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25 January, 04:06

The harmonic series from a long tube is given below. Isthis tube acting as an open-pipe resonator or a closed-piperesonator? Explain your answer. 203 Hz, 609 Hz, 1015 Hz, 1421 Hz

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  1. 25 January, 06:37
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    L=1.7604

    Explanation:

    Three successive harmonics in the pipe is given as

    f_1 = 435 Hzf1=435Hz

    f_2 = 725 Hzf2=725Hz

    f_3 = 1015 Hzf3=1015Hz

    now if we take the ratio of all frequency

    then we have

    f1 : f2 : f3 = 435 : 725 : 1015 = 3 : 5 : 7

    since the ratio of consecutive frequency is in ratio of odd

    numbers so this must be 3rd harmonics, 5th

    harmonics and 7 harmonics

    And as per the formula we have

    f_1 = / frac{3v}{4L}f1=4L3v

    435 = / frac{3 (340) }{4L}435=4L3 (340)

    L = 1.706 mL=1.706m
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