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One dimensional wave equation

  1. Oct 29, 2014 #1
    1. The problem statement, all variables and given/known data

    Reading the very first chapter of Weinberger's First Course in PDEs, I stumbled over the derivation of the tensile force in the horizontal direction. The question was posted already in this thread: https://www.physicsforums.com/threads/one-dimensional-wave-equation.531397/

    And the first answer provides the following solution:
    $$
    T(s,t)\cos\theta = T(s,t)\frac{\frac{\partial x}{\partial s}}{\sqrt{\left(\frac{\partial x}{\partial s}\right)^2+\left(\frac{\partial y}{\partial s}\right)^2}}
    \text{.}
    $$

    2. Relevant equations
    What leads to the above equation? Why write the the denominator in this form? Isn't it just equal to one?

    3. The attempt at a solution
    The slope of the string at the point ##s##is obviously given by ##\tan\theta = \frac{\partial y}{\partial x} = \left(\frac{\partial y}{\partial s}\right) \left(\frac{\partial x}{\partial s}\right)^{-1}##, which, if I understand that correctly, simply translates to ##\frac{\sin\theta}{\cos\theta} =\left(\frac{\partial y}{\partial s}\right) \left(\frac{\partial s}{\partial x}\right)##. Is that part correct so far?

    Now, this leads me directly to:
    $$
    T(s,t)\cos\theta = T(s,t) \frac{\sin\theta}{\tan\theta} = T(s,t) \frac{\partial y}{\partial s} \cdot \frac{\frac{\partial x}{\partial s}}{\frac{\partial y}{\partial s}} = T(s,t) \frac{\partial x}{\partial s}
    \text{.}
    $$

    Where exactly is the above-mentioned denominator coming from?

    I also notice that
    $$
    T(s,t)\cos\theta =T(s,t) \frac{\partial x}{\partial s} = T(s,t) \frac{\partial x}{\sqrt{\left(\partial x\right)^2+\left( \partial y\right)^2}}
    \text{,}
    $$
    where multiplication by ##\partial s## in the numerator and denominator leads to:
    $$
    T(s,t)\cos\theta = T(s,t)\frac{\frac{\partial x}{\partial s}}{\sqrt{\left(\frac{\partial x}{\partial s}\right)^2+\left(\frac{\partial y}{\partial s}\right)^2}}
    \text{.}
    $$

    But from the definition of cosine and sine we have:
    $$
    \left(\frac{\partial x}{\partial s}\right)^2+\left(\frac{\partial y}{\partial s}\right)^2 = \cos^2\theta + \sin^2\theta =1
    \text{,}
    $$
    and thus
    $$
    T(s,t)\cos\theta = T(s,t)\frac{\frac{\partial x}{\partial s}}{\sqrt{\left(\frac{\partial x}{\partial s}\right)^2+\left(\frac{\partial y}{\partial s}\right)^2}} = T(s,t) \frac{\partial x}{\partial s}
    \text{.}
    $$

    So what's the point of writing the expression in the way the author is doing in the book? I am sure there is something I am missing?
     
    Last edited: Oct 29, 2014
  2. jcsd
  3. Oct 30, 2014 #2

    stevendaryl

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    Staff Emeritus
    Science Advisor

    No, it's not 1 because the string stretches. As described in the original source, [itex]s[/itex] for a point on the string is the x-coordinate that that point would have had if the string were unstretched, straight, and parallel to the x-axis.

    If you have a section of string running from [itex]s=s_1[/itex] to [itex]s=s_2[/itex], then the unstretched length of that section would be

    [itex]L = |s_2 - s_1| = \delta s[/itex]

    The stretched length would be given by:
    [itex]L' = \sqrt{\delta x^2 + \delta y^2} = \sqrt{\frac{\partial x}{\partial s}^2 + \frac{\partial y}{\partial s}^2} \delta s[/itex]

    In general, [itex]L' > L[/itex].
     
  4. Nov 2, 2014 #3
    Thank you for your reply.

    So if ##L'## is the stretched length as described by you, then the needed equation above follows immediately. I am, however, still not sure about that ##L'##. Can you explain where your last equation comes from (as shown below; I have added parenthesis where I thought you wanted them)?
    $$
    L' = \sqrt{\left(\frac{\partial x}{\partial s}\right)^2 + \left(\frac{\partial y}{\partial s}\right)^2} \delta s
    $$

    Are ##\delta x^2## and ##\delta y^2## referring to the non-equilibrium, stretched state of the string? In other words, let ##\delta x'## and ##\delta s'## refer to the stretched state of the string; then:
    $$
    \cos \theta = \frac{\delta x'}{\delta s'}
    $$

    Now writing for ##\delta s'##
    $$
    \delta s' = \sqrt{\left(\delta x' \right)^2 + \left(\delta y' \right)^2} \frac{\delta s}{\delta s} = \sqrt{\left(\frac{\partial x'}{\partial s} \right)^2 + \left(\frac{\partial y'}{\partial s} \right)^2} \delta s
    $$

    Finally, renaming ##x' \rightarrow x## and ##y' \rightarrow y## I recover your equation.

    So is that all there is? Is it essentially just multiplying the RHS by ##\frac{\delta s}{\delta s}##, effectively turning the equation into a differential?

    I believe I was most confused about the notation. :)
     
  5. Nov 2, 2014 #4

    stevendaryl

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    Staff Emeritus
    Science Advisor

    You can think of it this way: Take the unstretched string and imagine holding it straight and drawing a black mark each millimeter. Then [itex]s[/itex] counts which mark you're talking about. [itex]x[/itex] and [itex]y[/itex] are not me If you have a piece of stretched string, then [itex]s_1[/itex] is the mark on one end, and [itex]s_2[/itex] is the mark on the other end. [itex]x_1[/itex] is the x-position of the first end, [itex]x_2[/itex] is the x-position of the other end. [itex]y_1[/itex] and [itex]y_2[/itex] are the y-positions of the two ends.

    string.jpg
    In the above figures, Figure 1 shows an unstretched piece of string, colored red. The unstretched length is [itex]L = 8 mm[/itex]. [itex]s[/itex] runs from [itex]s_1 = 0[/itex] to [itex]s_2 = 8[/itex].Now we stretch the piece of string, so that now the string runs from [itex](x=0, y=4)[/itex] to [itex](x=12, y=-4)[/itex], as shown in Figure 2. [itex]\delta x = 12[/itex], [itex]\delta y = -8[/itex]. That's the change in [itex]x[/itex] and [itex]y[/itex] from one end of the string to the other. The length of the string is now [itex]L' = \sqrt{\delta x^2 + \delta y^2} = \sqrt{12^2 + 8^2} = 14.42 mm[/itex].

    An alternative way to calculate the same [itex]L'[/itex] is to use [itex]L' = \sqrt{(\frac{dx}{ds})^2 + (\frac{dy}{ds})^2} \delta s[/itex].

    [itex]\frac{dx}{ds} = \frac{12}{8} = 1.5[/itex]
    [itex]\frac{dy}{ds} = 1[/itex]
    [itex]\sqrt{(\frac{dx}{ds})^2 + (\frac{dy}{ds})^2} = 1.802[/itex]
    [itex]L' = 1.802* 8 mm = 14.42 mm[itex]

    Note: [itex]s[/itex] is defined so that [itex]\delta s[/itex] doesn't change when you stretch it, but [itex]L'[/itex] does.
     
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