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

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

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

 

[paranormal]

The one-dimensional wave equation is a fundamental equation in physics and mathematics that describes the propagation of waves in one dimension. It is used to model a variety of phenomena such as vibrations in a string, sound waves, and electromagnetic waves.

In the context of the provided content, the one-dimensional wave equation is being used to derive the tensile force in a string. The equation is written in terms of the slope of the string (θ) and the partial derivatives of x and y with respect to s, which represent the horizontal and vertical displacements of the string.

The denominator in the equation is written in this form because it represents the length of the string at the point s. This length is given by the Pythagorean theorem as √(∂x/∂s)^2 + (∂y/∂s)^2, which is equivalent to 1 since the sine and cosine of θ satisfy the Pythagorean identity of sin^2θ + cos^2θ = 1.

Writing the equation in this form allows for a more general representation of the length of the string, rather than assuming it is equal to one. This is important because in more complex situations, the length of the string may not be constant or may depend on other variables.

In summary, the author is using a more general form of the equation to derive the tensile force, taking into account the varying length of the string at different points. This is a more accurate representation and can be applied to more complex situations.