Physics - SHM Question with Solution | TestHub

$x=A\sin \omega t$

$v=\frac{dx}{dt}=a \omega \cos \omega t$

$K.E.=\frac{1}{2}m v^2$

$=\frac{1}{2} m A^2{\omega }^2{\cos }^2\mathrm{ }\omega t$

$=\frac{1}{2}\mathrm{ }\mathrm{m}\mathrm{ }{\mathrm{A}}^2{\mathrm{\omega }}^2{\cos }^2\frac{2\mathrm{\pi }}{\mathrm{T}}\mathrm{ }\mathrm{t}$

At$t=0$,$K.E.$is maximum.

The period of$K.E.$graph is half of the period of displacement graph.

Comment:- The points on the graph of$x-$axis should be differentiable.

In all option mentioned, the graph has cusps at the points on$x-$axis.