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Wavelength^2 vs. Tension (graph/conceptual)

  1. Apr 17, 2010 #1
    hi,
    in our lab last Thursday we were doing standing waves on a string attached to a pulley and vibrator(at f=120Hz), and we produced loops by creating a tension force in the string.
    by calculating the wavelength(2*[distance from node to node]/#of loops), and the tension force(mass added to the end of the string*g), we are supposed to create a plot wavelength^2 vs. T, and then find the slope of the line, and use that to compare our experimental frequency vs. the known frequency.
    however, I have no idea what quantity the slope of the line of a v^2/T graph represents. the units would be, uh, m/kg*s, which doesn't really help me, either. =/
    any help would be much appreciated, thanks!
     
  2. jcsd
  3. Apr 18, 2010 #2
    The speed of the wave on the string is given by the formula
    v= √(T/μ)
    where T is the tension in the string, and μ is its mass per unit length.
    The speed of a wave is also given by v=fλ
    If you eliminate v between those two equations you will get a formula that relates the wavelength and the tension.
    This should tell you why a graph of wavelength squared against tension could be useful.
     
  4. Apr 18, 2010 #3
    so fλ=√(T/μ), or λ^2=(T/μ)*1/f?.. does that mean the slope of the line=1/f?
    sorry, I still don't really understand this... =/.
     
  5. Apr 18, 2010 #4
    You've made just a small mistake.

    [tex]\lambda ^2 =\frac{1}{\mu f^2} T[/tex]

    Since you're plotting the wavelength squared against the tension, your slope is [tex]\frac{\Delta (\lambda ^2)}{\Delta T}= \frac{1}{\mu f^2}[/tex]

    Since you know the frequency to a certain degree of accuracy, you can find the linear mass density from the slope, and compare it with an independent measurement (Weighing the rope and measuring its length).
     
    Last edited: Apr 18, 2010
  6. Apr 18, 2010 #5
    It gives λ²=(T/μ).(1/f²) You forgot to square the f.

    If you compare that with the equation of a straight line y=mx+c [m is the gradient]
    then for a graph of λ² against T, the gradient is (1/μ)(1/f²)
     
  7. Apr 18, 2010 #6
    so basically, m(slope)=1/μf², or f(experimental)=[tex]\sqrt{1/mu*m}[/tex]
    if so(and even if not), awesome, guys! thanks a lot. =]
     
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