A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is $\frac{2 \mathrm{~A}}{3}$. The new amplitude of motion is $\frac{\mathrm{nA}}{3}$. The value of $\mathrm{n}$ is____.
JEE MAIN 2024, Diffcult
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$v=\omega \sqrt{A^2-x^2}$
$\text { at } x=\frac{2 A}{3}$
$v=\omega \sqrt{A^2-\left(\frac{2 A}{3}\right)^2}=\frac{\sqrt{5} A \omega}{3}$
New amplitude $=\mathrm{A}^{\prime}$
$\mathrm{v}^{\prime}=3 \mathrm{v}=\sqrt{5} \mathrm{~A} \omega=\omega \sqrt{\left(\mathrm{A}^{\prime}\right)^2-\left(\frac{2 \mathrm{~A}}{3}\right)^2}$
$\mathrm{A}^{\prime}=\frac{7 \mathrm{~A}}{3}$
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