Two waves ${{\bf{E}}_1} = {{\bf{E}}_0}{\bf{sin\omega t}}$ and ${{\bf{E}}_2} = {{\bf{E}}_0}{\bf{sin}}\left( {{\bf{\omega t}} + {\bf{6}}{{\bf{0}}^0}} \right)$superimpose each other. Find out the initial phase of the resultant wave?
A. ${30^0}$
B. ${60^0}$
C. ${120^0}$
D. ${0^0}$

Hint: Phase of the wave represents the location of the point within a specific wave form and generally that location varies with respect to time and position too. Now if we want to view a wave we can do it by taking a snap that is by putting time as a fixed one and varying position or we can either fix a position and get the shape of the wave which varies with time.

Formula used:
$\operatorname{Sin} \left( A \right) + \operatorname{Sin} \left( B \right) = 2\operatorname{Sin} \left( {\dfrac{{A + B}}{2}} \right)\operatorname{Cos} \left( {\dfrac{{A - B}}{2}} \right)$

Complete step by step answer:
Usually most of the wave equations are of forms
$\eqalign{
  & {\text{Y = A}}\sin ( \pm {\text{wt}} \pm {\text{kx}}) \cr
  & {\text{Y = A}}\cos ( \pm {\text{wt}} \pm {\text{kx}}) \cr} $
Where $w$ is the angular frequency and $k$ is the angular number .
The part which is present inside the sin or cos functions is called the phase of the wave and if we want to find the phase difference of the waves we have to subtract one phase from the other to get it.
When two waves get super imposed then we get the resultant wave which will be having some other phase and amplitude.
The two waves which we have are
${E_1} = {E_0}\operatorname{Sin} \left( {\omega t} \right)$
${E_2} = {E_0}\operatorname{Sin} \left( {\omega t + {{60}^0}} \right)$
When we superimpose both the waves that is after adding them we get
${E_1} + {E_2} = {E_0}\operatorname{Sin} \left( {\omega t} \right) + {E_0}\operatorname{Sin} \left( {\omega t + {{60}^0}} \right)$
We have the formula
$\operatorname{Sin} \left( A \right) + \operatorname{Sin} \left( B \right) = 2\operatorname{Sin} \left( {\dfrac{{A + B}}{2}} \right)\operatorname{Cos} \left( {\dfrac{{A - B}}{2}} \right)$
$\eqalign{
  & {E_1} + {E_2} = {E_0}\operatorname{Sin} \left( {\omega t} \right) + {E_0}\operatorname{Sin} \left( {\omega t + {{60}^0}} \right) \cr
  & {E_1} + {E_2} = 2{E_0}[\operatorname{Sin} \left( {\dfrac{{\omega t + \omega t + {{60}^0}}}{2}} \right)\operatorname{Cos} \left( {\dfrac{{\omega t + {{60}^0} - \omega t}}{2}} \right)] \cr
  & {E_1} + {E_2} = 2{E_0}[\operatorname{Sin} \left( {\dfrac{{2\omega t + {{60}^0}}}{2}} \right)\operatorname{Cos} \left( {\dfrac{{{{60}^0}}}{2}} \right)] \cr
  & {E_1} + {E_2} = 2{E_0}[\operatorname{Sin} \left( {\omega t + {{30}^0}} \right)\operatorname{Cos} \left( {{{30}^0}} \right)] \cr} $
It’s clearly visible that if we put t=0 then initial phase we get as ${30^0}$

So, the correct answer is “Option A”.

Note:
Sometimes they might ask us to find the phase difference. In this case we get constant phase difference which doesn’t vary with position or time but sometimes we get phase difference which varies with only time or only position or with both time and position. Here due to superposition of the waves the resultant wave equation which we had got has a phase which varies with time.