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JEE MAIN 2022
27-06-2022 S2
Question
A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is ________ s
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Solution
$\mathrm{t}=\frac{\Delta \phi}{\omega}=\frac{\pi / 2-\pi / 6}{2 \pi / 6}=\frac{\pi / 3}{\pi / 3}=1 \mathrm{sec}$
Question Tags
JEE Main
Physics
Easy
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