Tension and frequency

  1. The G string of a guitar which should be 392 Hz is playing flat at 380 Hz. What percentage change in the tension of the wire is required so as to tune the string to 392 Hz?

    I just need to help getting started on this one, I don't know how to relate tension to frequency without knowing lamda
     
  2. jcsd
  3. dynamicsolo

    dynamicsolo 1,659
    Homework Helper

    The velocity of the waves on the string will be the product of the frequency and wavelength. Changing the tension in the string alters the velocity of the waves, but since the fundamental tone of the string is produced by a standing wave, that wavelength is constant. So you will not need to know it (or the length of the string itself).
     
    Last edited: Jun 14, 2008

  4. Okay, well where would I derive a formula to find the tension, or therefore the change of tension in this case?
     
  5. Hurkyl

    Hurkyl 16,090
    Staff Emeritus
    Science Advisor
    Gold Member

    You don't have to know lambda in order to manipulate it algebraically.
     
  6. How come? I don't get why we don't need lambda? What other equation is there? How do you do it "algebraically?"
    :cry:
     
  7. dynamicsolo

    dynamicsolo 1,659
    Homework Helper

    You have the wave relation

    v = f · lam.

    For the untuned string, then, you have

    v_1 = f_1 · lam_1

    and, for the properly tuned string,

    v_2 = f_2 · lam_2 .

    But the length of the G string doesn't change, so neither does the fundamental standing wavelength at which it vibrates. So we have

    lam_1 = lam_2 = lam .

    If we now compare the wave velocities, we get

    v_2/v_1 = (f_2 · lam)/(f_1 · lam) = f_2/f_1 .

    So we don't need to know the length of the string or the fundamental standing wavelength.

    You know the relationship between wave velocity, string tension, and string linear mass density, so you can work out the ratio of the tuned versus untuned string tension. (BTW, you don't need to know the linear mass density of the string either: that also will cancel out.)

    From the ratio of the tensions, you can find the fractional tension change, and from that, the percentage of tension change in the string.
     
  8. Okay

    Well from here i see that f2/f1 = 392/380 = 1.0316
    So the tuned string's tension is 1.0316 times more than the untuned strings tension.

    This is also seen as a 96.94%, or a 3.06% in tension.
     
  9. dynamicsolo

    dynamicsolo 1,659
    Homework Helper

    It is true that f2/f1 = v2/v1. But isn't the wave velocity equal to sqrt( T / mu ) ?
     
  10. Yeah..so we sub v for the sqrt( T / mu )

    so f2/f1= sqrt( T2 / mu )/sqrt( T1 / mu )

    1.0316=sqrt( T2 / mu )/sqrt( T1 / mu )

    1.0642=(T2/mu) / (T1/mu)
    1.0642=T2/T1

    So T1*1.0642 = T2

    This is a 93.98% difference between strings, where as the difference in percent is 6.02%
     
  11. dynamicsolo

    dynamicsolo 1,659
    Homework Helper

    That is close, but be careful about the definition. The question asks for the percentage change in the tension of the wire. This will be

    (delta_T / T) x 100% = [ (T2 - T1) / T1 ] x 100% .

    The fractional change is equivalent to

    (T2 - T1)/T1 = (T2/T1) - (T1/T1) = (T2/T1) - 1 = 1.0642 - 1 = 0.0642 .

    So your percentage change should be 6.42% . (Or is 6.02% just a typo?)
     
  12. Ah i See! thank you soo much!
     
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