To determine the new amplitude \( A' \) when the velocity of a particle in simple harmonic motion triples, we follow these steps:The velocity \( V \) of a particle in simple harmonic motion is defined as:\[V = \omega \sqrt{A^2 - x^2},\]where \( \omega \) is the angular frequency, \( A \) is the amplitude, and \( x \) is the displacement from equilibrium.Step 1: Calculate initial velocity at \( x = \frac{2A}{3} \)The initial velocity is:\[V = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\sqrt{5}A\omega}{3}.\]Step 2: Determine velocity when speed is tripledLet the new amplitude be \( A' \). The final velocity is \( 3V \), and it occurs at the same displacement \( x = \frac{2A}{3} \). Thus:\[3V = \omega \sqrt{A'^2 - \left(\frac{2A}{3}\right)^2}.\]Substituting the expression for \( V \):\[3 \left(\frac{\sqrt{5}A\omega}{3}\right) = \omega \sqrt{A'^2 - \frac{4A^2}{9}}.\]Simplifying gives:\[\sqrt{5}A\omega = \omega \sqrt{A'^2 - \frac{4A^2}{9}}.\]Step 3: Solve for the new amplitude \( A' \)Squaring both sides:\[5A^2 = A'^2 - \frac{4A^2}{9}.\]Rearranging to solve for \( A'^2 \):\[A'^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2}{9} + \frac{4A^2}{9} = \frac{49A^2}{9}.\]Taking the square root:\[A' = \frac{7A}{3}.\] Final Answer:\[\boxed{\frac{7A}{3}}.\]