## UnkleRhaukus one year ago Perpendicular Simple Harmonic Oscillators A particle is simultaneously subjected to two perpendicular simple harmonic oscillators of the same frequency, 112 Hz. If the amplitude and phase constant of one oscillator is 0.64mm and 21° respectively, and the other is 0.16mm and 83°, determine the resultant amplitude at 9.8 seconds.

1. IrishBoy123

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2. IrishBoy123

just like a phasor?

3. IrishBoy123

sorry they're orthogomal let's think

4. UnkleRhaukus

uhm

5. IrishBoy123

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6. UnkleRhaukus

¿it flipped

7. IrishBoy123

ie, i'd be looking to use $$A_1 e^{i (\omega t + \delta_1)} + i \ A_2 e^{i (\omega t + \delta_2)}$$ if i could make it work

8. IrishBoy123

because crunching $$\sqrt{ A_1^2 cos^2 (\omega t + \delta_1) + A_2^2 cos^2 (\omega t + \delta_2) }$$ would be a pain

9. UnkleRhaukus

is that the amplitude

10. UnkleRhaukus

oh , i see it flipted because they are perpendicular!

11. IrishBoy123

yes but it proves the idea is totally totally hair brained i think, and mixes up 2 ideas. the flip just adds a phase shift to the second shm but is modelling them in 1 dimension i have searched the web and no one is using phasors for shm, except in 1 dimension, which makes sense i suppose. http://www.wolframalpha.com/input/?i=sqrt+%28%280.16+cos+%282+*pi+*+112+*+9.8+++%2B+83%2Fpi+%29+%29%5E2%2B+%280.64+cos+%282+*pi+*+112+*+9.8+++%2B+21%2Fpi+%29%29%5E2%29

12. UnkleRhaukus
13. UnkleRhaukus

14. UnkleRhaukus

i solved it this way, with the aid of the phasor graph, i've assumed sines whereas you've got cosines,

15. IrishBoy123

holy moley that's interesting can't ope the attachment but i will definitely go back and study this again thx

16. UnkleRhaukus

i drew the ellipse

17. UnkleRhaukus

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