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Homework Help: Tuning fork frequency help

  1. Sep 24, 2008 #1
    6. A tuning fork is heard to resonate over a resonance tube when the
    air column is 48.5 cm long and again when it is 68.7 cm long. What is
    the frequency of the tuning fork if the temperature is 26˚C?

    I couldn't figure out how to work this. I thought it mattered if it were in a closed or open pipe, which is not mentioned.

    v = 346.6 m/s
  2. jcsd
  3. Sep 24, 2008 #2
    Re: Frequency

    I had a problem that was similar to this...if I recall correctly, lambda was .5(D2-D1). So lambda is half the distance between resonances.

    >_> However, I am not an expert and could very easily be wrong.
  4. Sep 27, 2008 #3
    Re: Frequency

    Don't I have to know if it's open or closed?
  5. Oct 1, 2008 #4
    Re: Frequency

    lambda = 2 (l2 - l2) - .404 m

    v = 346.4 ms^-1

    f = 857.4 Hz
  6. Oct 1, 2008 #5
    Re: Frequency

    First you ll need to get the v-air which is given by: v = (332.5)(squareroot(T/273))
    where T is the temperature in Kelvin. After that you know that v = f x lambda, and now i think: Ln+1 - Ln = 1/2 lambda.
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