1.

Two simple harmonic motions are represented by the equations y_(1) = 0.1 sin (100 pi t + (pi)/(3)) and y_(2) = 0.1 cos pi t. 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 :`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).


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