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(A) In linear simple harmonic motion $(SHM)$,the velocity $v$ of a particle at displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$.The magnitud

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$\frac{2 \pi}{\sqrt{3}} \ s$

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(A) In linear simple harmonic motion $(SHM)$,the velocity $v$ of a particle at displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$.The magnitude of acceleration $a$ is given by $a = \omega^2 x$.Given that the particle is midway between the mean position $(x=0)$ and the extreme position $(x=A)$,we have $x = \frac{A}{2}$.According to the problem,the magnitudes of velocity and acceleration are equal,so $v = a$.Substituting the expressions: $\omega \sqrt{A^2 - x^2} = \omega^2 x$.Substituting $x = \frac{A}{2}$: $\omega \sqrt{A^2 - (\frac{A}{2})^2} = \omega^2 (\frac{A}{2})$.$\omega \sqrt{A^2 - \frac{A^2}{4}} = \omega^2 \frac{A}{2} \Rightarrow \omega \sqrt{\frac{3A^2}{4}} = \omega^2 \frac{A}{2}$.$\omega \frac{\sqrt{3}A}{2} = \omega^2 \frac{A}{2}$.Dividing both sides by $\frac{\omega A}{2}$,we get $\omega = \sqrt{3}$.Since the periodic time $T = \frac{2\pi}{\omega}$,we have $T = \frac{2\pi}{\sqrt{3}} \ s$.

$A$ particle is performing a linear simple harmonic motion of amplitude $A$. When it is midway between its mean and extreme position,the magnitudes of its velocity and acceleration are equal. What is the periodic time of the motion?

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