If the two particles performs S.H.M. of same initial phase angle but different amplitudes of individuals, then the resultant motion initial phase angle depends on

Explanation

Let a particle be subjected to two linear SHMs of the same period, along the same line and having the same mean positon, represented by
`x_(1)=A_(1) sin (omegat-alpha)` and `x_(2)=A_(2) sin (omega+beta)`,
where `A_(1)` and `A_(2)` are the amplitudes and `alpha` and `beta` are the initial phases of the two SHMs.
According to the principle of superposition, the resultant displacement of the particle at any instant t is the algebraic sum `x=x_(1)+x_(2)`.
`:. x=A_(1) sin (oemgat+alpha)+A_(2) sin (omega+beta)`
`=A_(1) sin omegat cos alpha+A_(1) cos omegat sin alpha+A_(2) sin omegat cos beta+A_(2) cos omegat sin beta`
`=(A_(1) cos alpha+A_(2) cos beta) sin omegat+(A_(1) sin alpha+A_(2) sin beta) cos omegat`
Let `A_(1) cos alpha+A_(2) cos beta=R cos delta" "`.....(1)
and `A_(1) sin alpha+A_(2) sin beta=R sin delta" "`.......(2)
Equation (3), which gives the displacement of the particle, shows that the resultant motion is also linear simple harmonic, along the same line as the SHMs superposed, with amplitude |R| and inital phase `delta` but having the same mean position and the same period as the individual SHMs.
Amplitude of the resultant motion :
`R^(2)=R^(2) cos^(2)delta+R^(2) sin^(2)delta`
From Eqs. (1) and (2),
`R^(2)=(A_(1) cos alpha+A_(2) cos beta)^(2)+(A_(1) sin alpha+A_(2) sin beta)^(2)`
`=A_(1)^(2) cos^(2)alpha+A_(2)^(2) cos^(2)beta+2A_(1)A_(2) cos alpha cos beta+A_(1)^(2) sin^(2)alpha+A_(2)^(2) sin^(2) beta+2A_(2)A_(2) sin alpha sin beta`
`=A_(1)^(2)(cos^(2)alpha+sin^(2)alpha)+A_(2)^(2)(cos^(2)beta+sin^(2) beta)+2A_(1)A_(2)(cos alpha cos beta+sin alpha sin beta)`
`:. R^(2)=A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2)cos(alpha-beta)`
`:. |R|=sqrt(A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2) cos (alpha-beta))" "`......(4)
Initial phase of the resultant motion :
From Eqs. (1) and (2),
`(R sin delta)/(R cos delta)=tan delta=(A_(1) sin alpha+A_(2) sin beta)/(A_(1) cos alpha+A_(2) cos beta)`
`:. tan^(-1)((A_(1) sin alpha+A_(2) sin beta)/(A_(1) cos alpha+A_(2) cos beta))" "`......(5)
Now, consider Eq. (4) for |R|.
Case (1) : Phase difference, `alpha-beta=0^(@)`
`:. cos (alpha-beta)=1`
`:. |R|=sqrt(A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2))=A_(1)+A_(2)`
Case (2) : Phase difference, `alpha-beta=pi//3` rad
`:. cos (alpha-beta)=(1)/(2)" " :. |R|=sqrt(A_(1)^(2)+A_(2)^(2)+A_(1)A_(2))`
Case (3) : Phase difference, `alpha-beta=pi//2` rad
`:. Cos (alpha-beta)=0`
`:. |R|=sqrt(A_(1)^(2)+A_(2)^(2))`
Case (4) : Phase difference, `alpha-beta=pi` rad
`:. cos (alpha-beta)=-1`
`:. |R|=sqrt(A_(1)^(2)+A_(2)^(2)-2A_(1)A_(2))" " :. |R|=|A_(1)-A_(2)|`