Physics - SHM Question with Solution | TestHub

The velocity of a particle executing S.H.M. at a displacement $x$ from the mean position is given by $v=\omega \sqrt{A^2-x^2}$.

The maximum velocity of the particle is $v_{\text{max}}=A\omega$.

Given that $v=\frac{1}{3}{v}_{\text{max}}$, we have:

$\omega \sqrt{A^2-x^2}=\frac{1}{3}A\omega$

Squaring both sides:

$A^2-x^2=\frac{A^2}{9}$

$x^2=A^2-\frac{A^2}{9}=\frac{8A^2}{9}$

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

Answer: $\frac{2\sqrt{2}}{3}\mathrm{A}$