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A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance

$\frac{2}{3}$ A from the equilibrium position. The new amplitude of the motion is

$\frac{A}{3}\sqrt{41}$

B) 3A

$A\sqrt{3}$

$\frac{7}{3}A$

Option D

$v= \omega \sqrt{A^{2}-x^{2}}$

At $x=\frac{2A}{3}$

$v=\omega\sqrt{A^{2}-(\frac{2A^{}}{3})^{2}}=\frac{\sqrt{5}}{3}\omega A$

As, velocity is trebled , hence v'= $\sqrt{5}A\omega$

This leads to new amplitude A'

$\omega\sqrt{A'^{2}-(\frac{2A}{3})^{2}}=\sqrt{5}A\omega$

$\Rightarrow \omega^{2}[A'^{2}-\frac{4A^{2}}{9}]=5A^{2}\omega^{2}$

$\Rightarrow A'^{2}= 5A^{2}+\frac{4}{9}A^{2}=\frac{49}{9}A^{2}$

$A'=\frac{7}{3}A$

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