p36_025 - 5 0 sin − 45 ◦ = 4 so that y R = q y 2 h y 2...

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25. In adding these with the phasor method (as opposed to, say, trig identities), we mayse t t =0( see Sample Problem 36-3) and add them as vectors: y h =1 0 c o s 0 + 15 cos 30 +5 . 0cos( 45 )=26 . 5 y
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Unformatted text preview: + 5 . 0 sin( − 45 ◦ ) = 4 . so that y R = q y 2 h + y 2 v = 26 . 8 β = tan − 1 µ y v y h ¶ = 8 . 5 ◦ . Thus, y = y 1 + y 2 + y 3 = y R sin( ωt + β ) = 26 . 8 sin( ωt + 8 . 5 ◦ )....
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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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