JEE Main 2016 · 3 Apr · Shift 1, Physics question 10

At $x=\frac{2A}3$, $v=\omega\sqrt{A^2-\frac{4A^2}9}=\frac{\sqrt5}3\omega A$. Trebling it gives $v'=\sqrt5\,\omega A$, so
$$A'^2=x^2+\frac{v'^2}{\omega^2}=\frac{4A^2}9+5A^2=\frac{49A^2}9,\quad A'=\frac{7A}3.$$