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(B) Let the initial amplitude be $A$ and the new amplitude be $A'$. The velocity $v$ of a particle in simple harmonic motion at a distance $x$ from th

Vedclass · Accepted Answer

$\frac{7A}{3}$

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(B) Let the initial amplitude be $A$ and the new amplitude be $A'$. The velocity $v$ of a particle in simple harmonic motion at a distance $x$ from the equilibrium position is given by $v = \omega \sqrt{A^2 - x^2}$.At $x = \frac{2A}{3}$,the initial velocity $v$ is $v = \omega \sqrt{A^2 - (\frac{2A}{3})^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\omega A \sqrt{5}}{3}$.The new speed is $v' = 3v = \omega A \sqrt{5}$.Using the formula for the new amplitude $A'$ at the same position $x = \frac{2A}{3}$:$(v')^2 = \omega^2 (A'^2 - x^2)$$(\omega A \sqrt{5})^2 = \omega^2 (A'^2 - (\frac{2A}{3})^2)$$5A^2 = A'^2 - \frac{4A^2}{9}$$A'^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2 + 4A^2}{9} = \frac{49A^2}{9}$$A' = \sqrt{\frac{49A^2}{9}} = \frac{7A}{3}$.

$A$ 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 the equilibrium position. The new amplitude of the motion is

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