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(B) The velocity $V$ of a particle in simple harmonic motion at position $x$ is given by $V = \omega \sqrt{A^2 - x^2}$.The maximum velocity is $V_{\ma

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$\frac{2\sqrt{2}}{3}A$

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(B) The velocity $V$ of a particle in simple harmonic motion at position $x$ is given by $V = \omega \sqrt{A^2 - x^2}$.The maximum velocity is $V_{\max} = \omega A$.According to the problem,$V = \frac{V_{\max}}{3} = \frac{\omega A}{3}$.Equating the two expressions: $\frac{\omega A}{3} = \omega \sqrt{A^2 - x^2}$.Dividing both sides by $\omega$: $\frac{A}{3} = \sqrt{A^2 - x^2}$.Squaring both sides: $\frac{A^2}{9} = A^2 - x^2$.Rearranging for $x^2$: $x^2 = A^2 - \frac{A^2}{9} = \frac{8A^2}{9}$.Taking the square root: $x = \sqrt{\frac{8A^2}{9}} = \frac{2\sqrt{2}}{3}A$.

The amplitude of an oscillating particle is $A$. When the velocity of the particle is one-third of its maximum velocity,determine the position of the particle.

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