InterviewSolution
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InterviewSolution
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Two particles move parallel to x - axis about the origin with the same amplitude and frequency. At a certain instant they are found at distance `(A)/(3)` from the origin on opposite sides but their velocities are found to be in the same direction. What is the phase difference between the two ? |
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Answer» Let equation of two SHM be `x_(1)=A"sin"omegat" "…(i)` `x_(2)=A"sin"(omegat+phi)" "…(ii)` Given, `(A)/(3)=A"sin"omegat` and `-(A)/(3)=A"sin"(omegat+phi)` Which gives `"sin"omegat=(1)/(3)` `"sin"(omegat+phi)=-(1)/(3)` From Eq. (iv), `"sin"omegat" cos"phi+"cos"omegat" sin"phi=-(1)/(3)` `implies (1)/(3)"cos"phi+sqrt(1-(1)/(9))"sin"phi=-(1)/(3)` Solving this equation, we get or `"cos"phi=-1,(7)/(9)` `implies phi=pi` or `"cso"^(-1)((7)/(9))` Differentiating Eqs. (i) and (ii) we obtain `v_(1)=Aomega"cos"omegat` and `v_(2)=Aomega"cos"(omegat+phi)` If we put `phi=pi`, we find `v_(1)` and `v_(2)` are of opposite signs .Hence, `phi=pi` is not acceptance. `therefore phi="cos"^(-1)((7)/(9))` |
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