## The Shape of a String

Holding a string up with both ends at the same elevation causes the string to form a curve which not many really care to look into more than the first glance. At first, one may just assume that it is a parabola but upon closer look, it actually has a very interesting mathematical shape which is shown below.

Let’s analyze this mathematically. It is important to note that the horizontal tension on any differential segment of the rope is merely zero because if it was not, then it would be moving in either direction and it is static. The only tension that exists must be vertical which increases higher up on the curve to counteract gravity.

Say we call the line’s curve as a function of horizontal distance $\inline&space;u(x)$. The vertical tension divided by the horizontal tension $\inline&space;T_y/T_x$ must equal the slope of the line at that $\inline&space;u'(x)$ so $\inline&space;Cu'(x)=T_y/T_x$ where is C is some constant. Because $\inline&space;T_x$ is constant with respect to horizontal distance, it is absorbed into the constant yielding the equation $\inline&space;u'(x)$ so $\inline&space;Cu'(x)=T_y$.

Now for any differential segment of the curve $\inline&space;\textup{d}s$, two forces are acting on it: gravity and tension which must cancel each other out. It is known that the total length of the differential segment can be written in terms of differential arclength $\inline&space;\textup{d}x\sqrt{1+u'(x)^2}$ and if $\inline&space;\rho$ is taken to be its mass density, the gravity on each differential segment is $\inline&space;\rho&space;g&space;\textup{d}x\sqrt{1+u'(x)^2}$. Because the tension force on the segment is merely the difference between the force pulling up on the right and force pulling down on the left, the tension force can just be written as $\inline&space;\textup{d}T_y$. Setting these two equations equal to each other to ensure zero net force, the following differential equation forms after a little simplification.

$\frac{\mathrm{d}&space;{T_y}}{\mathrm{d}&space;x}=\rho&space;g&space;\sqrt{1+u'(x)^2}$

By using the earlier fact that $\inline&space;u'(x)$ so $\inline&space;Cu'(x)=T_y$., this equation is simplified down to the following.

$\inline&space;Cu''=\rho&space;g&space;\sqrt{1+u'^2}$

Solving this differential equation yields the following result with $\inline&space;A$ and $\inline&space;h$ as constants of integration.

$u(x)=\frac{C}{\rho&space;g}\textup{cosh}\frac{\rho}{Cg}(x+A)+h$

In its simplest form, this is simply a hyperbolic cosine function with the general form $\inline&space;\textup{cosh}(x)=\frac{e^{x}+e^{-x}}{2}$. It is interesting that the shape of a string would be characterized by something like the linear combination of exponentials or even hyperbolic cosine for that matter but it definitely starts some interesting talk.

If you want to learn more or see where I learned it from, read “Introduction to Classical Mechanics” by David Morin. It should be in the “Books” page of the site.