The phase difference between two waves represented by y 1 = 1 0 − 6 sin [ 100 t + ( x /50 ) + 0.5 ] m and y 2 = 1 0 − 6 cos [ 100 t + ( x /50 )] m, where x is expressed in metres and t is expresse

Question

The phase difference between two waves represented by $y_{1} = 1 0^{- 6} sin [100 t + (x /50) + 0.5]$ m and $y_{2} = 1 0^{- 6} cos [100 t + (x /50)]$ m, where $x$ is expressed in metres and $t$ is expressed in seconds, is approximately:

Options:

Answer:

B

Solution:

$y_{2} = 1 0^{- 6} sin [100 t + x /50 + π /2]$ . Phase difference = $π /2 - 0.5 = 1.5708 - 0.5 = 1.07$ rad.

TestHub · Accepted Answer

y 2 ​ = 1 0 − 6 sin [ 100 t + x /50 + π /2 ] . Phase difference = π /2 − 0.5 = 1.5708 − 0.5 = 1.07 rad.

Physics - Wave on String Question with Solution | TestHub

Options:

Doubts & Discussion