Particle performs simple harmonic motion with amplitude A. Its...

The velocity of a particle in simple harmonic motion is given by $v = \omega \sqrt{A^2 - x^2}$

At distance $x = \dfrac{2A}{3}$, the initial velocity is:

$v_1 = \omega \sqrt{A^2 - \left(\dfrac{2A}{3}\right)^2} = \omega \sqrt{A^2 - \dfrac{4A^2}{9}} = \dfrac{\sqrt{5}}{3}\omega A$

The speed is tripled at this instant, so the new velocity is:

$v_2 = 3v_1 = 3 \left(\dfrac{\sqrt{5}}{3}\omega A\right) = \sqrt{5}\omega A$

Let the new amplitude be $A'$. The position $x$ and angular frequency $\omega$ remain unchanged. Using the velocity formula for the new amplitude:

$v_2^2 = \omega^2 (A'^2 - x^2)$

Substituting the values of $v_2$ and $x$:

$(\sqrt{5}\omega A)^2 = \omega^2 \left(A'^2 - \left(\dfrac{2A}{3}\right)^2\right)<br><br>5A^2 = A'^2 - \dfrac{4A^2}{9}<br><br>$ A'^2 = 5A^2 + $\dfrac{4A^2}{9} = \dfrac{49A^2}{9}$

$A' = \dfrac{7A}{3}$

Answer: $\dfrac{7A}{3}$