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JEE PHYSICS MCQ: OSCILLATIONS

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Maximum speed of a particle in simple harmonic motion is vmax⁡.v_{\max }. Then average speed of a particle in SHM for complete cycle is equal to
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Correct option: D
2vmax π\frac{2 v_{\text {max }}}{\pi}
Option D is correct because the average speed for a complete cycle in SHM is calculated as total distance traveled divided by the time period, which results in 2vmax π\frac{2 v_{\text {max }}}{\pi}.

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