The phase difference between two waves, represented by
`y _(1) = 10 ^(-65) sin [100 t + (x //50) + 0.5]m`
`y_(2) = 10 ^(-6) cos [100 t + (x //50)]m`
(where x is expressed in metres and t is expressed in seconds is approximately .....

Options

A. `15` radian
B. `1.07` radian
C. `2.07` radian
D. `0.5` radian

Answer

B

Explanation

Here, `y _(1) = 10 ^(-6) sin (100 t + (x)/( 50) + 0.5) m " "…(1)`
and `y _(2) = 10 ^(-6) cos (100 t + (x)/(50) ) m " "…(2)`
`therefore y _(2) =10 ^(-6) sin ((pi)/(2) + 100 t + (x)/( 50)) " "…(3)`
`{because cos theta = sin ((pi)/(2) +theta)}`
From equation (1), phase of first wave at time t,
`theta _(1) = 100 t + (x)/(50) + 0.5`
From equation (3) phase of second wave at time, t,
`theta_(2) = (pi)/(2) + 100 t + (x)/(50)`
`implies `Phase difference between second and first wave at time t is,
`theta_(2) -theta_(1) = (pi)/(2) =0.5`
`= (3.14)/(2) -0.5`
`= 1.57 -0.5`
`= 1.07 rad`