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Using frequency to calculate resonances

  1. Apr 10, 2010 #1
    1. If the length of the tube used in our experiment is 4 m, how many resonances would you observe when a tuning fork of frequency 256HZ is used?



    2. v=f[tex]\lambda[/tex] ; L = (1/4)(2n+1)[tex]\lambda[/tex]



    3. 340m/s = 256Hz[tex]\lambda[/tex] [tex]\lambda[/tex] = 1.328125m
    4 = (1/4)(2n+1)(1.328125m)
    12.05 = 2n+1
    5.5 = n
     
  2. jcsd
  3. Apr 10, 2010 #2
    We need to know what kind of tube you used in lab (open, one end closed, etc.)
     
  4. Apr 10, 2010 #3
    one end of the tube is open. The other end has water at the end of it.
     
  5. Apr 10, 2010 #4
    Ok and you probably raised the water level while ringing the tuning fork and marked where you heard resonances, right?

    So this is a closed end tube. You are correct in using the equation [tex]L = \frac{2n+1}{4}\lambda[/tex]

    However, you are not correct in leaving L at 4 because you raised and lowered the water level. What you should do is evaluate the inequality [tex]4 \leq \frac{2n+1}{4}\lambda[/tex]
    and solve for n. Of course you are only allowed integer values of n.
     
  6. Apr 10, 2010 #5
    okay, that makes sense. so if my evaluation of the wavelength is correct, i would substitute it into the inequality and I should get n < 5? so at most there are 5 resonance structures.
     
  7. Apr 10, 2010 #6
    Yep, that is what I get as well.
     
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