Two particles A and B execute simple harmonic motion according to the equations ) $y_1=3\sin \omega t$ and $y_2=4\sin \left(\omega t+\frac{\pi }{2}\right)+3\sin \omega t$ The phase difference between them is

Representing a quantity with phase $\left(\omega \mathrm{t}\right)$ along x-axis any quantity with a phase $(\omega t+\varphi )$ can be represented by a vector making an angle $\varphi$ with X-axis in the anticlockwise sense and any quantity with a phase $(\omega t-\varphi )$ can be represented by a vector making an angle $\varphi$ with X-axis in the clockwise sense. 

$\text{ Phase difference }\varphi ={\tan }^{-1}\left(\frac{4}{3}\right)$

$\text{ Alternatively, given }{y}_1=3\sin \omega \mathrm{t}\text{. ......(i)}$

$\text{ and }{y}_2=4\sin \left(\omega t+\frac{\pi }{2}\right)+3\sin \omega t$

$\begin{array}{l}=4\cos \omega t+3\sin \omega t\\ =5\left[\frac{4}{5}\cos \omega t+\frac{3}{5}\sin \omega t\right]\\ =5[\sin \varphi \cos \omega t+\cos \varphi \sin \omega t]\\ =5\sin (\omega t+\varphi ) .........\left(\mathrm{ii}\right)\end{array}$

From (i) and (ii), the result follows.

$\cos \varphi =\frac{3}{5};\sin \varphi =\frac{4}{5}\text{ and }\tan \varphi =\frac{4}{3}$