A particle executes SHM with a frequency f, The frequency with which its KE oscillates is:

$y=A\sin \omega t$

$\frac{dy}{dt}=A\omega \cos \omega t$

$KE=\frac{1}{2}m{v}^2$

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

$=\frac{m{A}^2{\omega }^2(1+\cos 2\omega t)}{4}$

$=\frac{m{A}^2{\omega }^2}{4}+\frac{m{A}^2{\omega }^2\cos 2\omega t}{4}$

During one complete oscillation, the kinetic energy of the particle executing SHM will be twice. KE will oscillate with frequency double the frequency of SHM, Which is 2f