The two waves are represented by
`y_(1)= 10^(-6) sin(100t + (x)/(50)+ 0.5)m`
`Y_(2) =10^(-2) cos(100t + (x)/(50))m`
where x is ihn metres and t in seconds. The phase difference between the waves is approximately:

Options

A. `1.07 rad`
B. `2.07 rad`
C. `0.5 rad`
D. `1.5 rad`

Answer

A

Explanation

The given waves are
`y_1=10^-6sin[100t+(x//50)+0.5]m`
and `y_2=10^-6cos[100t+(x//5)]m`
`impliesy_2=10^-6sin[100t+(x//50)+(pi)/(2)]m`
`[:. sin((pi)/(2)+theta)=costheta]`
Hence, the phase difference between the waves is
`Delta phi=((pi)/(2)-0.5)rad=((3.14)/(2)-0.5)` rad
`=(1.57-0.5)rad=(1.07)rad`