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

Speed in SHM in terms of displacement is given by, $v = \omega\sqrt{{A}^{2} - {x}^{2}}$.

at $x = \dfrac{2A}{3}$, we get

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

Let the new amplitude be$= A '$

Then, 

$v ' = 3v ⇒ \omega\sqrt{{\left(A '\right)}^{2} - {\left(\dfrac{2A}{3}\right)}^{2}} = \sqrt{5}A\omega$

$⇒ A ' = \dfrac{7}{3}A$

Therefore, $n = 7$.