The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4 \mathrm{~m}, 2 \mathrm{~ms}^{-1}$ and $16 \mathrm{~ms}^{-2}$ at a certain instant. The amplitude of the motion is $\sqrt{\mathrm{x}} \mathrm{m}$ where $\mathrm{x}$ is. . . . . . .
JEE MAIN 2024, Diffcult
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$x=4 \mathrm{~m}, \mathrm{~V}=2 \mathrm{~m} / \mathrm{s}, \mathrm{a}=16 \mathrm{~m} / \mathrm{s}^2$
$|\mathrm{a}|=\omega^2 \mathrm{x}$
$\Rightarrow 16=\omega^2(4)$
$\omega=2 \mathrm{rad} / \mathrm{s}$
$\mathrm{v}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2}$
$A=\sqrt{\frac{\mathrm{v}^2}{\omega^2}+\mathrm{x}^2} \Rightarrow \mathrm{A}=\sqrt{\frac{4}{4}+16}$
$A=\sqrt{17} \mathrm{~m}$
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