Two waves are represented by the equations ` y _(1) = a sin (omega t + kx + 0.57)m and y_(2) =a cos (omega t + kx) m,` where x is in meter and t in sec. The phase difference between them is…….

Options

A. `1.0` radian
B. `1.25` radian
C. `1.57` radian
D. `0.57` radian

Answer

A

Explanation

Here `y _(1) = a sin (omega t + kx + 0.57) m " "…(1)`
`y _(2) =a cos (omega t + kx) m`
`therefore y _(2) = a sin ((pi)/(2) + omega t + kx) m " "…(2)`
`(because cos theta = sin ((pi)/(2) + theta ))))`
From (2), phase of second wave,
`theta_(2) = (pi)/(2) + omega t + kx`
`theta _(1) = omega t + kx + 0.57`
`therefore` Phase difference `= theta _(2) - theta_(1) (pi)/(2) - 0.57`
`= (3.14 )/(2) -0.57 =1.57 -0.57 =1 rad`