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Tension in string and oscillation

  1. Mar 5, 2015 #1
    We've got a 0,5m string attached to a frame and has its own fundamental frequency at 440Hz. We cool our system for 15°C. What is the proper frequency now?

    String length l=0,5m
    String section S=0,02 mm^2
    String density ρ=7800 kg/m^3
    Young module E=2*10^5 N/mm^2
    Fundamental frequency ν=440Hz
    ΔT= -15°C
    α(string)=1,2*10^-5 /K
    α(frame)=1,7*10^-5 /K


    2. Relevant equations
    Δl/l=F/(ES)+α⋅ΔT
    I assume that F(string)=F(frame) and Δl(string)+Δl(frame)=0, so:

    F/(ES) +α(string) ΔT + α(frame) ΔT=0

    From this equation I get F=1,74N, but I can't imagine what this number means?

    c=√(E/ρ) = 5063 m/s is this equal to √(F/(ρS))? What is this force? (4000N)
     
  2. jcsd
  3. Mar 5, 2015 #2

    Orodruin

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    Do you know how wave velocity depends on string tension?
     
  4. Mar 5, 2015 #3
    From this formula √(F/(ρS)) I would say that greater tension produces higher wave velocities.
     
  5. Mar 5, 2015 #4

    haruspex

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    It's hard to be sure from such a skeletal description, but I suspect you have some errors in that calculation.
    For a start, I would have thought Δl(string)=Δl(frame).
    Please post the details.
     
  6. Mar 5, 2015 #5
    Half a meter long steel string with a cross-section 0,02 mm2 with a density of 7800 kg /m3 and Young module 2*105 N/mm2 is embedded in the massive brass frame and has its fundamental frequency at 440 Hz. What is the new frequency, when the string and the frame is cooled for ΔT= 15°C?
    Temperature coefficient of length expansion are αs = 1.2*10-5 / K for the string and αf = 1.7*10-5 / K for the frame.

    So:
    Δls/l = F/(ES)+αs ΔT
    Δlf/l = αf ΔT

    Δls = Δlf
    F/(ES)+αs ΔT= αf ΔT
    F=(αf - αs) ΔT E S
    F= -0,3N --> What does this mean?
     
  7. Mar 5, 2015 #6

    haruspex

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    If I rewrite your equation as F/(ES)= αf ΔT-αs ΔT, the right hand side is the change in the stretch of the wire. Does that help?
    (I agree with -0,3N.)
     
  8. Mar 8, 2015 #7
    So if I understand this correctly, it means, that there is -0,3N less force in the string than at the beginning?

    For the force at the beginning I use the following formula:
    f1=√(F1/(ρS))/2l, where f =440Hz
    and get F1=120,8N

    Is it correct now, that I use F2=F1-0,3N=120,5N for the final force?
    ...and get f2=√(F2/(ρS))/2l2=879Hz?
    Do I even have to take into account that l2=l-(αf ΔT-αs ΔT), because it is negligibly small?
    This result is an octave higher which seems a little bit exaggerated for me at only ΔT=15°C? :)
     
  9. Mar 8, 2015 #8

    haruspex

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    That's double what it should be. Did you take l as 1 instead of .5?
    It certainly shouldn't be much changed.
    The change in length is not (αf ΔT-αs ΔT). But you are right, it does not have much effect.
     
  10. Mar 9, 2015 #9
    Ups, yes I took l=1m but if I use the correct value I get 30,2N which is 4 times smaller?

    So my thought that new force is 30,2N-0,3N=29,9N is right?
    And I get final frequency f2=437 Hz?
     
    Last edited: Mar 9, 2015
  11. Mar 9, 2015 #10

    haruspex

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    Sorry for the delay, I went looking for my scribbled answer but couldn't find it. I think that's what I got.
     
  12. Mar 9, 2015 #11
    Ok, thank you very much for your time and help. :)
     
  13. Mar 11, 2015 #12
    Hi, I've got one more question for the same exercise:
    How does the energy of oscillation change, if the amplitude is 2mm at the beginning and at the end if the cooling is the same?

    So E=1/2 μ A2ω2λ
    ΔE=E1-E2=1/2 2l μ A2(2Π)2(f12-f22)=3*10^-9J

    Is this correct? It seems very small. :)
     
  14. Mar 11, 2015 #13

    haruspex

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    Is that right? I thought it would be 1/4 μ A2ω2l = (1/8) μ A2ω2λ.
    Yes, that does seem much too small. Using your equation, I get 3*10-5, so I suspect a powers-of-ten error. What do you get for E1?
    By the way, because the expression involves the difference of two close numbers (f12-f22) there will be a loss of precision. You can either calculate the new frequency to several decimal places or, by keeping everything symbolic until the final step, avoid taking a difference of two large numbers.
     
  15. Mar 12, 2015 #14
    Uf, yes it is 1/4, I searched the formula on the internet, but overlooked that formula that I used is for one whole wavelength.
    So:
    μ = m/l = ρ*V/l = S*ρ
    A = 2*10^-3 m
    l = 0,5m
    ω = 2Πf
    f=√(F1/(ρS))/2l; f2=(F1/(ρS))/4l2
    F1-F2=0,3N

    ΔE=E1-E2=1/4 μ A2 l (ω12 - ω22)
    =1/4 μ A2 l (2Π)2 ((F1/(ρS))/4l2-(F2/(ρS))/4l2)
    =1/4 S ρ A2 l 4Π2 * 1/(ρS 4l2) * (F1-F2)
    =(AΠ)2/(4 l) (F1-F2)
    =(2*10-3m*Π)2/(4*0,5m) * 0,3N
    =5,9*10-6 J
     
  16. Mar 12, 2015 #15

    haruspex

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    Yes, that looks better. Haven't checked the numbers in detail, but it's about right.
     
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