A particle performs S.H.M. with amplitude A. | vedclass

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(B) The velocity of a particle in $S.H.M.$ at displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$.At $x = \frac{2A}{3}$,the velocity is $v = \o

Vedclass · Accepted Answer

$\frac{7A}{3}$

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(B) The velocity of a particle in $S.H.M.$ at displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$.At $x = \frac{2A}{3}$,the velocity is $v = \omega \sqrt{A^2 - (\frac{2A}{3})^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\sqrt{5}}{3} \omega A$.When the speed is tripled,the new velocity $v' = 3v = 3 	imes \frac{\sqrt{5}}{3} \omega A = \sqrt{5} \omega A$.The total energy of the $S.H.M.$ remains constant at any point,so $E' = K' + U$.The new total energy is $E' = \frac{1}{2} m \omega^2 A'^2$.The potential energy at $x = \frac{2A}{3}$ is $U = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 (\frac{2A}{3})^2 = \frac{1}{2} m \omega^2 \frac{4A^2}{9}$.The new kinetic energy is $K' = \frac{1}{2} m v'^2 = \frac{1}{2} m (\sqrt{5} \omega A)^2 = \frac{1}{2} m \omega^2 (5A^2)$.Equating the total energy: $\frac{1}{2} m \omega^2 A'^2 = \frac{1}{2} m \omega^2 (\frac{4A^2}{9}) + \frac{1}{2} m \omega^2 (5A^2)$.$A'^2 = \frac{4A^2}{9} + 5A^2 = A^2 (\frac{4+45}{9}) = \frac{49A^2}{9}$.Taking the square root,$A' = \frac{7A}{3}$.

$A$ particle performs $S.H.M.$ with amplitude $A$. Its speed is tripled at the instant when it is at a distance of $\frac{2A}{3}$ from the mean position. The new amplitude of the motion is

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