The phase difference between two waves, represented by y1=10-6 sin [100 t+(x / 50)+0.5] m y2=10-6 cos [100 t+(x / 50)] m where x is expressed in metres and t is expressed in seconds, is approximately.

Question

The phase difference between two waves, represented by y1=10-6 sin [100 t+(x / 50)+0.5] m y2=10-6 cos [100 t+(x / 50)] m where x is expressed in metres and t is expressed in seconds, is approximately. (A) 1.07 radians (B) 2.07 radians (C) 0.5 radians (D) 1.5 radians

Tardigrade Admin · Accepted Answer

y1=10-6 sin [100 t+(x / 50)+0.5] y2=10-6 cos [100 t+(x / 50)] =10-6 sin [100 t+(x / 50)+π / 2] =10-6 sin [100 t+(x / 50)+1.57] [ using cos x= sin (x+π / 2)] The phase difference =1.57-0.5=1.07 [or using sin x= cos (π / 2-x) . We get the same result.]

Q. The phase difference between two waves, represented by y1​=10−6sin[100t+(x/50)+0.5]m y2​=10−6cos[100t+(x/50)]m where x is expressed in metres and t is expressed in seconds, is approximately.

1.07 radians

2.07 radians

0.5 radians

1.5 radians

y1​=10−6sin[100t+(x/50)+0.5] y2​=10−6cos[100t+(x/50)] =10−6sin[100t+(x/50)+π/2] =10−6sin[100t+(x/50)+1.57] [ using cosx=sin(x+π/2)] The phase difference =1.57−0.5=1.07 [or using sinx=cos(π/2−x). We get the same result.]

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