An open organ pipe (i.e., a pipe open at both ends) of length l0 has a fundamental frequency f0. part a if the organ pipe is cut in half, wh

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An open organ pipe (i.e., a pipe open at both ends) of length l0 has a fundamental frequency f0. part a if the organ pipe is cut in half, what is the new fundamental frequency?

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Nguyệt Ánh 5 years 2021-08-10T10:33:45+00:00 1 Answers 135 views 0

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  1. King
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    2021-08-10T10:35:37+00:00
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    Answer:

    The fundamental frequency is 2f_0

    Explanation:

    For an open-open pipe, the wavelength of the fundamental mode of vibration is equal to twice the length of the pipe:

    \lambda=2L

    where L is the length of the pipe.

    Moreover, from the wave equation we know that the frequency of the fundamental mode is inversely proportional to the wavelength, according to:

    f=\frac{v}{\lambda}

    where

    f is the frequency

    v is the speed of the wave

    \lambda is the wavelength

    Combining the two equations, we get

    f_0=\frac{v}{2L}

    this means that the fundamental frequency in an open organ pipe is inversely proportional to the length of the pipe.

    Here, the pipe is cut in a half, so the new length is

    L'=\frac{L}{2}

    Therefore, since the speed of the sound wave does not change (it depends only on the medium), the new fundamental frequency will be:

    f'=\frac{v}{2L'}=\frac{v}{2(L/2)}=\frac{v}{L}=2f_0

    So, the fundamental frequency has doubled.

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