Answer
The amplitude of the resultant wave is $~~9.4~mm$
Work Step by Step
In part (a), we found that $\phi_1 = \pi$
In part (c), we found that $\phi_2 = 0$
Then $\frac{\phi_1-\phi_2}{2} = \frac{\pi}{2}$
We can find the horizontal component of the resultant wave's amplitude:
$y_{mh}' = (y_{m1}) cos ~(0) + (y_{m2}) cos~(\frac{\pi}{2})$
$y_{mh}' = (5.0~mm) cos ~(0) + (8.0~mm) cos~(\frac{\pi}{2})$
$y_{mh}' = 5.0~mm$
We can find the vertical component of the resultant wave's amplitude:
$y_{mv}' = (y_{m1}) sin ~(0) + (y_{m2}) sin~(\frac{\pi}{2})$
$y_{mv}' = (5.0~mm) sin ~(0) + (8.0~mm) sin~(\frac{\pi}{2})$
$y_{mv}' = 8.0~mm$
We can find the amplitude of the resultant wave:
$y_m' = \sqrt{(5.0~mm)^2+(8.0~mm)^2}$
$y_m' = 9.4~mm$
The amplitude of the resultant wave is $~~9.4~mm$
