Two simple harmonic motions are represented by equations y₁=0.5 [200 t+ 3 ] and y₁=0.5 t . The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is:

Step-by-step solution

The first equation is y₁ = 0.5 (200 t + 3 ) . The velocity v₁ is given by the derivative dy₁ dt = 0.5 200 (200 t + 3 ) = 100 (200 t + 3 + 2 ) . The second equation is y₂ = 0.5 ( t) = 0.5 ( t + 2 ) . The velocity v₂ is given by the derivative dy₂ dt = 0.5 ( t + 2 ) = 0.5 ( t + 2 + 2 ) = 0.5 ( t + ) . The phase of velocity v₁ is ₁ = 200 t + 3 + 2 = 200 t + 5 6 . The phase of velocity v₂ is ₂ = t + . The phase difference = ₁ - ₂ = (200 t + 5 6 ) - ( t + ) = 199 t - 6 . At t=0 , the phase difference is - 6 . Answer: -