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(A) The velocity of a particle in simple harmonic motion is given by $v = \omega \sqrt{A^2 - x^2}$.At displacement $x = \frac{2A}{3}$,the velocity is

Vedclass · Accepted Answer

$7$

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(A) The velocity of a particle in simple harmonic motion is given by $v = \omega \sqrt{A^2 - x^2}$.At displacement $x = \frac{2A}{3}$,the velocity is $v = \omega \sqrt{A^2 - (\frac{2A}{3})^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\sqrt{5}A\omega}{3}$.When the speed is increased to three times,the new velocity $v' = 3v = 3 	imes \frac{\sqrt{5}A\omega}{3} = \sqrt{5}A\omega$.Let the new amplitude be $A'$. The new velocity at the same position $x = \frac{2A}{3}$ is $v' = \omega \sqrt{(A')^2 - x^2}$.Equating the expressions: $\sqrt{5}A\omega = \omega \sqrt{(A')^2 - (\frac{2A}{3})^2}$.Squaring both sides: $5A^2 = (A')^2 - \frac{4A^2}{9}$.$(A')^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2 + 4A^2}{9} = \frac{49A^2}{9}$.Taking the square root: $A' = \frac{7A}{3}$.Comparing this with $\frac{nA}{3}$,we get $n = 7$.

$A$ particle performs simple harmonic motion with amplitude $A$. Its speed is increased to three times at an instant when its displacement is $\frac{2A}{3}$. The new amplitude of motion is $\frac{nA}{3}$. The value of $n$ is . . . . . . .

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