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Musical Tube and String

  1. Jun 23, 2010 #1
    1. The problem statement, all variables and given/known data

    A brass tube of mass 23 kg and length 1.5 m is closed at one end. A wire of mass 9.9 g and length 0.39 meters is stretched near the open end of the tube. When the wire is plucked, it oscillates at its fundamental frequency. By resonance, it sets the air column in the tube oscillating at the column's fundamental frequency.

    a) What is the frequency of oscillation of the air in the tube?

    b) What is the tension in the wire?

    2. Relevant equations

    f=nv/4L

    3. The attempt at a solution

    I've been pretty much stuck on this one. When I asked my teacher, she wrote:

    "A plucked wire in part (A) will oscillate at it's fundamental frequency. In f=nv/4L, you know v is the speed of sound in air, n=1, and they give you L. Plug n chug to get f."

    I tried this and it didn't work, my answer was 219.8717Hz. And I plugging something in wrong?
     
  2. jcsd
  3. Jun 23, 2010 #2
    Which frequency were you looking for, frequency of the air column or frequency of the wire? Did you plug in the right number corresponding to that frequency? :)
    (you have 2 lengths here!)
     
  4. Jun 23, 2010 #3
    Oh I see! I put in the wrong length. So I did speed of sound divided by 4, divided by the length of the tube.

    Now for part b I know the equation v=sqrt(T/mu).
    I've always struggled with solving for mu. Is it M/L? Of what?
     
  5. Jun 23, 2010 #4
    What do you think? Whose sound speed is that? :)
     
  6. Jun 23, 2010 #5
    It's the string's...? So I'd use v=343m/s.
    But when I solve using the string's mass divided by the string's length, it doesn't work.
     
  7. Jun 23, 2010 #6
    v=343m/s? It's the speed of sound in the air. Be careful! :)
    So how would you find sound speed of the string? Look at the problem again. You miss some important detail about the vibration of the string :)
     
  8. Jun 23, 2010 #7
    Oh okay. So I see that it goes at it's fundamental frequency.
    Do we use L=.25(lambda)? (We use the L from the string right? Or the tube?)
    How do we get the velocity from that??
     
  9. Jun 23, 2010 #8
    From the string of course, because you are finding the speed in the string.
    L = 0.25 lambda = 0.25 v/f. You have L and f then you can solve for v. What do you think about f (frequency in the string)? How is it related to the frequency in the air column? Notice that the vibration in the air column is forced vibration - it is excited by the vibration of the string!
     
  10. Jun 23, 2010 #9
    The frequencies are the same, right? Since the problem says, "By resonance, it sets the air column in the tube oscillating at the column's fundamental frequency"?

    So for v I get .027288 m/s.
    When I try plugging it into the equation v=sqrt(T/(M/L)), I get an incredibly small number (1.89e-5) that is incorrect. What the heck! What else could I possibly be missing?
     
  11. Jun 23, 2010 #10
    How did you calculate v?
     
  12. Jun 23, 2010 #11
    ((length of string)/.25)*(Answer from part a) = v.
     
  13. Jun 23, 2010 #12
    Answer from part a: f = 57 Hz (did you get the same answer?)
    Is f = nv/4L still right for the string? For the air in the tube, whose one end is closed and the other is open, the closed end is always a node and the open one is always an antinode. On the other hand, for the string whose both ends are firmly held, the two ends are always nodes. You see the difference? How would the equation be for the string? :)
     
    Last edited: Jun 23, 2010
  14. Jun 23, 2010 #13
    Yes I did get that answer.
    Oh, so for the string it's f=nv/2L, correct?
    But when I plug this value in to the same equation as above, it doesn't work.
     
  15. Jun 23, 2010 #14
    v = 44m/s, do you get the same answer?
     
  16. Jun 23, 2010 #15
    Yes!
    I think I got it!
    I did the frequency from a times 2 times the length of the string to get the velocity.
    Then I plugged it into the equation v=sqrt(T/mu)!
    Thanks!!
     
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