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…………

Options

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

Answer

D

Explanation

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`.