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Math Help - Frequency Resonation, Waves & Sounds

  1. #1
    Member realintegerz's Avatar
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    Frequency Resonation, Waves & Sounds

    I have no clue how to solve these problems, the only formulas I know relating to any of this at all are:

    frequency (n) = n ( v/2L ) for pipes with both ends open

    frequency (n) = n( v/4L ) for pipes with one end open

    Any help would be great

    1) A pipe open at one end and closed at the other is 76.4 cm long. What are three lowest frequencies to which it will resonate? Draw the wave within the tube for each frequency. Repeat for a pipe open at both ends. (112.2 Hz, 336.6 Hz, 561 Hz, 224.5 Hz, 448.9 Hz, 673.4 Hz)

    2) A man wishes to find out how far down the water level is in an iron pipe leading into an old well. Being blessed with perfect pitch, he merely hums musical sounds at the mouth of the pipe and notices that the lowest frequency resonance is about 81 Hz. About how far from the top of the pipe is the water level? (1.05 m)

    3) The lincoln tunnel under the Hudson River in NYC is about 2630 m long. To what sound frequencies does it resonate? What practical importance, if any, do you think this has? ( 0.065 Hz, 0.13 Hz, 0.195 Hz)
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  2. #2
    MHF Contributor
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    f = \frac{v}{\lambda}

    1) for the pipe open at one end only, find the fundamental frequency for the wavelength \lambda such that 76.4 \, cm = \frac{\lambda}{4} , then determine the 3rd and 5th harmonic frequencies.

    for a pipe open at both ends, the fundamental frequency corresponds to the wavelength \lambda such that 76.4 \, cm = \frac{\lambda}{2}, then determine the second and third harmonic frequencies.

    2) find the wavelength for an 81 Hz tone ... distance to the water's surface is \frac{\lambda}{4}

    3) treat the tunnel as an open-ended pipe ... fundamental frequency has a wavelength equal to twice the tunnel's length.
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  3. #3
    Member realintegerz's Avatar
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    For #1 & #2, I still don't know the velocity.... And by using that 76.4 cm = v/4 formula I get wavelength is equal to 305.6
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  4. #4
    MHF Contributor
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    Quote Originally Posted by realintegerz View Post
    For #1 & #2, I still don't know the velocity.... And by using that 76.4 cm = v/4 formula I get wavelength is equal to 305.6
    you should know (or at least been told) that the average speed of sound in air used in these types of problems is v = 343 m/s
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