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A particle is simultaneously subjected to three simple harmonic motions, all of the same frequency and in the x direction. The amplitudes are 0.25, 0.20, 0.15, and the face difference between the first and second is 45

^{0}and between the second and third is 30

^{0}. Find the amplitude of the resultant displacement and its phase relative to the first component.

I made a diagram

The 0.25, 0.20, and 0.15 components are represented by Z

_{1}=A

_{1}e

^{jwt}, Z

_{2}=A

_{2}e

^{j(wt+a)}, and Z

_{3}=A

_{3}e

^{j(wt+b)}, adding them all together to give me the resultant component, and simplifying gives me: Z=e

^{jwt}[A

_{1}+A

_{2}e

^{ja}+A

_{3}e

^{jb}]...so what I would think I am suppose to do is find the x component of the resultant vector and the y component, and use Pythagorean to find the resultant vector and then do arctan(y/x) to get the phase angle. The answers are 0.52 and 33.5

^{0}.

I can get 0.52 for the amplitude, but that is only by taking the x component of the resultant vector and ignoring the rest, if I do it with Pythagorean I get [0.52

^{2}+0.216

^{2}]

^{1/2}=0.56, and the angle is arctan (0.216/0.52)=22.5......so apparently I am wrong, can someone help please?