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JEE MAIN 2024
27-01-2024 SHIFT-1
Question
A particle executes simple harmonic motion with an amplitude of 4 cm . At the mean position, velocity of the particle is $10 \mathrm{~cm} / \mathrm{s}$. The distance of the particle from the mean position when its speed becomes $5 \mathrm{~cm} / \mathrm{s}$ is $\sqrt{\alpha} \mathrm{cm}$, where $\alpha=$
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