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Discuss the composition of two S.H.M.s along the same path having same period. Find the resultant amplitude and initial phase. A sonometer wire is in unison with a tuning fork of frequency 125Hz when it is stretched by a weight. When the weight is completely immersed in water, 8 beats are heard per second. Find the sepecific gravity of the material of the weight. |
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Answer» Solution :Consider two S.H.M.s having same period and travelling along the same straight line but the they are having different initial PHASES and different AMPLITUDES Equations of DISPLACEMENT of two S.H.M.s travelling along same straight line (`X`-axis) are `x_(1)=a_(1)sin(omegat+alpha_(1))`............`(i)` and `x_(2)=a_(2)sin(omegat+alpha_(2))` ...........`(ii)` Here `a_(1)` and `a_(2)` are amplitudes and `alpha_(1)` and `alpha_(2)` are initial phases of two S.H.M.s. Resultant displacement at any instant is equal to vector sum of its displacement due to both S.H.M.s at that instant. Resultant displacement, `x=x_(1)+x_(2)` `:. x=a_(1)sin(omegat+alpha_(1))+a_(2)sin(omegat+alpha_(2))` By using `sin(A+B)=sinAcosB+cosAsinB` and expanding above equations, we get `x=a_(1)[sinomegat*cosalpha_(1)+cosomegat*sinalpha_(1)]+a_(2)[sinomegat*cosalpha_(2)+cosomegat*sinalpha_(2)]` `:.x=sinomegat[a_(1)cosalpha_(1)+a_(2)cosalpha_(2)]+cosomegat[a_(1)sinalpha_(1)+a_(2)sinalpha_(2)]`........`(iii)` Let, `a_(1)sinalpha_(1)+a_(2)sinalpha_(2)=Rsindelta` .............`(iv)` and `a_(1)cosalpha_(1)+a_(2)cosalpha_(2)=Rcosdelta`.........`(v)` where `R` is resultant amplitude of S.H.M. and `delta` is resultant phase of S.H.M. Substitute this value in equation `(iii)`, we get `x=sinomegat*Rcosdelta+cosomegat*Rsindelta` `x=R(sinomegat*cosdelta+cosomegat*sindelta)` `:. x=Rsin(omegat+delta)`..........`(vi)` This equation shows that resultant motion is also S.H.M. whose amplitude is `R` and initial phase angle is `delta` and having same period as that of individual S.H.M.'s. Resultant amplitude (R ) of the resultant motion : SQUARING and adding equation `(iv)` and `(v)`, `R^(2)sin^(2)delta+R^(2)cos^(2)delta=(a_(1)sinalpha_(1)+a_(2)sinalpha_(2))^(2)+(a_(1)cosalpha_(1)+a_(2)cosalpha_(2))^(2)` `:. R^(2)[sin^(2)delta+cos^(2)delta]` `=a_(1)^(2)sin^(2)alpha_(1)+a_(2)^(2)sin^(2)alpha_(2)+2a_(1)a_(2)sinalpha_(1)sinalpha_(2)+a_(1)^(2)cos^(2)alpha_(1)+a_(2)^(2)cos^(2)alpha_(2)+2a_(1)a_(2)cosalpha_(1)cosalpha_(2)` `:. R^(2)=a_(1)^(2)[sin^(2)alpha_(1)+cos^(2)alpha_(1)]+a_(2)^(2)[sin^(2)alpha_(2)+cos^(2)alpha_(2)]+2a_(1)a_(2)`[sinalpha_(1)sinalpha_(2)+cosalpha_(1)csalpha_(2)]` But `sin^(2)alpha_(1)+cos^(2)alpha_(1)=1`, `sin^(2)alpha_(2)+cos^(2)alpha_(2)=1` By using `cos(A-B)=sinA*sinB+cosA*cosB` in the above equation, we get `R^(2)=a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2)cos(alpha_(1)-alpha_(2))` Taking square root on both sides, we get `R=sqrt(a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2)cos(alpha_(1)-alpha_(2)))`........`(viii)` This expression of resultant amplitude of the S.H.M. Divide equation `(iv)` by equation `(v)`, we get, `(Rsindelta)/(Rcosdelta)=(a_(1)sinalpha_(1)+a_(2)sinalpha_(2))/(a_(1)cosalpha_(1)+a_(2)cosalpha_(2))` `tandelta=(a_(1)sinalpha_(1)+a_(2)sinalpha_(2))` `a_(1)cosalpha_(1)+a_(2)cosalpha_(2))` `delta=tan^(-1)((a_(1)sinalpha_(1)+a_(2)sinalpha_(2))/(a_(1)cosalpha_(1)+a_(2)cosalpha_(2)))` This is the expression of initial phase angle `delta`. Numerical : Given : `n_(1)=125Hz`, Beats per second `=8` `:. n_(2)=125-8=117Hz` `(n_(1))/(n_(2))=(sqrt((sigma)/(sigma-1)))` `(125)/(117)=sqrt((sigma)/(sigma-1))` `1.141=(sigma)/(sigma-1)` `sigma=8.09` |
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