The phase difference between two waves, represented byy _(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 .....

The phase difference between two waves, represented byy _(1) = 10 ^(-6

1.	The phase difference between two waves, represented byy _(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 .....
Answer» `15` RADIAN `1.07` radian `2.07` radian `0.5` radian Solution :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`