As particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance $\frac{2\mathrm{A}}{3}$ from equilibrium position. The new amplitude of the motion is

We know that $\mathrm{V}=\mathrm{\omega }\sqrt{{{\mathrm{A}}_0}^2-{\mathrm{x}}^2} {\mathrm{A}}_0 \mathrm{is} \mathrm{the} \mathrm{initial} \mathrm{amplitude}$ 

Initially $\mathrm{V}=\mathrm{\omega }\sqrt{{{\mathrm{A}}_0}^2-{\left(\frac{2{\mathrm{A}}_0}{3}\right)}^2}$ Finally $3\mathrm{V}=\mathrm{\omega }\sqrt{{\mathrm{A}}^2-{\left(\frac{2\mathrm{A}}{3}\right)}^2}$

Where A =  Final amplitude (Given at $\mathrm{x}={\frac{2{\mathrm{A}}_0}{3}}_{ }$, velocity to trebled)

On dividing we get

$\frac{3}{1}=\frac{\sqrt{{\mathrm{A}}^2-{\left({\displaystyle \frac{2{\mathrm{A}}_0}{3}}\right)}^2}}{\sqrt{{{\mathrm{A}}_0}^2-{\left({\displaystyle \frac{2{\mathrm{A}}_0}{3}}\right)}^2}}$

$9\left[{{\mathrm{A}}_0}^2-\frac{4{{\mathrm{A}}_0}^2}{9}\right]={\mathrm{A}}^2-\frac{4{{\mathrm{A}}_0}^2}{9}$

$\Rightarrow \mathrm{A}=\frac{7{\mathrm{A}}_0}{3}$