1.

Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. Two simple harmonic motions are represented by the equations y_(1)=0.15m(100pit+pi//3) and y = 0.1 cos pit. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is :

Answer»

`-PI//6`
`pi//3`
`-pi//3`
`pi//6`

Solution :`v_(1)=(dy_(1))/(dt)=0.1xx100pi cos (100pi t+pi//3)`
`v_(2)=(dy_(2))/(dt)=-0.1xx pi SIN pi t=0.1xx pi cos (pi t+pi//2)`
PHASE DIFF. `Delta phi_(12)=(pi)/(3)-(pi)/(2)=-(pi)/(6)`
Thus correct choice is (a).


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