A particle performs simple harmonic motion with amplitude $A$. Its speed is trebled at the instant that it is at a distance $\frac{{2A}}{3}$ from equilibrium position. The new amplitude of the motion is
JEE MAIN 2016, Diffcult
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Let new amplitude is $A'$
initial velocity
$v^{2}=\omega^{2}\left(A^{2}-\left(\frac{2 A}{3}\right)^{2}\right)$ $...(1)$
Where $A$ is initial amplitude $\&\, \omega$ is angular frequency.
Final velocity
$(3 v)^{2}=\omega^{2}\left(A^{\prime 2}-\left(\frac{2 A}{3}\right)^{2}\right)$ $...(2)$
From equation $\&$ equation $( 2)$
$\frac{1}{9}=\frac{A^{2}-\frac{4 A^{2}}{9}}{A^{2}-\frac{4 A^{2}}{9}}$
$A^{\prime}=\frac{7 A}{3}$
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