=8T⇒β=8