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30 October, 18:27

A tube, open at only one end, is cut into two shorter (non-equal) lengths. The piece open at both ends has a fundamental frequency of 425 Hz, while the piece open only at one end has a fundamental frequency of 675 Hz, both in room temperature air. What is the fundamental frequency of the original tube?

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  1. 30 October, 19:34
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    f = 158.6Hz

    Explanation

    Given Data

    f₀=425 Hz ... Open at both ends has a fundamental frequency

    f₁=675 Hz ... open only at one end has a fundamental frequency

    To find

    Fundamental frequency of the original tube

    Solution

    Fundamental frequency with open ends given as

    f = v / (2·L)

    where v speed of sound,

    and L is length

    And

    Fundamental frequency of a tube closed at one end is given as

    f = v / (4·L)

    For the piece with two open ends:

    f₀ = v / (2·L₀)

    L₀ = v / (2·f₀)

    For the other piece

    f₁ = v / (4·L₁)

    L₁ = v / (4·f₁)

    The fundamental frequency of the original tube is that of a closed end tube of length L₀ + L₁.

    f = v / (4· (L₀ + L₁)

    f = v / (4· (v / (2·f₀) + v / (4·f₁))

    f = 1 / (2/f₀ + 1/f₁)

    f = 1 / (2 / 425Hz + 1 / 625Hz)

    f = 158.6Hz
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