A particle executes simple harmonic motion with an amplitude of $4 \mathrm{~cm}$. 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 $\alpha=$____________.
JEE MAIN 2024, Diffcult
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$ \mathrm{V}_{\mathrm{at} \text { mann position }}=\mathrm{A} \omega \Rightarrow 10=4 \omega $
$ \omega=\frac{5}{2} $
$ \mathrm{~V}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} $
$ 5=\frac{5}{2} \sqrt{4^2-\mathrm{x}^2} \Rightarrow \mathrm{x}^2=16-4 $
$ \mathrm{x}=\sqrt{12} \mathrm{~cm}$
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