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JEE PHYSICS MCQ: OSCILLATIONS

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The ratio of K.E. of the particle executing S.H.M. at mean position to the K.E. at point whose distance is half of amplitude is:
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Correct option: C
43\frac{4}{3}
The correct answer is C (43\frac{4}{3}) because the kinetic energy at the mean position is maximum, and at half the amplitude, it's 34\frac{3}{4} of the maximum, resulting in a ratio of 43\frac{4}{3}.

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