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Wave Pulse - Finding Average Transverse Acceleration of Segment

  1. Sep 26, 2013 #1
    1. The problem statement, all variables and given/known data
    A travelling wave pulse is shown in figure 1 below, travelling at v=6 m/s across a string. In figure 2, a short segment of the string is shown zoomed in. The angle of this string goes from θ1 = 17o to zero within a small horizontal distance Δx = 3 mm.

    wave%20concavity.jpg

    2. Relevant equations
    ∂^2y/∂x^2= 1/v2 * (∂^2y/∂t2^2)



    3. The attempt at a solution
    I know using that equation I want to solve for the acceleration, which is the second derivative of position (y), so I should solve for (∂^2y/∂t2^2) in terms of v, but when I do that I am still left with an equation in terms of ∂^2y/∂x^2, the concavity, which I am not sure what do do with / what it equals.

    So next I tried using my intuition:


    |average transverse acceleration| = |Δv|/Δt = initial transverse velocity / Δt.

    And since initial transverse velocity = the slope of the graph at point 1 (where Δx begins) = rise/run = tan(θ), and Δt = Δx/v, the above eqn becomes:

    ( v * tan(θ) ) / Δx

    But that isn't right
     
  2. jcsd
  3. Sep 26, 2013 #2
    I believe you are supposed to approximate ##y(x, t) ## and ## \frac {\partial^2 y} {\partial x^2} ## as a power series in the vicinity the crest and then use the given values to estimate whatever coefficients you will end up with.
     
  4. Sep 26, 2013 #3
    You know that dy/dx varies from 0 to tan(17) over a distance Δx=3 mm.
    You can approximate the derivative of f(x)=dy/dx as Δf/Δx. And this will be the left term of your equation.
     
  5. Sep 26, 2013 #4
    Thanks guys! Approximating the derivative of f(x)=dy/dx as Δf/Δx did the trick.

    Also, for reference purposes if anyone is reading this thread in search of an answer to a similar problem, I forgot to explicitly mention in my OP that the problem question states: "What is the (average) transverse acceleration of this string segment?"
     
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