Correct Answer - A::B::C::D
(a) ` y= y_1 + y_2`
`= 0.2 sin (x-3.0t) + 0.2 sin (x-3.0t+ pi/2)`
`=A sin (x- 3.0t + theta)`
Here, A `= sqrt((0.2)^2 + (0.2)^2)`
and `theta = pi/4`
`:. y=0.28 sin (x-3.0t + pi/4)`
(b) Since, the amplitude of the resulting wave is
`0.32 m and A = 0.2 m`, we have
` 0.32 = sqrt((0.2)^2 + (0.2)^2 + (2)(0.2)(0.2) cos phi)`
Solving this, we get
`phi= +- 1.29` rad .