Determine the resultant of two waves given by `y_(1) = 4 sin(200 pi t) and y_(2) = 3 sin(200 pi t + pi//2)`.

Explanation

In general, resultant of two waves given by `y_(1) = A_(1) sin omega t` and `y_(2) = A_(2) sin(omega t + phi)` can be obtained by the principle of superposition, i.e.,
`y = y_(1) + y_(2) = A_(1) sin omega t + A_(2) sin(omega t + phi)`
`= A sin(omega t + alpha)`
where
`A = [A_(1)^(2) + A_(2)^(2) + 2A_(1) A_(2) cos phi]^(1//2)`
`alpha = tan^(-1)[(A_(2) sin phi)/(A_(1) + A_(2) cos phi)]`
In the present case,
`y_(1) = 4 sin 200 pi t and y_(2) = 3 sin (200 pi t + (pi)/(2))`
Here`A_(1) = 4, A_(2) = 3`, and `phi = pi//2`.
Resultant wave will by `y = A sin(200 pi t + alpha)` with
`A = [A_(1)^(2) + A_(2)^(2) + 2A_(1) A_(2) cos phi]^(1//2)`
` implies A = [4^(2) + 3^(2) +2 (4) (3) cos pi//2]^(1//2)`
`implies A = 5`
`tan alpha = (A_(2) sin phi)/(A_(1) + A_(2) cos phi) = (3 sin pi//2)/(4 + 3 cos pi//2) = (3)/(4)`
or `alpha = 37^(@) ~~ 0.2 pi` rad
Hence, the resultant wave is `y = 5 sin(200 pi t + 0.2 pi)`.