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

D

Explanation

`y_(1) = 0.1 sin (100 pi t + pi//3)`
`therefore v_(1) = (dy_(1))/(dt) = 0.1 cos (100 pi t + pi//3 ) . 100 pi `
`y_(2) = 0.1 cos pi t = 0.1 sin ((pi)/(2) + pi t ) `
`v_(2) = (dy_(2))/(dt ) = 0.1 cos ((pi)/(2) + pi t).pi`
`therefore` initial phase deff. between velocity of Ist particle and 2nd particle is = `phi_(1) - phi_(2) = ((pi)/(3) - pi//2 ) = - (pi)/(6) `
so correct choice is (d).