A body is executing simple harmonic motion with frequency ‘n’, the frequency of its potential energy is :

The correct option is 2n.

Underlying Concept:

To find the frequency of the potential energy of a body executing simple harmonic motion (SHM), we can write the expression for its displacement and use it to derive the potential energy as a function of time.

Let the displacement $x$ of a body executing simple harmonic motion as a function of time $t$ be given by:

$x=A\sin \left(\omega t\right)$

where:
• $A$ is the amplitude of the motion.
• $\omega$ is the angular frequency, which is related to the frequency $n$ by $\omega =2\pi n$.

The potential energy ($U$) of a particle executing simple harmonic motion at any displacement $x$ is given by:

$U=\frac{1}{2}k{x}^2$

where $k$ is the force constant of the restoring force ($k=m{\omega }^2$).

Substituting the expression for displacement $x$ into the potential energy formula:

$U=\frac{1}{2}k{A}^2{\sin }^2\left(\omega t\right)$

Using the trigonometric identity ${\sin }^2\left(\theta \right)=\frac{1-\cos \left(2\theta \right)}{2}$, we can rewrite the potential energy expression as:

$U=\frac{1}{2}k{A}^2\left(\frac{1-\cos \left(2\omega t\right)}{2}\right)$

$U=\frac{1}{4}k{A}^2(1-\cos (2\omega t\left)\right)$

From the above equation, the potential energy oscillates with a new angular frequency, say ${\omega }^{\prime }$, given by:

${\omega }^{\prime }=2\omega$

Since the relation between angular frequency and frequency is linear ($\omega =2\pi n$), the frequency of the potential energy $n^{\prime }$ is:

$2\pi n^{\prime }=2\left(2\pi n\right)$

$n^{\prime }=2n$

Conclusion:

While the displacement completes one full cycle, the potential energy attains its maximum value twice (once at each extreme position). Hence, the frequency of the potential energy is twice that of the displacement, which is 2n.