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

C

Explanation

`y_1=0.1sin(100pit+pi//3)`
`v_1=(dy_1)/(dt)=0.1xx100picos(100pit+pi//3)`
`=10picos(100pit+pi//3)`
`=10pisin(pi/2+100pit+pi/3)`
`y_2=0.1cospit`
`v_2=(dy_2)/(dt)=-0.1pisinpit=0.1pisin(pi+pit)`
At `t=0`, phase difference of `v_1` w.r.t. `v_2`
`=(pi/2+pi/3)-pi=-pi//6`