A particle is performing simple harmonic motion with frequency $f$. What is the frequency $f^{\prime }$ of its kinetic energy oscillation?

Solution

1. The displacement of a particle in simple harmonic motion is given by $x(t)=A\sin (\omega t)$, where the angular frequency is $\omega =2\pi f$.
2. The velocity is the derivative of displacement, $v(t)=A\omega \cos (\omega t)$, so the kinetic energy $K$ is proportional to $v^2$, or ${\cos }^2(\omega t)$.
3. Using the trigonometric identity ${\cos }^2(\theta )=\frac{1+\cos (2\theta )}{2}$, the kinetic energy expression involves a term with frequency $2\omega$.
4. Since the angular frequency doubles to $2\omega$, the oscillation frequency $f^{\prime }$ becomes $2f$.
5. Hence the answer is (C).