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

B

Explanation

`y_(1)=0.1 si (100 pi t+ (pi)/(3))`
`therefore v_(1)=(dy_(1))/(dt)=0.1 (100 pi) "cos" (100 pi t+(pi)/(3))`
and `because v_(2)=(dy_(2))/(dt)= -0.1 pi "cos" (pi t+ (pi)/(2))`
`therefore` At time t =0, the phase difference between `v_(1) and v_(2)` is `[(pi)/(3)-(pi)/(2)]= -(pi)/(6)`