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Derivation of wave equation and wave speed

  1. Mar 2, 2015 #1
    Hi,
    I'm trying to wrap my head around the derivation of the wave equation and wave speed.
    For starters I'm looking at the derviation done on this site: http://www.animations.physics.unsw.edu.au/jw/wave_equation_speed.htm

    I could maybe explain what I understand at this point

    Given a string with the linear density μ with and applied tension of T. dx represents the length of the string and the angle θ1 and θ2 is the slope of the angle of the string as far as I have understood (correct me if I'm wrong) and they are appoximated through the small angle approximation such that sin θ ≅ θ, cos θ ≅ 1 and due to the definition of tangent tan θ would also approximate to be θ.
    By the definition of Newton's law F = ma and we desire the force acting in the vertical direction the force acting would be the mass multiplied by the vertical acceleration.

    The net force would then be the sum of the horizontal components of the tension (Instead of taking the difference I interpret it as up being the positive axis and down negetive therefore the sum would be equivalent of the net force).
    For these reasons we oculd write the net horizontal force as:

    Fy = Tsinθ1 + Tsinθ2

    As stated above theta is considered the slope of the angle of the string and for that reason could be written in as a first derivative with respect to x.

    Fy = T(∂y/∂x)1 + T(∂y/∂x)2

    This basically means that the resultant force is dependent on the difference in the slope of the two ends of the segment.

    Carrying on with further use of Newton's second law. If I'm honest I am not sure what dm represents but what it is equated to makes sense. The linear density μ multiplied with the length of the string dx would be equivalent to the mass of the segment. The vertical acceleration would also be the rate of change of the vertical velocity. As stated in the link it would be ∂y2/∂t2. These two things would then be identical to mass times acceleration (μ dx (∂y2/∂t2). Then equating it to the previous equation for vertical force it would be possible to write;

    Fy = T(∂y/∂x)1 + T(∂y/∂x)2

    as

    μ dx (∂y2/∂t2) = T(∂y/∂x)1 + T(∂y/∂x)2

    Factoring out the tension and solving for acceleration;
    (∂y2/∂t2 = T / μ ((∂y/∂x)1 + (∂y/∂x)2) / dx

    The first derivative for position x and x+dx divided by dx is the rate of change of the frist derivative which then again by definition is the second derivative. Therefore it can be written as;

    ∂y2/∂t2 = T / μ ∂y2/∂x2

    Now I think I got evertyhing up to this point

    More specifically I fell off a bit o nthe segment with "A solution to the wave equation". What I generally want to do is to derive the equation for wavespeed (v = (T / μ)1/2) Though I have issues "understanding the equations etc. and also the idea of wave number, I've read that the angular frequency divided by the wave number is equal to the speed or the "wave velocity" but thats about it, and I do not really get how the equations are equated at the very end right before the final equation is derived:

    I know this is a long post but I hope someone could explain the missing pieces for me.
     
  2. jcsd
  3. Mar 2, 2015 #2

    Svein

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    The two forces are acting in different directions. Since [itex]\frac{d}{d\theta}sin(\theta)=cos(\theta) [/itex], you get [itex]Fcos(\theta)d\theta=\mu ds \frac{\partial^{2}y}{\partial t^{2}} [/itex]. Now you need to connect up [itex]cos(\theta)d\theta [/itex] to [itex]\frac{\partial y}{\partial x} dx[/itex].
     
  4. Mar 2, 2015 #3
    I seem to have written horizontal instead of vertical, my bad.
    I still seem to miss the point you're trying to make here or in what direction you are trying to steer me. I'm sorry if I seem oblivious but could you please elaborate.
     
  5. Mar 3, 2015 #4

    Svein

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