A particle undergoes simple harmonic motion at a frequency . The...

Let the displacement of the particle in simple harmonic motion be $x = A \sin(2\pi \nu t + \phi)$.

The velocity of the particle is $v = \dfrac{dx}{dt} = A(2\pi \nu) \cos(2\pi \nu t + \phi)$.

The kinetic energy is given by $K = \dfrac{1}{2}mv^2 = \dfrac{1}{2}mA^2(2\pi \nu)^2 \cos^2(2\pi \nu t + \phi)$.

Using the trigonometric identity $\cos^2 \theta = \dfrac{1 + \cos(2\theta)}{2}$, the kinetic energy can be written as:

$K = \dfrac{1}{4}mA^2(2\pi \nu)^2 [1 + \cos(4\pi \nu t + 2\phi)]$

The time-varying part of the kinetic energy is $\cos(2\pi(2\nu)t + 2\phi)$.

This shows that the kinetic energy oscillates with a frequency of $2\nu$.