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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.
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.

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442320340503:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1just like a phasor?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1sorry they're orthogomal let's think

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442320936419:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1ie, 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

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1because crunching \(\sqrt{ A_1^2 cos^2 (\omega t + \delta_1) + A_2^2 cos^2 (\omega t + \delta_2) }\) would be a pain

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0is that the amplitude

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0oh , i see it flipted because they are perpendicular!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1yes 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

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0i solved it this way, with the aid of the phasor graph, i've assumed sines whereas you've got cosines,

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1holy moley that's interesting can't ope the attachment but i will definitely go back and study this again thx

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0i drew the ellipse

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442404085179:dw
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