Two simple harmonic motion are represented by the equation 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…………

Two simple harmonic motion are represented by the equation y_(1)= 0.1

1.	Two simple harmonic motion are represented by the equation 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 :Differentiate w.r.t. to .t. `y_(1)= 0.1 sin [100pi t +(pi)/(3)]` `there v_(1) = 0.1xx 100 COS [100 pi t +(pi)/(3)]` and `y_(2)= 0.1 cos pi t` `therefore v_(2)= -0.1 sin(pi t+(pi)/(3))` The PHASE difference of velocity of particle-2 RELATIVE to particle-1 `delta= (pi t+(pi)/(3))-(pit +(pi)/(2))` `therefore delta = (pi)/(3)- (pi)/(3)= (2pi -3pi)/(6)= -(pi)/(6)rad`.