The phase difference between two waves represented by `y_(1) = 10^(-6) sin [ 100 t + (x)/(50) + 0.5 ] ` m
`y_(2) = 10^(-6) co [ 100 t + (x)/(50) ] `m
Where x is in metre and t in seconds, is :

Options

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

Answer

B

Explanation

`y_(1) = 10^(-6) sin [ 100 t + (x)/(50) + 0.5 ]`
`y_(2) = 10^(-6) sin [ 100 t + (x)/(50) ] = 10^(-6) sin [ (pi)/(2) + 100 t + (pi)/(2) ]`
`therefore` phase diff. Between two waves = `(pi)/(2) - 0.5`
= `(3.14)/(2) - 0.5`
= 1.07 rad.
Correct choice is (b).