Hi, this isn't homework but a practice problem.(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Homework Help: Resultant component of adding Three simple harmonic motions together help Diagram!

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