Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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

$3n$

$4n$

$n$

$2n$

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).
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.
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 $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$ : $U = \frac{1}{2}kA^2 \left[ \frac{1 - \cos(2\omega t)}{2} \right] = \frac{1}{4}kA^2 - \frac{1}{4}kA^2 \cos(2\omega t)$
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 $2n$ .
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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