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Emspak

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## Homework Statement

We are deriving the PDE that models a hanging chain of Length L.

The x-axis is placed vertically. Positive direction points upwards.

The fixed end of the chain is at x=L.

Let [itex]u(x,t)[/itex] denote the deflection of the chain. We assume the deflection is in the x,u plane. Let [itex]\rho[/itex] denote the mass density in units of mass per length.

A) show that in equilibrium position the tension at a point [itex]x[/itex] is [itex]\tau(x)= \rho gx [/itex] where g is the acceleration due to gravity.

B) show that [tex]\rho \Delta x \frac{\partial ^2 u}{\partial t^2} = \frac{1}{ \Delta x}\left[\tau(x+ \Delta x) \frac{\partial u}{\partial x}( x+ \Delta x,t) - \tau(x) \frac{\partial u}{\partial x}(x,t)\right][/tex]

C) Let [itex] \Delta x \rightarrow 0[/itex] and obtain [tex] \rho \Delta x \frac{\partial ^2 u}{\partial t^2}= \frac{\partial}{\partial x}\left[\tau(x)\frac{\partial u}{\partial x}\right][/tex]

This is all forced vibrations, I think(???), but I am never sure what anyone means by that, I am just presenting what's in the book, you know? I had one person tell me it's unforced, hence the title of the post. The top of the chain is attached to the ceiling or something like it.

## The Attempt at a Solution

I approached it like this:

We know F=ma. That means [itex]\tau=ma[/itex] since [itex]a=g[/itex]

The toal mass of the chain between the end and any point x is [itex]\rho x[/itex]

Therefore [itex]\tau=\rho g x [/itex]

For part B). The component of the tension in the y-axis is going to be [itex]-\tau \sin \alpha + \tau \sin \beta[/itex] because we are superposing the tension at both ends of the chain. β and α are both the angles that the hanging chain is "bent" through at either end when there's any force applied to it. [itex]-\tau \sin \alpha + \tau \sin \beta = ma = \rho g x = \rho g L = \rho g x \Delta x[/itex]. Since g is acceleration [itex] g= \frac{\partial ^2 u}{\partial x^2}[/itex] and that gets us:

[itex] -\tau \sin \alpha + \tau \sin \beta =\rho \frac{\partial ^2 u}{\partial x^2} \Delta x[/itex]

Now, the interesting thing here is that for most angles [itex]\tau \sin \alpha [/itex] and [itex] \tau \sin \beta[/itex] will be pretty close to [itex[\tau \tan \alpha [/itex] and [itex] \tau \tan \beta[/itex]. And the tangent on the bent string is going to describe the velocity of the bit of string at that point.

So we can plug in the tangents to our earlier expression for the tension.

[itex] -\tau \tan \alpha + \tau \tan \beta =\rho \frac{\partial ^2 u}{\partial t^2} \Delta x[/itex]

Now let's look at [itex]u(x,t)[/itex] in terms of x only. The slope of the tangent to the hanging chain – that's the graph of [itex]u(x,t)[/itex] – is [itex] \frac{\partial u}{\partial x}(x,t)[/itex] which means

[itex] \tan \alpha = \frac{\partial u}{\partial x}(x,t)[/itex] and [itex] \tan \beta = \frac{\partial u}{\partial x}(x+\Delta x,t)[/itex]

Plug these back into the equation with the sines of alpha and beta

[itex]-\tau \tan \alpha + \tau \tan \beta = \left[\rho \frac{\partial ^2 u}{\partial x^2} \Delta x \right] \Rightarrow \tau \left[-\frac{\partial u}{\partial x}(x+\Delta x,t) - \frac{\partial u}{\partial x}(x,t)\right] = \rho \frac{\partial ^2 u}{\partial x^2} \Delta x[/itex]

The last step I am a bit fuzzy on. If I move the variables around a bit:

[tex]\frac{\left[-\frac{\partial u}{\partial x}(x+\Delta x,t) - \frac{\partial u}{\partial x}(x,t)\right]}{\Delta x} = \frac{\rho}{\tau} \frac{\partial ^2 u}{\partial x^2} [/tex]

I get a zero in the denominator as [itex]\Delta x[/itex] approaches zero. But of course I am assuming I did the rest of the problem right to begin with. I know that

[itex]\frac{\partial}{\partial x}\left[\tau(x)\frac{\partial u}{\partial x}\right] = \left[\frac{\partial \tau(x)}{\partial x}\frac{\partial u}{\partial x} + \tau(x)\frac{\partial^2 u}{\partial x^2}\right][/itex]

But again I feel like I am just missing the very last and probably dead simple bit at the very end.

Anyhow sorry to make you all slog through this.

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