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Vibration problem given Force,length and mass

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data
    A flexible wire 80 cm long has a mass of 0.40g. It is stretched across stops on a sonometer that are 50cm apart by a force of 500N. The wire may vibrate at the following frequencies except one. WHich one?
    a. 4000vib/s
    b. 1000 vib/s
    c. 3000 vib / s
    d. 1500 vib/s


    2. Relevant equations
    searched google for formula but none so far

    no formula in my notes.
    probably this one

    [itex]a = -\frac{k}{m}s[/itex]

    s = displacement
    m = mass
    k = spring constant
    a = acceleration

    or this one
    [itex]T=2\pi\sqrt{\frac{m}{k}}[/itex]

    T = period
    m = mass
    k = constant
    3. The attempt at a solution
    i dont know
    can you help me start?
     
    Last edited: Jan 24, 2012
  2. jcsd
  3. Jan 24, 2012 #2

    BruceW

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    You've written down the equations for simple harmonic motion. But I don't think that is what the question is about. If you look up a sonometer on google, it looks very much like the situation of standing waves on a string. (since it is fixed at two ends). So I think you should be using the equations for standing waves on a string.
     
  4. Feb 24, 2013 #3
    http://hyperphysics.phy-astr.gsu.edu/hbase/waves/string.html


    i see now but the answer is not given in the multiple choice
    [tex]f=\frac{v}{\lambda}[/tex]
    [tex]f=\frac{v}{2L}\mbox{ since }L=\frac{\lambda}{2}[/tex]
    [tex]v=\sqrt{\frac{T}{\rho}}=\sqrt{\frac{T}{\frac{m}{L}}}[/tex]
    [tex]f=\frac{1}{2L}\sqrt{\frac{T}{\frac{m}{L}}}[/tex]
    [tex]f=625Hz=625vib/s[/tex]
    the question might be a typo or is it not?
     
    Last edited: Feb 24, 2013
  5. Feb 24, 2013 #4

    BruceW

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    I think you might be using incorrect values for L or m. The entire wire is 80cm, with mass 0.4g. But only 50cm of that wire is within the sonometer. So what should you use for L and m in the equation?
     
  6. Feb 24, 2013 #5
    i see, so [tex]\rho=\frac{m}{L_{string}}[/tex] L_string = 80cm is different from
    the length of the string apart which is the wavelength/2 = 50cm

    1000vib/s
    thanks
     
  7. Feb 25, 2013 #6

    BruceW

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    Homework Helper

    yep. no worries :)
     
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