# Homework Help: One dimensional wave equation

1. Oct 29, 2014

### mSSM

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. Oct 30, 2014

### stevendaryl

Staff Emeritus
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, $L' > L$.

3. Nov 2, 2014

### mSSM

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. :)

4. Nov 2, 2014

### stevendaryl

Staff Emeritus
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 $L' = \sqrt{(\frac{dx}{ds})^2 + (\frac{dy}{ds})^2} \delta s$.

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