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 Post subject: Catenary problem with varying densityPosted: Wed, 12 Sep 2012 02:10:30 UTC
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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...)

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 Post subject: Re: Catenary problem with varying densityPosted: Wed, 12 Sep 2012 03:49:26 UTC
 Moderator

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6007
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
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 , where is the (linear) density at length on the curve .

For the curve , we find , 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.)

_________________

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 Post subject: Re: Catenary problem with varying densityPosted: Thu, 13 Sep 2012 04:25:38 UTC
 Member

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

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

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 along the entire curve...I could attack the part a) in the same way, using such that

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

-Nate

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 Post subject: Re: Catenary problem with varying densityPosted: Thu, 13 Sep 2012 06:02:24 UTC
 Member

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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