1.

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 :

Answer»

0.5 radian
1.07 radian
1.5 radian
2.07 radian.

Solution :`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).


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