When two progressive waves `y_(1) = 4 sin (2x - 6t)` and `y_(2) = 3 sin (2x - 6t - (pi)/(2))` are superimposed, the amplitude of the resultant wave is

Options

A. 4 units
B. 5 units
C. 8 units
D. 10 units

Answer

B

Explanation

`y_(1) = 4 sin (2x - 6t)`
and `y_(2) = 3 sin (2x - 6t - (pi)/(2)) = - 3 cos (2x - 6t)`
When they are superimposed, the resultant wave `y = y_(1) + y_(2) = 4 sin (2x - 6t) -3 cos (2x - 6t)`
Let us use `R cos theta = 4, R sin theta = 3 and (2x - 6t) = alpha` then `y = R (cos theta sin alpha) - (R sin theta cos alpha)`
`= R sin (alpha - theta) = R sin (2x - 6t - theta)`
where `R = sqrt(R^(2) cos^(2) theta + R^(2) sin^(2) theta)`
`= sqrt(4^(2) + 3^(2)) = 5`
Thus the resultant is a wave of amplitude 5