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Temperature and Frequency?

  1. Jun 23, 2008 #1
    1. The problem statement, all variables and given/known data
    A piano is to be tuned at 20.0°C. However, when the temperature has lowered to 6.32°C, the piano will be off tune. By what percentage will the piano be off key at this low temperature?

    2. Relevant equations
    V = Square Root (3RT/M)

    3. The attempt at a solution
    T must be in Kelvins, so 20--> 293 K and 6.32--> 279.47 K

    Because 3R/M is constant for both temperatures, you can ignore that in the equation so it is just the square root of temperature.
    V1 = Sqrt (293)
    V2 = Sqrt (279.47)

    To find the percent difference, I did (1 - V2/V1) = 0.0234. Thus, the percent difference is 2.34%, but that wasn't the right answer.
  2. jcsd
  3. Jun 23, 2008 #2
    I just tried doing it a different way, but that still didn't give me the right answer.

    V = 331 + .6(T)
    V1=343 m/s
    V2=334.792 m/s

    1-v2/v1 = 0.0239 = 2.39%

    I am really confused on this problem if anyone knows what I am doing wrong.
  4. Jun 23, 2008 #3


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    Where did you get the equation from ?
    Are you sure T isn't tension?
  5. Jun 23, 2008 #4
    I'm no expert, I'm in grade 11 just like you but I'm a little better at this stuff than the rest... I'm not sure if I'm right, but use the V = 331 + .6(T) formula, and divide the first number (original), by the new number? And you might get a percentage?

    T is the temperature, it is also tension in different equations.
    The equation that I repeated (v=) is from the Grade 11 textbook... it is a valid formula.
    Last edited: Jun 23, 2008
  6. Jun 24, 2008 #5
    I flipped the equation so that it was V1/V2, which gave a percent of 2.45, but that didn't work either. I don't think this question is supposed to be that difficult, so could it just be a problem with the answer, since I'm pretty sure everything I did was right.

    Does anyone notice anything wrong though with the work I did earlier?
    Last edited: Jun 24, 2008
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