A 20g particle is subjected to two simple harmonic motions simultaneously which are given by $x_1=2\sin 10t$ and $x_2=4\sin (10t+\frac{\pi }{3})$, where $x_1,x_2$  are in metre and t in second. Choose the correct statement(s).

Phase difference is  $\delta =\frac{\pi }{3}={60}^0$
 Resultant amplitude is  $A=\sqrt{2^2+4^2+\mathrm{2.2.4.}\cos {60}^0}=\sqrt{28}m$
Phase difference between resultant SHM and first SHM
$$\begin{gathered} \epsilon ={\tan }^{-1}\left(\frac{A_2\sin \delta }{A_1+A_2\cos \delta }\right) \\ \therefore \epsilon ={\tan }^{-1}\left(\frac{4\sin {60}^0}{2+4\cos {60}^0}\right)={\tan }^{-1}\left(\frac{2\sqrt{3}}{4}\right)={\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right) \end{gathered}$$

$\therefore$Resultant SHM is $x=\sqrt{28}\sin (10t+{\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right))$

At  $t=0,x=\sqrt{28}\sin \left({\tan }^{-1}\frac{\sqrt{3}}{2}\right)$

$\therefore x=\sqrt{28}\sqrt{\frac{3}{7}}=2\sqrt{3}m$