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JEE PHYSICS MCQ: OSCILLATIONS

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A simple pendulum is oscillating with amplitude AA and angular frequency ω\omega. At displacement xx from mean position, the ratio of kinetic energy to potential energy is:
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Correct option: C
A2−x2x2\frac{A^2-x^2}{x^2}
Option C (A2−x2x2\frac{{A^2-x^2}}{{x^2}}) is correct because it represents the ratio of kinetic energy to potential energy in simple harmonic motion at a given displacement x.

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