Answer
The amplitude of the resultant wave is $~~3.0~mm$
Work Step by Step
In part (a), we found that the required phase difference is $(\pi) rad$
We can find the horizontal component of the resultant wave's amplitude:
$y_{mh}' = (y_{m1}) cos ~(0) + (y_{m2}) cos~(\pi)$
$y_{mh}' = (5.0~mm) cos ~(0) + (8.0~mm) cos~(\pi)$
$y_{mh}' = -3.0~mm$
We can find the vertical component of the resultant wave's amplitude:
$y_{mv}' = (y_{m1}) sin ~(0) + (y_{m2}) sin~(\pi)$
$y_{mv}' = (5.0~mm) sin ~(0) + (8.0~mm) sin~(\pi)$
$y_{mv}' = 0$
We can find the amplitude of the resultant wave:
$y_m' = \sqrt{(-3.0~mm)^2+(0)^2}$
$y_m' = 3.0~mm$
The amplitude of the resultant wave is $~~3.0~mm$
