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)=0.1sin(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 :

Options

A. `-pi//6`
B. `pi//3`
C. `-pi//3`
D. `pi//6`

Answer

A

Explanation

`v_(1)=(dy_(1))/(dt)=0.1xx100pi cos (100 pi 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).