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Wave equation and wave speed

  1. Nov 6, 2013 #1
    The wave on the string could be described with wave equation.

    Wave equation has a factor v^2 = Tension/linear density.

    It has dimensions of speed, but from where exactly does it follow that this is actually speed of propagation of the wave?
     
  2. jcsd
  3. Nov 6, 2013 #2

    mfb

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    v^2 has dimensions of speed squared. You can calculate the forces in the string based on the deflection in the string, and get this formula as a result. I'm sure there are textbooks where this derivation is done.
     
  4. Nov 6, 2013 #3

    HallsofIvy

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    That can be interpreted as either (or both) of two questions:
    1) How do we know that
    [tex]\dfrac{\partial^2 y}{\partial x^2}= \frac{1}{c^2}\dfrac{\partial^2y}{\partial t^2}[/tex]
    (the "wave equation) does describe wave motion with wave speed "c"?

    2) How do we show that the movement of a string, under tension T and with linear density [itex]\delta[/itex] satisfies the wave equation with [itex]c= T/\delta[/itex]?

    To answer the first one, suppose that f(x, t) satisfies the wave equation. Let g(x, t)= f(x- ct_0, t- t_0). Show that g(x, t) also satisfies the wave equation. Of course, g(x, t) is just f(x, t) with time shifted by [itex]t_0[/itex] and distance shifted by [itex]c t_0[/itex] so that the difference in distances is [itex]ct_0[/itex] for a change in time [itex]t_0[/itex] so that the speed is [itex]ct_0/t_0= c[/itex].

    Before answering the second, I have to admit it isn't exactly true! It is an approximation but a very good approximation for small waves. If you were to pull on the string or wire really hard, you might well permanently stretch or even break it!

    Imagine a small part of the string between [itex](x_0, y(x_0))[/itex] and [itex]x_0+ \Delta x, y(x_0+\Delta x))[/itex]. Let the angle the string makes at [itex]x_0[/itex] be [itex]\theta_{x_0}[/itex]. With tension, T, the upward force at [itex]x_0[/itex] is [itex]T sin(\theta_{x_0}[/itex]. Let the angle at [itex]x_0+ \Delta x[/itex] be [itex]\theta_{x_0+ \Delta x}[/itex]. The upward force at [itex]x_0+ \Delta x[/itex] is [itex]T sin(\theta_{x_0+ \Delta x}[/itex] so the net force is [itex]T(sin(\theta_{x_0+ \Delta x})- sin(\theta_{x_0}))[/itex]. For small angles sine is approximately tangent so we can approximate that by [itex]T(tan(\theta_{x_0+ \Delta x}- tan(\theta_{x_0}))[/itex].

    The tangent of the angle of a curve, at any point, is the derivative with respect to x so can write that as
    [tex]T(\left[\partial y/\partial x\right]_{x_0+\Delta x}- \left[\partial y/\partial x\right]_{x_0}[/tex]

    If that section of string has length s and density [itex]\delta[/itex] then it has mass [itex]\delta \Delta s[/itex]. For small angles, we can approximate [itex]\Delta s[/itex] by [itex]\Delta x[/itex]. "Mass times acceleration" is
    [tex]\delta \Delta x \dfrac{\partial y^2}{\partial t^2}[/tex]
    so that "force equals mass times acceration" becomes
    [tex]T(\left[\partial y/\partial x\right]_{x_0+\Delta x}- \left[\partial y/\partial x\right]_{x_0})= \delta \Delta x \dfrac{\partial^2 y}{\partial t^2}[/tex]

    Dividing both sides by [itex]T \Delta x[/itex], that becomes
    [tex]\dfrac{\left[\partial y/\partial x\right]_{x_0+\Delta x}- \left[\partial y/\partial x\right]_{x_0}}{\Delta x}= \dfrac{\delta}{T}\dfrac{\partial^2 y}{\partial t^2}[/tex]

    Now we can recognize that fraction on the left as a "difference quotient" which, in the limit as [itex]\Delta x[/itex] goes to 0, becomes the derivative with respect to x. Since the function in the difference quotient is the first derivative of y with respect to x, the limit gives the second derivative:
    [tex]\dfrac{\partial^2y}{\partial x^2}= \dfrac{\delta}{T}\dfrac{\partial^2 y}{\partial t^2}[/tex]

    the wave equation with [itex]1/c^2= \delta/T[/itex] so that [itex]c^2= T/\delta[/itex].
     
    Last edited: Nov 6, 2013
  5. Nov 6, 2013 #4
    Convince yourself that for an arbitrary function ##f(x)## the function ##y(x, t) = f(x - vt)## solves the wave equation. Then convince yourself that this solution describes a wave whose shape is given by ##f## and which propagates in the positive ##x## direction at a speed ##v##.
     
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