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Homework Help: Traverse Wave on a String

  1. Dec 7, 2011 #1

    GGD

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    1. The problem statement, all variables and given/known data

    In a demonstration, a 0.45 kg horizontal rope is fi xed in place at its two ends (x = 0 and x = 5.0 m) and made to oscillate up and down in the fundamental mode, at frequency 4.0 Hz. At t = 0, the point at x = 2.5 m has zero displacement and is moving upward in the positive direction of a y-axis with a transverse velocity of 5.0 m/s.

    (a) What is the amplitude of the motion of that point?
    (b) What is the wave speed in the rope?
    (c) What is the tension in the rope?


    2. Relevant equations

    w=2*pi*f
    k=2*pi*λ
    y(x,y)=Asin(wt-kx)
    v=sqrt(T/μ)
    v=λ*f
    3. The attempt at a solution

    I already found to amplitude to be .199 m by taking the derivite of the wave equation and solving for A.

    For part B, though the wave speed would simply be wavelength*frequency, but 20m/s is not the correct answer. I tried to mess around with a few other equations but they led me nowhere.

    And for part C I know T is simply (v^2)*μ, but I obviously need to first find velocity.

    Any help would be grealty appreciated, thanks!
     
  2. jcsd
  3. Dec 7, 2011 #2
    What value did you use for the wavelength. The vibration is the fundamental which means the string is half a wavelength long .
     
  4. Dec 7, 2011 #3

    GGD

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    Alright so the speed should equal ((wavelength*freq)/2), or (5*4)/2 , which is equal to 10 m/s? That answer doesn't work for me either.
     
  5. Dec 7, 2011 #4
    First of all : you know the frequency is 4Hz and the wavelength is (2 x 5) =10m.... this gives you the speed (v = fλ)
    Speed = sqrt (T/μ) .....you should be able to get T ???
    The amplitude is given by a SHM equation. Max velocity in SHM = ωA = 2∏fA
    The midpoint of the string is doing SHM with max amplitude....
    Hope this helps you to finish it.
     
  6. Dec 7, 2011 #5

    GGD

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    Alrigh I finally got it, I was dividing by 2 instead of multiplying by it. Thanks for your help!
     
  7. Dec 7, 2011 #6
    Great... well done
     
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