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Speed of Sound from resonance

  1. Feb 16, 2010 #1
    1. The problem statement, all variables and given/known data
    One end of a plastic tube, open at both ends, is placed into a large container of water. A 256 Hz tuning fork, continuously vibrating, is held over the end of the tube in the air and the tube is raised until maximum loudness is observed. The plastic tube is then raised until the next position of maximum loudness is found. This new position is 65 cm higher than the first. Calculate the speed of sound.


    2. Relevant equations
    v = f x λ
    λ = 4 x L (L = length of tube)


    3. The attempt at a solution
    Since this is a closed ended column of air, only the odd harmonics are capable of resonating. Thus the two positions must correspond to a frequency difference of 512 HZ (ie. the difference between successive odd harmonics).

    The 65 cm difference in tube length must correspond to the wavelength. The first situation λ1 = 4 x L. In the second situation, λ2 = 4(L+.65)

    Anyway, here is where I keep getting bogged down.

    The velocity has to be the same in both situations, so

    v = f1λ1 = f2λ2

    or f1(4L) = f2 (4L + 2.60)

    Also

    f1 - f2 = 512

    3 variables and 2 equations...seems like I am over-complicating this problem, but if I can figure out either of the sets of frequency and wavelength in the two situations I should be good to go. What am I missing?
     
  2. jcsd
  3. Feb 17, 2010 #2
    So, this seems to boil down to that 65 cm corresponding to 1/2 wavelength in the long tube situation...but I do not understand why this is so...
     
  4. Feb 17, 2010 #3
    The frequency is the same in both cases. So it's the wavelength.
    The resonance conditions are different. The length of the empty tube must fit an odd number of lambda/4.
    In the first case the length of the empty region can be for example 5*lambda/4 and in the second case 7*lambda/4.
    The difference (known) can be used to calculate lambda and then v (because you know f already.
     
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