A bugle can be thought of as an open pipe. If a bugle were straightened out, it would be 2.65 mlong.a.) If the speed of sound is 343m/????,

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A bugle can be thought of as an open pipe. If a bugle were straightened out, it would be 2.65 mlong.a.) If the speed of sound is 343m/????, find the lowest frequency that is resonant for a bugle (ignoring end corrections)b.) Find the next two resonant frequencies for the bugle.

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MichaelMet 5 years 2021-09-02T20:07:00+00:00 1 Answers 10 views 0

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  1. Maris
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    2021-09-02T20:08:51+00:00
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    Answer:

    (a). The lowest frequency is 64.7 Hz.

    (b). The next two resonant frequencies for the bugle are 129.4 Hz and 194.2 hz.

    Explanation:

    Given that,

    Length = 2.65 m

    Speed of sound = 343 m

    We need to calculate the wavelength

    Using formula of wavelength

    \lambda=2l

    Put the value into the formula

    \lambda=2\times2.65

    \lambda=5.3\ m

    (a). We need to calculate the lowest frequency

    Using formula of frequency

    f_{1}=\dfrac{v}{\lambda_{1}}

    Put the value into the formula

    f_{1}=\dfrac{343}{5.3}

    f_{1}=64.7\ Hz

    (b). We need to calculate the next two resonant frequencies for the bugle

    Using formula of resonant frequency

    f_{2}=\dfrac{v}{\lambda_{2}}

    f_{2}=\dfrac{v}{l}

    Put the value into the formula

    f_{2}=\dfrac{343}{2.65}

    f_{2}=129.4\ Hz

    For third frequency,

    f_{3}=\dfrac{v}{\lambda_{3}}

    f_{3}=\dfrac{3v}{2l}

    Put the value into the formula

    f_{3}=\dfrac{3\times343}{2\times2.65}

    f_{3}=194.2\ Hz

    Hence, (a). The lowest frequency is 64.7 Hz.

    (b). The next two resonant frequencies for the bugle are 129.4 Hz and 194.2 hz.

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