A particle executes simple harmonic motion with a frequency $f$. The frequency with which its kinetic energy oscillates is?

Hint: Equation of position of a particle in simple harmonic motion is noted. Velocity of the particle in simple harmonic motion is calculated. Finally, the equation of kinetic energy of the particle in simple harmonic motion is derived, from which, frequency of oscillation is easily determined.

Formula used:
$1)x=A\sin ft$
where
$x$ is the position of the particle at time $t$
$A$ is the amplitude of oscillation
$f$ is the frequency of oscillation
$2)v=\dfrac{dx}{dt}$
where
$v$ is the velocity of the particle
$3)K=\dfrac{1}{2}m{{v}^{2}}$
where
$K$ is the kinetic energy of the particle
$m$ is the mass of the particle.

Complete step by step answer:
The basic idea is to derive the equation of kinetic energy of a particle undergoing simple harmonic motion. Suppose that a particle is moving in simple harmonic motion. It is given that the frequency of oscillation of the particle is $f$. Let the position of a particle at a time $t$ be $x$. The position of the particle is given by
$x=A\sin ft$
where $A$ is the amplitude of oscillation.
Now, if $v$ is the velocity of the particle, it is given by
$v=\dfrac{dx}{dt}$
where $dx$ is the change in position and $dt$is the change in time.
Let us substitute the value of $x$ in this formula. We get
$v=\dfrac{dx}{dt}=\dfrac{d(A\sin ft)}{dt}=Af\cos ft$
Now, let us derive the equation of kinetic energy. Kinetic energy is given by
$K=\dfrac{1}{2}m{{v}^{2}}$
where $m$is the mass of the particle.
Substituting the value of$v$in the above equation, we have
$K=\dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}m{{(Af\cos ft)}^{2}}=\dfrac{1}{2}m{{A}^{2}}{{f}^{2}}{{\cos }^{2}}ft$
We know that ${{\cos }^{2}}ft=\dfrac{1+\cos 2ft}{2}$
So,
$\dfrac{1}{2}m{{A}^{2}}{{f}^{2}}{{\cos }^{2}}ft=\dfrac{1}{2}m{{A}^{2}}{{f}^{2}}\left( \dfrac{1+\cos 2ft}{2} \right)=\dfrac{1}{4}m{{A}^{2}}{{f}^{2}}(1+\cos 2ft)$
Therefore, kinetic energy of a particle undergoing simple harmonic motion with frequency $f$ is given by
 $K=\dfrac{1}{4}m{{A}^{2}}{{f}^{2}}(1+\cos 2ft)=\dfrac{1}{4}m{{A}^{2}}{{f}^{2}}(1+\cos {{f}_{k}}t)$
From this equation, it is clear that the frequency of oscillation of kinetic energy is $2f$. It can be represented as
${{f}_{k}}=2f$
Therefore, the kinetic energy of the particle oscillates with a frequency double the frequency with which the particle oscillates.

Note:
It can easily be noted that velocity of the particle oscillates with the same frequency with which the particle oscillates.
$v=Af\cos ft=Af\cos {{f}_{v}}t$
where
${{f}_{v}}$ is the frequency with which the velocity oscillates. Clearly, ${{f}_{v}}=f$.