The hanging chain

Posted by: christian on 26 Feb 2018

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The description of the oscillations of a hanging chain was a problem in mechanics solved by Daniel Bernoulli in the 18th century and which led to the discovery of the class of functions known as Bessel functions.

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The chain is taken to be a narrow, continuous body of length $L$ with linear mass density $\mu$, suspended from a fixed point. Take the origin to be at the free end of the chain in its equilibrium configuration (hanging vertically downwards!), and the $z$ axis to point up; displacements occur in the $xz$ plane (the chain is attached at $(0,L)$) and assume that displacements from the equilibrium configuration are small so that we do not need to distinguish between distances measured along the chain and distances measured along $z$.

With such small displacements, the tension in the chain acts mainly in the vertical direction: $T(z) = \mu g z$. Any net transverse force on an infinitessimal mass element bent through an angle $\alpha$ is due to the difference in the $x$ components of the tension above and below it, $$ \mathrm{d}T_x = \mathrm{d}(\mu g z \sin\alpha) \approx \mathrm{d}(\mu g z \tan\alpha) = \mathrm{d}\left(\mu g z \frac{\mathrm{d}x}{\mathrm{d}z}\right). $$

By Newton's second law, this net force accelerates the mass element: $\mathrm{d}T_x = \mathrm{d}m \frac{\mathrm{d}^2x}{\mathrm{d}t^2}$. Therefore (changing to partial derivatives in recognition of the fact that $x$ depends on time, $t$, and vertical coordinate $z$,

$$ \frac{\mathrm{d}m}{\mathrm{d}z} \frac{\partial^2x}{\partial t^2} = \frac{\partial}{\partial z}\left(\mu g z \frac{\partial x}{\partial z}\right). $$

But $\mathrm{d}m/\mathrm{d}z = \mu$, and so the equation governing the chain's motion becomes

$$ \frac{\partial^2x}{\partial t^2} = g\frac{\partial}{\partial z}\left(z \frac{\partial x}{\partial z}\right). $$

We try a separable solution, $x(z,t) = F(z)G(t)$:

$$ F\frac{\mathrm{d}^2G}{\mathrm{d} t^2} = gG\frac{\mathrm{d}}{\mathrm{d} z}\left(z \frac{\mathrm{d} F}{\mathrm{d} z}\right) = gG\left[ z \frac{\mathrm{d}^2F}{\mathrm{d}z^2} + \frac{\mathrm{d} F}{\mathrm{d} z} \right]. $$ Hence, $$ \frac{1}{G}\frac{\mathrm{d}^2G}{\mathrm{d} t^2} = \frac{g}{F}\left[ z \frac{\mathrm{d}^2F}{\mathrm{d} z^2} + \frac{\mathrm{d} F}{\mathrm{d} z} \right] = -\omega^2, $$ where each equation in $G$ and $F$ must be equal to the same constant, which we call $-\omega^2$. The equation for $G$ is simply that of simple harmonic motion and we may take $G(t) = \cos(\omega t + \phi)$ and set $\phi=0$ since the maximum displacement of the chain will occur at $t=0$. The equation for $F$ can be transformed by the substitution $u^2 = 4z/g$ into $$ u^2\frac{\mathrm{d}^2F}{\mathrm{d}u^2} + u\frac{\mathrm{d}F}{\mathrm{d}u} + u^2\omega^2F = 0, $$ which is one form of Bessel's differential equation of zeroth order. Its solution is $$ F(u) = AJ_0(\omega u), $$ where $J_0$ is the zeroth order Bessel function of the first kind and $A$ is the maximum amplitude of the oscillation.

The full solution to the motion of the chain is therefore $$ x(z,t) = AJ_0\left(2\omega\sqrt{\frac{z}{g}}\right)\cos\omega t. $$ The fixed top of the chain adds the constraint $x(L,t) = 0$ and so the possible values of $\omega$ are determined by the zeros of the Bessel function: $J_0(2\omega\sqrt{L/g}) = 0$: these allowed values are the modes.

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