The phase difference between two waves represented by `y_(1)=10 ^(-6) sin [100 t+(x//50)+0.5]m, y_(2)=10^(-6) cos[100t+(x//50)]m`where x is expressed in metres and t is expressed in seconds, is approximately

Options

A. 1.5 rad
B. 1.07 rad
C. 2.07 rad
D. 0.5 rad

Answer

B

Explanation

`y_(1)=10^(-6)sin[100t+(x)/(50)+0.5]`
`y_(2)=10^(-6)sin[100t+(x)/(50)+(pi)/(2)]`
`Deltaphi=[100t+(x)/(50)+1.57]-[100t+(x)/(50)+0.5]`
`=1.07rad//sec`.