How Does the One-Dimensional Wave Equation Model Tensile Forces in a String?

[mSSM]

Homework Statement

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:
$$
T(s,t)\cos\theta = T(s,t)\frac{\frac{\partial x}{\partial s}}{\sqrt{\left(\frac{\partial x}{\partial s}\right)^2+\left(\frac{\partial y}{\partial s}\right)^2}}
\text{.}
$$

Homework Equations

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

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?

 

[stevendaryl]

Homework Statement

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:
$$
T(s,t)\cos\theta = T(s,t)\frac{\frac{\partial x}{\partial s}}{\sqrt{\left(\frac{\partial x}{\partial s}\right)^2+\left(\frac{\partial y}{\partial s}\right)^2}}
\text{.}
$$

Homework Equations

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

No, it's not 1 because the string stretches. As described in the original source, s 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 s=s_1 to s=s_2, then the unstretched length of that section would be

L = |s_2 - s_1| = \delta s

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

In general, L' > L.

 

[mSSM]

No, it's not 1 because the string stretches. As described in the original source, s 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 s=s_1 to s=s_2, then the unstretched length of that section would be

L = |s_2 - s_1| = \delta s

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

In general, L' > L.

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. :)

 

[stevendaryl]

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

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

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

\frac{dx}{ds} = \frac{12}{8} = 1.5
\frac{dy}{ds} = 1
\sqrt{(\frac{dx}{ds})^2 + (\frac{dy}{ds})^2} = 1.802
L' = 1.802* 8 mm = 14.42 mm<br /> <br /> Note: s is defined so that \delta s doesn't change when you stretch it, but L' does.