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(C) The velocity $v$ of a particle in simple harmonic motion at displacement $y$ is given by $v = \omega \sqrt{A^2 - y^2}$.The maximum velocity is $v_

Vedclass · Accepted Answer

$A\sqrt{3}/2$

Vedclass · Answer

(C) The velocity $v$ of a particle in simple harmonic motion at displacement $y$ is given by $v = \omega \sqrt{A^2 - y^2}$.The maximum velocity is $v_{	ext{max}} = A\omega$.Given that the velocity is half of the maximum velocity,we have $v = \frac{A\omega}{2}$.Substituting this into the velocity equation:$\frac{A\omega}{2} = \omega \sqrt{A^2 - y^2}$Squaring both sides:$\frac{A^2}{4} = A^2 - y^2$Rearranging for $y^2$:$y^2 = A^2 - \frac{A^2}{4} = \frac{3A^2}{4}$Taking the square root:$y = \frac{\sqrt{3}A}{2}$.

For a simple harmonic motion with amplitude $A$ and time period $T$,at what position is the velocity half of the maximum velocity?

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