Two waves are represented by the equationsy _(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…….

Two waves are represented by the equationsy _(1) = a sin (omega t + kx

1.	Two waves are represented by the equationsy _(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…….
Answer» `1.0` RADIAN `1.25` radian `1.57` radian `0.57` radian SOLUTION :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`