Catenary problem with varying density

nateb · **Joined:** Wed, 12 Sep 2012 01:49:22 UTC **Posts:** 12

G'day. A bit of help on this one would be greatly appreciated.

A string hangs between x=-2 and x=2 and has the equation y=19+cosh x. (All lengths in metres)

Calculate the mass of the string given the following linear densities:

a) density = 10 grams/metre
b) density = 11-|x| grams/metre

I know how to calculate the length of the string, so here's where I am so far:

Given cosh^2(x) - sinh^2(x) = 1
and sinh(-x) = -sinh(x)
and arclength = int(a,b) sqrt[1+y'(x)^2] dx
= int(a,b) sqrt[1+sinh^2(x)] dx
= int(a,b) sqrt[cosh^2(x)] dx
= int(a,b) cosh(x) dx
= sinh(b) - sinh(a)

So, I reposition the catenary so it's minimum is at (0,1) ie y=coshx

now arclength = int(-2,2) cosh(x) dx
= sinh(2) - sinh(-2)
= sinh(2) + sinh(2)
= 2sinh(2)
~ 7.254m
For a) density = 10 g/m, mass of string is 10*7.254 = 72.54 grams (4 sig figs)

And then I'm stuck. I understand that for b), the density changes with respect to x but I'm not sure how to apply this to the problem. Can anyone assist? (and please let me know if my original working is flawed somewhere - I'm learning this at my own pace and without tuition (big thanks to MIT OCW and Wolfram Alpha!) and I don't know any maths geniuses...)

outermeasure · **Posted:** Wed, 12 Sep 2012 03:49:26 UTC

nateb wrote:

G'day. A bit of help on this one would be greatly appreciated.

A string hangs between x=-2 and x=2 and has the equation y=19+cosh x. (All lengths in metres)

Calculate the mass of the string given the following linear densities:

a) density = 10 grams/metre
b) density = 11-|x| grams/metre

I know how to calculate the length of the string, so here's where I am so far:

Given cosh^2(x) - sinh^2(x) = 1
and sinh(-x) = -sinh(x)
and arclength = int(a,b) sqrt[1+y'(x)^2] dx
= int(a,b) sqrt[1+sinh^2(x)] dx
= int(a,b) sqrt[cosh^2(x)] dx
= int(a,b) cosh(x) dx
= sinh(b) - sinh(a)

So, I reposition the catenary so it's minimum is at (0,1) ie y=coshx

now arclength = int(-2,2) cosh(x) dx
= sinh(2) - sinh(-2)
= sinh(2) + sinh(2)
= 2sinh(2)
~ 7.254m
For a) density = 10 g/m, mass of string is 10*7.254 = 72.54 grams (4 sig figs)

And then I'm stuck. I understand that for b), the density changes with respect to x but I'm not sure how to apply this to the problem. Can anyone assist? (and please let me know if my original working is flawed somewhere - I'm learning this at my own pace and without tuition (big thanks to MIT OCW and Wolfram Alpha!) and I don't know any maths geniuses...)

Total mass is $\int_\gamma\rho(s)\,\mathrm{d}s$ , where $\rho(s)$ is the (linear) density at length

on the curve $\gamma$ .

For the curve $y=19+\cosh(x)$ , we find $\mathrm{d}s=\cosh(x)\,\mathrm{d}x$ , so ...

(Of course, the varying density means your string doesn't actually hang, but there must be some other forces involved or the catenary is an approximation.)

nateb · **Joined:** Wed, 12 Sep 2012 01:49:22 UTC **Posts:** 12

Thanks for the quick reply!

Okay, so if I've got this straight, it's just

$\int_{-2}^2\rho(x)\cosh(x)\mathrm{d}x = \int_{-2}^2 (11- |x|)\cosh(x)\mathrm{d}x\approx 70.81$

I have a feeling that I really should have figured this out for myself! What the integral "says" is that it is the sum of the density (with respect to its displacement from x=0) multiplied by the length of $\mathrm{d}s$ along the entire curve...I could attack the part a) in the same way, using $\rho = 10$ such that

$\int_{-2}^2\rho(x)\cosh(x)\mathrm{d}x = \int_{-2}^2 10\cosh(x)\mathrm{d}x\approx 72.54$

Thanks heaps for the push in the right direction. Please let me know if I have stuffed this up somehow...

-Nate

nateb · **Joined:** Wed, 12 Sep 2012 01:49:22 UTC **Posts:** 12

Oh, and I should credit you with helping me learn about how to write maths problems properly! Cheers.

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Catenary problem with varying density

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