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22 January, 03:05

A tube, open at only one end, is cut into two shorter (nonequal) lengths. The piece that is open at both ends has a fundamental frequency of 422 Hz, while the piece open only at one end has a fundamental frequency of 666 Hz. What is the fundamental frequency of the original tube?

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  1. 22 January, 06:01
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    159.1 Hz

    Explanation:

    The formula for the fundamental frequency of an open pipe is given as,

    f' = v/2l' ... Equation 1

    Where f' = fundamental frequency of the open pipe, v = velocity of sound in air, l' = length of the open pipe.

    make l' the subject of the equation

    l' = v/2f' ... Equation 2

    Given: f' = 422 Hz, v = 343 m/s

    Substitute into equation 2

    l' = 343 / (2*422)

    l' = 0.41 m.

    Also, for the closed pipe

    f = v/4l

    Where f = fundamental frequency of the closed pipe, l = length of the closed pipe.

    make l the subject of the equation

    l = v/4f ... Equation 4

    Given: f = 666 Hz, v = 343 m/s.

    Substitute into equation 4

    l = 343 / (4*666)

    l = 0.129 m.

    But,

    Length of the original tube = l+l' = 0.41+0.129 = 0.539 m.

    L = 0.539 m.

    Note: The tube is a closed pipe.

    F = v/4L ... Equation 5

    Where F = Fundamental frequency of the original tube.

    F = 343 / (4*0.539)

    F = 159.1 Hz.

    Hence the fundamental frequency of the original tube = 159.1 Hz
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