Two simple harmonic are represented by the equation `y_(1)=0.1 sin (100pi+(pi)/3) and y_(2)=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)/(3)`
B. `(pi)/(6)`
C. `(-pi)/(6)`
D. `(pi)/(3)`

Answer

C

Explanation

`v_(1) = (dy_(1))/(dt) = - 0.1 xx 100pi cos(100 pi t + (pi)/(3))`
`v_(2) = (dy_(2))/(dt) = - 0.1 pi sin pi t = 0.1 pi cos(pi t + (pi)/(2))`
phase difference of velocity of first particle with respect to the velocity of `2^(nd)` particle at `t = 0` is
`Delta phi = phi_(1) - phi_(2)= (pi)/(3) - (pi)/(2)= -(pi)/(6)`