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

We have
`y_(1) = 10^(-6)sin [100 t + (x//50) + 0.5]`
`y_(2) = 10^(-6) sin [100 t + ((x)/(50)) = ((pi)/(2))]`
Phase difference `phi`