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Two particles move parallel to x - axis ...

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 ?

Text Solution

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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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