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YumYum247

  • one year ago

Yelp!!!!

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  1. YumYum247
    • one year ago
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    @IrishBoy123 help check my work please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

  2. YumYum247
    • one year ago
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    i tink i know my mistake!!!!

  3. YumYum247
    • one year ago
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    it should have been f3=5((346m/sec)/2(0.54m) )

  4. YumYum247
    • one year ago
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    never mind i take that back!!! that's for the closed harmonic!!! LOL :"D

  5. YumYum247
    • one year ago
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    f3=3((346m/sec)/2(0.54m) ) this is right...!!!

  6. YumYum247
    • one year ago
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    in a closed air column each harmonic occurs at 1 3 5 7 pattern where as the open doesn't

  7. YumYum247
    • one year ago
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    @IrishBoy123 are you there?!?!?!?!!?!? O-O

  8. IrishBoy123
    • one year ago
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    go back to basics and draw the wave shapes starting from the given that v = const and v = \(f.\lambda\) we get v = \(f_n.\lambda_n\) and, for the harmonics for an open air column: \(L = \lambda_1/2 = 2\lambda_2/2=3\lambda_3/2 = 4\lambda_4/2 = \ ...n\lambda_n/2\) for the 3rd harmonic, \(f_3 = v / \lambda_3 = \frac { v }{\frac{2L}{3}} = \frac{3v}{2L}\) \(f_3 = \frac{3*346}{2*0.54} = 961Hz\)

  9. IrishBoy123
    • one year ago
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    harmonics for open ended go in halves: http://hyperphysics.phy-astr.gsu.edu/hbase/waves/opecol.html#c3 everything else follows from that

  10. YumYum247
    • one year ago
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    LOL yah sorry i kinda confused my self but that's what i got!!!! :"D i watched a really nice video on this topic and it did a great deal of help!!!!

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