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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 |
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Answer» 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)|` |
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