Answer:
0.5 m
Explanation:
Givens:
ym1 = 2.5 mm
ym2 = 4.5 mm
Ф_1=π / 4
Ф_2=π / 2
We have 2 ways to solve this problem. The first one given that the 2 waves have the frequency then we know that the resultant wave amplitude is
Ym = (ym1 + ym2)cos(Ф_2/2)
By substitution we have
Ym= (0.025 + 0.045)cos(π/4) = 0.496 m
The second one is it treat them as Phasors where the phase between them is Ф_2=π / 2 Therefore
Ym^2=(ym1^2+ym2^2)
So we have Ym=√0.025^2+0.045^2
= 0.5 m