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24 January, 11:11

Let's now apply our equations for the normal modes of open and closed pipes. On a day when the speed of sound is 345 m/s, the fundamental frequency of an open organ pipe is 690 Hz. If the n=2 mode of this pipe has the same wavelength as the n=5 mode of a stopped pipe, what is the length of each pipe?

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  1. 24 January, 14:54
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    L = 0.5 m, L ' = 0.625 m

    Explanation:

    The equations that describe the resonant wavelength in tube are

    Tube or open at both ends

    λ = 2L / n n = 1, 2, 3 ...

    Tube with one end open and one closed

    λ = 4L / m m = 1, 3, 5 ...

    Let's raise our case

    In the first part the fundamental frequency is f = 690 Hz

    Use the relationship

    v = λ f

    λ = v / f

    The fundamental frequency occurs for n = 1 in the first equation

    L = n λ / 2

    L = n v / 2f

    L = 2 345 / (2 690)

    L = 0.5 m

    In the second part they say. The n = 2 of the open pipe

    λ = 2L / 2

    The m = 5 of the closed pipe

    λ = 4L ' / 5

    They indicate that the two wavelengths are equal, so we can match the equations

    2L / 2 = 4L ' / 5

    L ' / L = 5/4

    L ' = 5/4 L

    L ' = 5/4 0.5

    L ' = 0.625 m
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