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A particle performs simple harmonic motion with amplitude A. Its speed is tripled by supplying energy at the instant when it is a distance $\frac{2 A}{3}$ from equilibrium position. The new amplitude of the motion is $\frac{x A}{3}$ . The value of x is

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answer is 7.

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Detailed Solution

We know that $V = ω \sqrt{A^{2} - x^{2}}$

Initially $V = ω \sqrt{A^{2} - {(\frac{2 A}{3})}^{2}} - (1)$

Finally $3 V = ω \sqrt{A'^{2} - {(\frac{2 A}{3})}^{2}} - - (1)$

$\frac{(2)}{(1)} \Rightarrow \frac{3}{1} = \frac{\sqrt{A'^{2} - {(\frac{2 A}{3})}^{2}}}{\sqrt{A^{2} - {(\frac{2 A}{3})}^{2}}}$

$9 (A^{2} - \frac{4 A^{2}}{9}) = (A^{' 2} - \frac{4 A^{2}}{9})$

$∴ A^{'} = \frac{7 A}{3}$

$x = 7$

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