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Help with Superposition

  1. Aug 31, 2010 #1
    1. The problem statement, all variables and given/known data
    A uniform tube, 60.0cm long, stands vertically with its lower end dipping into water. When the lenght above water is 14.8cm, and again when it is 48.0cm. the tube resounds to a vibrating fork of frequency 512Hz. Find the lowest frequency to which the tube will resound when it is open at both ends


    2. Relevant equations



    3. The attempt at a solution

    When the tube is open at both end, the fundamental frequency f1 is given by:
    f1=v/(2x0.6)
    From the given condition
    512 = (2n1-1)v/(4x0.148)
    512 = (2n2-1)v/(4x0.48)

    How can I find the velocity from the abve two equations?
     
  2. jcsd
  3. Aug 31, 2010 #2

    Chi Meson

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    Re: Superposition

    These problems can be a lot easier if you draw a diagram of tube resonance (using transverse waves, nearly all textbooks do this). If you keep the medium and frequency the same, and increase the length of the tube, you will see that the difference in tube length between any two successive harmonics is half a wavelength. This is true for open and closed tubes.
     
  4. Sep 1, 2010 #3
    Re: Superposition

    As Chi Meson stated "difference in tube length between any two successive harmonics is half a wavelength", using this u can easily find the wavelength and then u can find velocity. U will find that the lowest frequency for the tube will be in the second harmonic.
     
  5. Sep 1, 2010 #4
    Re: Superposition

    Btw is the answer 566.7 Hz ??
     
  6. Sep 1, 2010 #5

    Chi Meson

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    Re: Superposition

    The lowest frequency will be the fundamental, or first harmonic.

    The data for the closed-end tube is taking the "end-correction" into account. The length of the tube for the first harmonic should be 1/4 wavelength, and ideally this should be the same as half the difference of the tube lengths as mentioned above. These numbers are not the same because the point of reflection is actually some distance beyond the tube end. Ths distance is determined by the geometry and width/diameter of the tube.

    I would apply the end-correction to both ends of the open tube to determine the fundamental wavelength that resonates in it.
     
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