1.

The phase difference between two waves, represented by y_(1) = 10^(-6) sin [100t +(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.

Answer»

1.07 radians
2.07 radians
0.5 radians
1.5 radians

Solution :`y_(1) = 10^(-6) sin[100t + (x//50) + 0.5]`
and `y_(2) = 10^(-6) COS [100t + (x//50)]`
`=10^(-6)sin[100t + (x//50) + 1.57]` [ USING `cos x = sin(x + pi//2)]`
The PHASE difference `=1.57 -0.5 = 1.07` radius. [Also using `sinx = cos(pi//2 -x)`, we can get the same result]


Discussion

No Comment Found

Related InterviewSolutions