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22 April, 06:08

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 416 Hz, while the piece open only at one end has a fundamental frequency of 678 Hz. What is the fundamental frequency of the original tube?

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  1. 22 April, 06:19
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    The fundamental frequency of the original tube is 159.169 Hz.

    Explanation:

    From the orientation of a wave in a tube open at both ends, we have at the fundamental frequency there is an anti-node at both ends.

    Therefore the length of the tube = λ/2

    and we have v = fλ where v = speed of sound in air = 343 m / s

    f = frequency, therefore

    Therefore λ = v/f = 343/416 = 0.824 m, therefore the tube opened at both ends is

    λ/2 or 0.824/2 m = 0.4123 m long

    The other tube with one end open we have a node at the closed end and an anti-node at the opened end which gives a fundamental frequency with the wavelength = λ/4

    Therefore from v = fλ we have λ = v/f = 343/678 = 0.5059 m

    Therefore the length of the tube = λ/4 = 0.5059:4 = 0.1265 m

    The length of the original tube is then given by the sum of the lengths of the two split tubes, that is = 0.1265 + 0.4123 = 0.5387 m

    Note that when combined the original tube still has only one end opened

    Therefore the length of the tube = λ/4 or the wavelength = 4*0.5387

    λ = 2.1549 m

    Therefore we have f = v/λ = 343/2.1549

    Frequency, f = 159.169 Hz
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