P17_056 - p − 1 66 2 4 11 2 = 4 4 mm(b And the phase...

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56. We orient one phasor along the x axis with length 4 . 0 mm and angle 0 and the other at 0 . 8 π rad = 144 (in the second quadrant) with length 7 . 0 mm. Adding the components, we obtain 4 . 0+7 . 0 cos(144 )= 1 . 66 mm along x axis 7 . 0 sin(144 )=4 . 11 mm along y axis . (a) The amplitude of the resultant wave is consequently
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Unformatted text preview: p ( − 1 . 66) 2 + 4 . 11 2 = 4 . 4 mm . (b) And the phase constant (an angle, measured counterclockwise from the + x axis) is 180 ◦ + tan − 1 µ 4 . 11 − 1 . 66 ¶ = 112 ◦ ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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