If a body is executing simple harmonic motion with frequency $n$, then the frequency of its potential energy i

Question

If a body is executing simple harmonic motion with frequency $n$, then the frequency of its potential energy is:

Accepted Answer

1.  **Displacement Equation:** For a body executing Simple Harmonic Motion (SHM) with frequency $n$, the angular frequency is $\omega = 2\pi n$. The displacement $x$ varies with time $t$ as $x = A \sin(\omega t)$ (or a cosine function).
2.  **Potential Energy Formula:** The potential energy ($U$) of the oscillator is given by $U = \frac{1}{2}kx^2$, where $k$ is the force constant.
3.  **Substitution and Trigonometric Identity:** Substituting the displacement equation:
    $$ U = \frac{1}{2}k (A \sin(\omega t))^2 = \frac{1}{2}kA^2 \sin^2(\omega t) $$
    Using the trigonometric identity $\sin^2 	heta = \frac{1 - \cos(2	heta)}{2}$:
    $$ U = \frac{1}{2}kA^2 \left[ \frac{1 - \cos(2\omega t)}{2} ight] = \frac{1}{4}kA^2 - \frac{1}{4}kA^2 \cos(2\omega t) $$
4.  **Frequency Analysis:** The term $\cos(2\omega t)$ indicates that the potential energy oscillates with an angular frequency of $2\omega$. Since frequency is proportional to angular frequency ($f = \frac{\Omega}{2\pi}$), the frequency of the potential energy is $2n$. 
5.  **Conclusion:** Both kinetic and potential energies in SHM vary periodically with double the frequency of the displacement ($2n$), although the total energy remains constant.

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