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Frequency of a standing wave on a string

  1. Jun 3, 2012 #1
    1. The problem statement, all variables and given/known data

    a 50m length of string with mass 0.0175 kg has waves traveling at 112 m/s. There is a mass attached to the string, 0.45 kg that is creating a 4.41 N tension. What is the frequency of the string vibrator?

    2. Relevant equations
    v=f*λ
    f= [√T/(m/L)]/2L

    3. The attempt at a solution
    I can't for the life of me get to the right answer, i know that it is 187 Hz, but I'm just not doing the math right or something like that.
     
  2. jcsd
  3. Jun 3, 2012 #2

    TSny

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    The problem statement appears to be inconsistent. The speed of a wave on a string is

    √(T/(m/L)) which yields 159 m/s using the data given. Yet the problem states that the speed

    is 112 m/s. :confused:

    [OOPS, the speed does in fact come out to be 112 m/s. Sorry!]

    But something still seems strange. The formula you give for f is for the fundamental frequency of a standing wave (right?). Using L = 50 m yields 1.12 Hz. This is a very low frequency due to the string being very long. Is there any information given that would allow you to determine which harmonic the string is vibrating in? The numbers seem a little weird to me.
     
    Last edited: Jun 3, 2012
  4. Jun 3, 2012 #3
    I'm not getting that same answer. T=4.41 m=0.0175 kg and L=50. which comes to 112.25. This is a worksheet from a teacher, I am a tutor, and the teacher for this student teaches wildly away from the book, so I am trying to find the simplest way to do this. His answer sheet says 112 for the speed, he rounded down. then knowing all of that information, asks to find the frequency. I can't find, even going backwards, any equation that will get 187 Hz.
     
  5. Jun 3, 2012 #4
    I get a speed of 112m/s using the data but there is something not correct about the data.
     
  6. Jun 3, 2012 #5
    Here is the problem in question. I can't seem to explain it without that picture of the string vibrator. on that worksheet i know that 8d is 4.41 N and 8e is 112 m/s. I know that 8f is 187 Hz but don't understand how to get that answer.
     

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    Last edited: Jun 3, 2012
  7. Jun 3, 2012 #6
    If I read it correctly a BALL of string 50m long has a mass of 0.0175kg.
    The length of string vibrating is 1.2m (I think) from the diagram Diagram is not very clear
     
  8. Jun 3, 2012 #7
    That is my issue, I'm not sure whether 8f is referring to the diagram, or the previously gained data in the problem...
     
  9. Jun 3, 2012 #8
    Correct me if I'm wrong (which I think I am):

    With the supposed answer of f = 187 Hz

    and

    v=f*λ

    λ = 1/2 L

    I get v = 112.2

    This seems wildly incorrect but I can't see any fault in it.

    Edit: Yeah turns out it was the diagram length that should have been used, it makes sense now.
     
    Last edited: Jun 3, 2012
  10. Jun 3, 2012 #9
    if the string on the diagram is 1.2m long can you see that it is 2 wavelengths long?
    If you know the wavelength and the speed then you can find frquency.
     
  11. Jun 3, 2012 #10
    After playing around with it. By using the aforementioned speed and the equation f=v/2L and using L=1.2. I find that 1.87. which is a few decimals off, but I think I'm close to it. v in that equation being equal to the square root of T/(m/L)
     
  12. Jun 3, 2012 #11
    Magical, that was the issue. It really was just that simple but with the diagram. Thank you!
     
  13. Jun 3, 2012 #12
    With 2 full wavelengths on the string, shouldn't λ = 1/2 L?
     
  14. Jun 3, 2012 #13
    if l = 1.2m and it is 2 wavelenths then wavelength = 0.6m
    v = fλ 112 = f x 0.6 gives f = ???????
    this is all you need to do now
     
  15. Jun 3, 2012 #14
    for f=v/lamda I got the 187 Hz. Speed being 112, and with Wavelength being .6.
     
  16. Jun 3, 2012 #15
    Alright yeah so the issue was the length being confused in the diagram and the density calculation.
     
  17. Jun 3, 2012 #16
    Yes, i thought the length was the 50, but it was the 1.2 from the diagram.
     
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