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A 42.9cm long tube has a 42.9cm long insert that can be pulled in and

  1. Apr 17, 2013 #1
    1. The problem statement, all variables and given/known data

    A 42.9cm long tube has a 42.9cm long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. As the insert is slowly pulled out, the sound from the tuning fork creates standing waves in the tube when the total length L is 46.8cm, 79.2cm, and 111.6cm. What is the frequency of the tuning fork? Take the speed of sound in air to be 343m/s


    2. Relevant equations

    open-closed end
    (2n-1)/ 4 lamda = L


    3. The attempt at a solution

    so I know this question has been posted before, but I dont really understand the solution.
    79.2 - 46.8 = 32.4 cm or 0.324m

    Why do you need to multiply 0.324 by 2 to get the wavelength?

    Since it is open-closed, isn't it 1/4 lamda = L , which is times it by 4?
     
  2. jcsd
  3. Apr 17, 2013 #2
    A standing wave is an effect of constructive/deconstructive interference as the wave enters the tube, reflects, and crosses paths with itself. In this case, the differences between these measurements are the distances from antinode to antinode. And as we know, the distance from antinode to antinode is one half the wavelength. This is why you must double the distances to get the wavelength.

    79.2-46.8 = 32.4
    111.6-79.2 = 32.4

    32.4 cm = distance from antinode to antinode.

    2*(distance from antinode to antinode) = wavelength


    Hope this helps
     
  4. Apr 17, 2013 #3
    sorry I dont really understand constructive and destructive interference..
    all i know for destructive is that it is 1/2 a integer, constructive is something about whole numbers...
    and how do u know it is from antinode to antinode?
     
  5. Apr 17, 2013 #4
  6. Apr 17, 2013 #5
    ok thanks!
     
  7. Apr 17, 2013 #6
    your site actually gives quite different information from what my notes said.
    My notes said in the open end, you will get antinode, but the site said it opposite..
     
  8. Apr 17, 2013 #7

    haruspex

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    A problem with discussing sound waves in pipes is that you can think of the wave in terms of displacement or in terms of pressure. AFAIK, it's most common to use displacement. In that case, there is zero displacement at the closed end and maximum at the open end. A node always refers to where the amplitude is a minimum, so in this view the node would be at the closed end.
    The site jdibble links discusses the wave in terms of pressure. The pressure changes most where the displacement is least, and v.v. So now the node is at the open end, etc.
    In principle, the same confusion could arise with waves in strings, etc., but I'm not aware that it ever does.
     
  9. Apr 17, 2013 #8
    ok thanks haruspex
    so my notes express waves in terms of dispalcement
    in the website, it express waves in terms of pressure
    is that correct?
     
  10. Apr 17, 2013 #9

    haruspex

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