$y_1 = \sin \left( {\omega t + \frac{\pi }{3}} \right)$ and $y_2 = \sin \omega t$ is :
$a_{y}=a_{1} \sin \frac{\pi}{3}=1 \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$
$\therefore A=\sqrt{a_{x}^{2}+a y^{2}}=\sqrt{\left(\frac{3}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}$
$=\sqrt{\frac{9}{4}+\frac{3}{4}}=\sqrt{3}$
Then the values of $A$ (in $m$) and $T$ (in $sec$) are :
