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)/(6)`
B. `(pi)/(3)`
C. `(-pi)/(3)`
D. `(pi)/(6)`

Answer

A

Explanation

`y_(1) = 0.1 sin(100 pit + pi/3)`
`v_(1) = (dy_(1))/(dt) = 0.1 xx 100picos(100 pit + (pi)/(3))`
`= 10 picos(100 pit + (pi)/(3))`
`y_(2) = 0.1 cos100pit`
`rArr v_(2) = -0.1 pisin 100 pit = 0.1pi cos(100 pit + (pi)/(2))`
Phase difference between `v_(1)` and `v_(2)`
`= (100 pit + (pi)/(3)) - (100 pit (pi)/(2)) = - (pi)/(6)`