Finding the Resultant Amplitude & Phase: A Practice Problem

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The discussion focuses on solving a practice problem involving three simple harmonic motions with given amplitudes and phase differences. The user attempts to find the resultant amplitude and phase angle but encounters discrepancies in their calculations. They initially calculate the x component as 0.52 but struggle with the y component, leading to confusion about the correct angle. After some back-and-forth, the user realizes a mistake in their phase angle for the third component, suggesting that using 75 degrees instead of 30 degrees yields the correct results. The conversation highlights the importance of accurate phase angle representation in harmonic motion calculations.
mmmboh
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Hi, this isn't homework but a practice problem.
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 450 and between the second and third is 300. Find the amplitude of the resultant displacement and its phase relative to the first component.

I made a diagram
i74bcm.jpg


The 0.25, 0.20, and 0.15 components are represented by Z1=A1ejwt, Z2=A2ej(wt+a), and Z3=A3ej(wt+b), adding them all together to give me the resultant component, and simplifying gives me: Z=ejwt[A1+A2eja+A3ejb]...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 33.50.
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 [0.522 +0.2162]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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mmmboh said:
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 [0.522 +0.2162]1/2=0.56, and the angle is arctan (0.216/0.52)=22.5...so apparently I am wrong, can someone help please?

I can only get 0.52 for the amplitude if I use both the x and the y component of the resultant vector. I also don't get 0.216 for the y component. If you want to know
what's wrong show your calculation
 


Ok well for the x component I did 0.25+0.20cos45+0.15cos30=0.52...and for the y component I did 0.20sin45+0.15sin30=0.216...what am I doing wrong?
 


Hm I guess if i use 75 degrees for the third one instead of 30 degrees it works, that's how I originally did it, but in the book they made it seem like that's not the way :S

Edit: I think I see my mistake
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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