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. 2.07 radians
B. 0.5 radians
C. 1.5 radians
D. 1.07 radians

Answer

D

Explanation

`y_1=10^(-6) sin { 100 t + x/50 +0.5 }`
`y_2=10^(-6) cos {100t+x/50}`
`=10^(-6) sin { pi/2 + 100t +x/50}`
Phase difference between `y_1` & `y_2 =pi/2 -0.5` =1.58-0.5 =1.08 radians